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#### Page No 76:

#### Question 1:

Does a gas have just two specific heat capacities or more than two? Is the number of specific heat capacities of a gas countable?

#### Answer:

No, a gas doesn't have just two specific heat capacities, as the heat capacities depend on the process followed. There are infinite processes; therefore, there can be infinite number of specific heat capacities.

#### Page No 76:

#### Question 2:

Can we define specific heat capacity at constant temperature?

#### Answer:

Specific heat capacity, $s=\frac{\mathrm{\Delta}Q}{m\mathrm{\Delta}T}$, where $\Delta $*Q/m* is the heat supplied per unit mass of the substance and $\mathrm{\Delta}$*T* is the change in temperature produced. At constant temperature, $\mathrm{\Delta}$*T* = 0; therefore, *s* = infinity. So, we cannot define specific heat capacity at constant temperature.

#### Page No 76:

#### Question 3:

Can we define specific heat capacity for an adiabatic process?

#### Answer:

Specific heat capacity, $s=\frac{\mathrm{\Delta}Q}{m\mathrm{\Delta}T}$, where$\u2206$*Q**/m* is the heat supplied per unit mass of the substance and $\u2206$*T* is the change in temperature produced. In an adiabatic process, no heat exchange is allowed; so, $\u2206$*Q =* 0 and hence, *s* = 0. Therefore, in an adiabatic process, specific heat capacity is zero.

#### Page No 76:

#### Question 4:

Does a solid also have two kinds of molar heat capacities *C _{p}* and

*C*? If yes, is

_{v}*C*>

_{p}*C*? Or is

_{v}*C*−

_{p}*C*=

_{v}*R*?

#### Answer:

Yes, a solid also has two kinds of molar heat capacities, *C _{p}_{ }*and

*C*. In a solid, expansion coefficient is quite small; therefore dependence of heat capacity on the process is negligible. So,

_{v}*C*

_{p}_{ }>*C*

_{v}

_{ }with just a small difference, which is not equal to

*R*.

#### Page No 76:

#### Question 5:

In a real gas, the internal energy depends on temperature and also on volume. The energy increases when the gas expands isothermally. Examining the derivation of *C _{p}* −

*C*=

_{v}*R*, find whether

*C*−

_{p}*C*will be more than R, less than

_{v}*R*or equal to

*R*for a real gas.

#### Answer:

In a real gas, as the internal energy depends on temperature and volume, the derived equation for an ideal gas $(\mathrm{d}Q{)}_{\mathrm{p}}=(\mathrm{d}Q{)}_{\mathrm{v}}+nR\mathrm{d}T$ will change to $(\mathrm{d}Q{)}_{\mathrm{p}}=(\mathrm{d}Q{)}_{\mathrm{v}}+nR\mathrm{d}T$ + *k, *where *k* is the change in internal energy (positive) due to change in volume when pressure is kept constant. So, in the case of a real gas, for *n*=1 mole (say),

${C}_{\mathrm{p}}-{C}_{\mathrm{v}}=R+\frac{k}{\mathrm{d}T}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{p}}-{C}_{\mathrm{v}}R,$

where* C _{p}_{ }*and

*C*

_{v}

_{ }are the specific heat capacities at constant pressure and volume, respectively.

#### Page No 76:

#### Question 6:

Can a process on an ideal gas be both adiabatic and isothermal?

#### Answer:

According to the first law of thermodynamics, change in internal energy, $\u2206$*U* is equal to the difference between heat supplied to the gas, $\u2206$*Q* and the work done on the gas,$\u2206$*W, *such that $\mathrm{\Delta}Q=\mathrm{\Delta}U+\mathrm{\Delta}W$. In an adiabatic process, $\u2206$*Q = *0 and in an isothermal process, change in temperature, $\u2206$*T* = 0. Therefore,

$\mathrm{\Delta}Q=\mathrm{\Delta}U+\mathrm{\Delta}W\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\Delta}Q=n{C}_{v}\mathrm{\Delta}T+\mathrm{\Delta}W\phantom{\rule{0ex}{0ex}}\Rightarrow 0=n{C}_{v}\left(0\right)+\mathrm{\Delta}W\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\Delta}W=0,$

where *C _{v}* is the heat capacity at constant volume.

This shows that if the process is adiabatic as well as isothermal, no work will be done. So, a process on an ideal gas cannot be both adiabatic and isothermal.

#### Page No 76:

#### Question 7:

Show that the slope of the *p*−*V* diagram is greater for an adiabatic process compared to an isothermal process.

#### Answer:

In an isothermal process,*
PV = k * ...(i)

On differentiating it w.r.t

*V*, we get

$V\frac{\mathrm{d}P}{\mathrm{d}V}+P=0\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}P}{\mathrm{d}V}=-\frac{P}{V}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}P}{\mathrm{d}V}=-\frac{k}{{V}^{2}}[\text{U}\mathrm{sing}(\mathrm{i}\left)\right],\text{k=constant}$

*k*= constant

In an adiabatic process,

$P{V}^{\gamma}=K....\left(\mathrm{ii}\right)$

On differentiating it w.r.t

*V,*we get

${V}^{\gamma}\frac{\mathrm{d}P}{\mathrm{d}V}+\gamma P{V}^{\gamma -1}=0\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}P}{\mathrm{d}V}=-\frac{\gamma P{V}^{\gamma -1}}{{V}^{\gamma}}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}P}{\mathrm{d}V}=-\frac{\gamma K}{{V}^{\gamma +1}}[\text{U}\mathrm{sing}(\mathrm{ii}\left)\right],\gamma 1\text{and}$

*K*is constant

$\gamma \text{and}\frac{\mathrm{d}P}{\mathrm{d}V}$ are the slope of the curve and the ratio of heat capacities at constant pressure and volume, respectively;

*P*is pressure and

*V*is volume of the system.

By comparing the two slopes and keeping in mind that $\gamma >1$, we can see that the slope of the

*P-V*diagram is greater for an adiabatic process than an isothermal process.

#### Page No 76:

#### Question 8:

Is a slow process always isothermal? Is a quick process always adiabatic?

#### Answer:

For an isothermal process, *PV = K*, where *P* is pressure, *V* is volume of the system and *K* is constant. In an isothermal process, a small change in *V* produces only a small change in *p*, so as to keep the product constant. On the other hand, in an adiabatic process, $P{V}^{\gamma}=k,\gamma =\frac{{C}_{\mathrm{P}}}{{C}_{\mathrm{V}}}$ > 1 is the ratio of heat capacities at constant pressure and volume, respectively, and *k* is a constant. In this process, a small increase in volume produces a large decrease in pressure. Therefore, an isothermal process is considered to be a slow process and an adiabatic process a quick process.

#### Page No 76:

#### Question 9:

Can two states of an ideal gas be connected by an isothermal process as well as an adiabatic process?

#### Answer:

For two states to be connected by an isothermal process,

${P}_{1}{V}_{1}={P}_{2}{V}_{2}$ ...(i)

For the same two states to be connected by an adiabatic process,

${P}_{1}{{V}_{1}}^{\gamma}={P}_{2}{{V}_{2}}^{\gamma}$ ...(ii)

If both the equations hold simultaneously then, on dividing eqaution (ii) by (i) we get

${{V}_{1}}^{\gamma -1}={{V}_{2}}^{\gamma -1}$

Let the gas be monatomic. Then,

$\gamma =\frac{5}{3}$

$\Rightarrow {{V}_{1}}^{\frac{2}{3}}={{V}_{2}}^{\frac{2}{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{1}={V}_{2}$

If this condition is met, then the two states can be connected by an isothermal as well as an adiabatic process.

#### Page No 76:

#### Question 10:

The ratio *C _{p}* /

*C*for a gas is 1.29. What is the degree of freedom of the molecules of this gas?

_{v}#### Answer:

For the molecules of a gas, $\gamma =\frac{{C}_{\mathrm{p}}}{{C}_{\mathrm{v}}}=1+\frac{2}{f}$,

where *f* is the degree of freedom.

$\mathrm{Given}:\gamma =1.29\phantom{\rule{0ex}{0ex}}\Rightarrow 1+\frac{2}{f}=1.29=\frac{9}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{f}=\frac{9}{7}-1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{f}=\frac{2}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow f=7\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Therefore, the molecules of this gas have 7 degrees of freedom.

But in reality, no gas can have more than 6 degrees of freedom.

#### Page No 76:

#### Question 1:

Work done by a sample of an ideal gas in a process *A* is double the work done in another process *B*. The temperature rises through the same amount in the two processes. If *C _{A}* and

*C*be the molar heat capacities for the two processes,

_{B}(a)

*C*=

_{A}*C*

_{B}(b)

*C*<

_{A}*C*

_{B}(c)

*C*>

_{A}*C*

_{B}(d)

*C*and

_{A}*C*cannot be defined.

_{B}#### Answer:

(c) *C _{A }> C_{B}*

According to the first law of thermodynamics, $\mathrm{\Delta}Q=\mathrm{\Delta}U+\mathrm{\Delta}W$, where $\u2206$

*Q*is the heat supplied to the system when $\Delta $

*W*work is done on the system and $\u2206$

*U*is the change in internal energy produced. Since the temperature rises by the same amount in both processes, change in internal energies are same, i.e. $\u2206$

*U*$\u2206$

_{A}=*U*

_{B}.

But as, $\u2206$

*W*$\u2206$

_{A}=*W*

_{B}

_{ }this gives $\Delta $

*Q*= 2$\Delta $

_{A}*Q*.

_{B}Now, molar heat capacity of a gas, $C=\frac{\mathrm{\Delta}Q}{n\mathrm{\Delta}T}$, where $\u2206$

*Q/n*is the heat supplied to a mole of gas and $\u2206$

*T*is the change in temperature produced. As $\u2206$

*Q*= 2$\u2206$

_{A}*Q*

_{B}

_{,}

_{ }*C*.

_{A }> C_{B}#### Page No 76:

#### Question 2:

For a solid with a small expansion coefficient,

(a) *C _{p}* −

*C*=

_{v}*R*

(b)

*C*=

_{p}*C*

_{v}(c)

*C*is slightly greater than

_{p}*C*

_{v}(d)

*C*is slightly less than

_{p}*C*

_{v}#### Answer:

c) *C _{p}* is slightly greater than

*C*

_{v}For a solid with a small expansion coefficient, work done in a process will also be small. Thus, the specific heat depends slightly on the process. Therefore,

*C*is slightly greater than

_{p}*C*

_{v}_{.}#### Page No 76:

#### Question 3:

The value of *C _{p}* −

*C*is 1.00

_{v}_{ }*R*for a gas sample in state

*A*and 1.08

*R*in state

*B*. Let

*p*and

_{A}*p*denote the pressures and

_{B}*T*and

_{A}*T*denote the temperatures of the states

_{B}*A*and

*B,*respectively. It is most likely that

(a)

*p*<

_{A}*p*and

_{B}*T*>

_{A}*T*

_{B}(b)

*p*>

_{A}*p*and

_{B}*T*<

_{A}*T*

_{B}(c)

*p*=

_{A}*p*and

_{B}*T*<

_{A}*T*

_{B}(d)

*p*>

_{A}*p*and

_{B}*T*=

_{A}*T*

_{B}#### Answer:

(a) *p _{A}* <

*p*and

_{B}*T*>

_{A}*T*

_{B}*C*−

_{p}*C*for the gas in state A, which means it is acting as an ideal gas in that state, whereas

_{v }= R*C*−

_{p}*C*= 1.08

_{v}*R*in state B, i.e. the behaviour of the gas is that of a real gas in that state. To be an ideal gas, a real gas at STP should be at a very high temperature and low pressure. Therefore,

*P*

_{A }<

*P*

_{B}and

*T*

_{A}>

^{ }T_{B }where

*P*

_{A }and

*P*

_{B}denotes the pressure and

*T*

_{A}and

*T*

_{B}denotes the temperature of system A and B reepectively.

#### Page No 76:

#### Question 4:

Let *C _{v}* and

*C*denote the molar heat capacities of an ideal gas at constant volume and constant pressure respectively. Which of the following is a universal constant?

_{p}(a) $\frac{{C}_{p}}{{C}_{v}}$

(b)

*C*

_{p}C_{v}(c)

*C*−

_{p}*C*

_{v}(d)

*C*+

_{p}*C*

_{v}#### Answer:

(c) *C _{p} *−

*C*

_{v}For an ideal gas,

*C*−

_{p}

*C*=

_{v}*R*, where

*C*and

_{v}*C*denote the molar heat capacities of an ideal gas at constant volume and constant pressure, respectively and R is the gas constant whos value is 8.314 J/K. Therefore,

_{p}*C*−

_{p}*C*is a constant. On the other hand, the ratio of these two varies as the atomicity of the gas changes. Also, their sum and product are not constant.

_{v}#### Page No 76:

#### Question 5:

70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is

(a) 30 calories

(b) 50 calories

(c) 70 calories

(d) 90 calories

Figure

#### Answer:

(b) 50 calories

It is given that 70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. Also, specific heat at constant pressure,

${C}_{\mathrm{p}}=\frac{\u2206Q}{n\u2206T}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{p}}=\frac{70}{2\times (35-30)}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{p}}=\frac{70}{2\times 5}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{p}}=7\mathrm{calories}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

For an ideal gas,

${\mathrm{C}}_{\mathrm{p}}-{\mathrm{C}}_{\mathrm{v}}=\mathrm{R}=8.314\mathrm{J}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\simeq 2\mathrm{calories}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{v}}={C}_{\mathrm{p}}-R\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{v}}=(7-2)\mathrm{calories}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{v}}=5\mathrm{calories}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\phantom{\rule{0ex}{0ex}}\Rightarrow {C}_{\mathrm{v}}=\frac{\u2206Q}{n\u2206T}\phantom{\rule{0ex}{0ex}}\Rightarrow 5=\frac{\u2206Q}{2\times (35-30)}\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206\mathrm{Q}=5\times 2\times (35-30)\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206\mathrm{Q}=5\times 2\times 5\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206\mathrm{Q}=50\mathrm{calories}\phantom{\rule{0ex}{0ex}}$

Therefore, 50 calories need to be supplied to raise the temperature of 2 moles of gas from 30-35 ^{o}C at constant volume.

#### Page No 76:

#### Question 6:

The figure shows a process on a gas in which pressure and volume both change. The molar heat capacity for this process is *C*.

(a) *C* = 0

(b) *C* = *C _{v}*

(c)

*C*>

*C*

_{v}(d)

*C*<

*C*

_{v}#### Answer:

(c) *C > C _{v}*

Consider two processes

*AB*and

*ACB*; let

*W*be the work done.

*C*is the molar heat capacity of process AB. Process

*ACB*can be considered as the sum of the two processes,

*AC*and

*CB*. The molar heat capacity of process AC is

*C*

_{p}

_{,}as pressure is constant in this process and the molar heat capacity of process

*CB*is

*C*

_{v}, as volume is constant in it.

Internal energy,

*U*, is a state function, i.e. it doesn't depend on the path followed. Therefore,

${U}_{\mathrm{AB}}={U}_{\mathrm{ACB}}\phantom{\rule{0ex}{0ex}}{W}_{\mathrm{AB}}{W}_{\mathrm{ACB}}$

Work done in the

*p-V*diagram is the area enclosed under the curve.

$\Rightarrow {W}_{\mathrm{AB}}+{U}_{\mathrm{AB}}>{W}_{\mathrm{ACB}}+{U}_{\mathrm{ACB}}\phantom{\rule{0ex}{0ex}}\Rightarrow C>{C}_{\mathrm{v}}+{C}_{\mathrm{p}}$

Molar heat capacity is the heat supplied per mole to change the temperature by a degree Kelvin and according to the first law of thermodynamics, d

*Q*

*=*d

*U +*d

*W*

*,*where d

*Q*is the heat supplied to the system in a process.

$\Rightarrow C>{C}_{\mathrm{v}}$

_{}

#### Page No 76:

#### Question 7:

The molar heat capacity for the process shown in the figure is

(a) *C* = *C _{p}*

(b)

*C*=

*C*

_{v}(c)

*C*>

*C*

_{v}(d)

*C*= 0.

Figure

#### Answer:

(d) *C* = 0.

The defined process is

$p=\frac{k}{{V}^{\mathrm{g}}}\phantom{\rule{0ex}{0ex}}\Rightarrow p{V}^{\mathrm{g}}=k,$

such that the process is adiabatic in which there's no heat supplied to the system, i.e. *Q* = 0. Molar heat capacity is the amount of heat supplied to the system per mole to produce a degree change in temperature. Also, in an adiabatic process, no heat exchange is allowed. So, molar heat capacity equals zero, i.e. *C* = 0.

#### Page No 76:

#### Question 8:

In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about

(a) 0.25%

(b) 0.5%

(c) 0.7%

(d) 1%.

#### Answer:

(b) 0.5%

Let *p* and *p'* be the initial and final pressures of the system and *V* and *V'* be the initial and final volumes of the system. *p'* is 0.5% more than *p* and the process is isothernal. So, *pV = k = p'V'* = constant. Therefore,

$pV=p\text{'}V\text{'}\phantom{\rule{0ex}{0ex}}\Rightarrow pV=\left(p+\frac{0.5}{100}p\right)V\phantom{\rule{0ex}{0ex}}\Rightarrow pV=\frac{100.5}{100}pV\text{'}\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}=\frac{100}{100.5}V\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}-V=\frac{100}{100.5}V-V\phantom{\rule{0ex}{0ex}}=-\frac{0.5}{100.5}\phantom{\rule{0ex}{0ex}}=-0.49\%\phantom{\rule{0ex}{0ex}}$

So, volume *V'* decreases by about 0.5% of *V*.

#### Page No 76:

#### Question 9:

In an adiabatic process on a gas with γ = 1.4, the pressure is increased by 0.5%. The volume decreases by about

(a) 0.36%

(b) 0.5%

(c) 0.7%

(d) 1%

#### Answer:

(a) 0.36 %

Let *p* and *p*^{,} be the initial and final pressures of the system and *V* and *V*^{, }be the initial and final volumes of the system. p^{,} is 0.5% more than *p* and the process is adiabatic. So,

$p{V}^{\mathit{\gamma}}=p\text{'}V{\text{'}}^{\gamma},\mathit{}\gamma =1.4=\frac{7}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{V{\text{'}}^{\gamma}}{{V}^{\gamma}}=\frac{p}{p\text{'}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{V\text{'}}{V}\right)}^{\gamma}=\frac{p}{p+{\displaystyle \frac{0.5p}{100}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{V\text{'}}{V}\right)}^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}=\frac{100}{100.5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{V\text{'}}{V}={\left(\frac{100}{100.5}\right)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}=(\frac{200}{201}{)}^{\frac{5}{7}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{V\text{'}}{V}=0.99644\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}=0.99644V\phantom{\rule{0ex}{0ex}}\Rightarrow V\text{'}=V-0.00356V$

Therefore, *V*^{,} is 0.36 % less than *V*.

#### Page No 76:

#### Question 10:

Two samples *A* and *B* are initially kept in the same state. Sample *A* is expanded through an adiabatic process and the sample *B* through an isothermal process. The final volumes of the samples are the same. The final pressures in *A *and *B* are *p _{A}* and

*p*respectively.

_{B}(a)

*p*>

_{A}*p*

_{B}(b)

*p*=

_{A}*p*

_{B}(c)

*p*<

_{A}*p*

_{B}(d) The relation between

*p*and

_{A}*p*cannot be deduced.

_{B}#### Answer:

(c) *p _{A}* <

*p*

_{B}Let the initial states of samples

*A*and

*B*be

*i*and

*the final states of samples*

*B*and

*A*be

*f*and

*f'*

*,*respectively. Let the final volumes of both be

*V*. As sample

_{o}*A*is expanded through an adiabatic process, its curve in the

*p-V*diagram is steeper than that of sample

*B*, which is expanded through an isothermal process. Therefore, from the

*p-V*diagram,

*p*<

_{A}*p*.

_{B}#### Page No 76:

#### Question 11:

Let *T _{a}* and

*T*be the final temperatures of the samples

_{b}*A*and

*B,*respectively, in the previous question.

(a)

*T*<

_{a}*T*

_{b}(b)

*T*=

_{a}*T*

_{b}(c)

*T*>

_{a}*T*

_{b}(d) The relation between

*T*

_{a}_{ }and

*T*cannot be deduced.

_{b}#### Answer:

(a) *T _{a}* <

*T*

_{b}As sample

*B*is undergoing expansion through an isothermal process, its initial and final temperatures will be same, i.e.

*T*. On the other hand, sample

_{b}*A*is at the same initial state as

*B*, such that the initial temperature of

*A*is

_{}

*T*and it is expanding through an adiabatic process in which no heat is supplied. Therefore, sample

_{b}*A*will expand at the cost of its internal energy and its final temperature will be less than its initial temperature.

This implies that

*T*<

_{a}*T*.

_{b}#### Page No 76:

#### Question 12:

Let ∆*W _{a}*

_{ }and ∆

*W*be the work done by the systems

_{b}*A*and

*B,*respectively, in the previous question.

(a) ∆

*W*

_{a}_{ }> ∆

*W*

_{b}(b) ∆

*W*= ∆

_{a}*W*

_{b}(c) ∆

*W*< ∆

_{a}*W*

_{b}(d) The relation between ∆

*W*and ∆

_{a}*W*cannot be deduced.

_{b}#### Answer:

(c) ∆*W _{a}* < ∆

*W*

_{b}In the

*p-V*diagram, the area under the curve w.r.t the

*V*axis is equal to the work done by the system. Since the area under the isotherm is greater than that under the adiabat, the work done by system

*A*is less than that done by system

*B*. Hence, ∆

*W*< ∆

_{a}*W*.

_{b}#### Page No 77:

#### Question 13:

The molar heat capacity of oxygen gas at STP is nearly 2.5 *R*. As the temperature is increased, it gradually increases and approaches 3.5 *R*. The most appropriate reason for this behaviour is that at high temperatures

(a) oxygen does not behave as an ideal gas

(b) oxygen molecules dissociate in atoms

(c) the molecules collide more frequently

(d) molecular vibrations gradually become effective

#### Answer:

(d) molecular vibrations gradually become effective

Molar specific heat capacity has direct dependence on the degree of freedom of gas molecules. As temperature is increased, the gas molecules start vibrating about their mean position, leading to change (increase) in the degree of freedom and, hence, increasing molar heat capacity.

#### Page No 77:

#### Question 1:

A gas kept in a container of finite conductivity is suddenly compressed. The process

(a) must be very nearly adiabatic

(b) must be very nearly isothermal

(c) may be very nearly adiabatic

(d) may be very nearly isothermal

#### Answer:

(c) may be very nearly adiabatic

(d) may be very nearly isothermal

Due to sudden compression, the gas did not get sufficient time for heat exchange. So, no heat exchange occurred. Therefore, the process may be adiabatic. For any process to be isothermal, its temperature should remain constant, i.e. pressure and volume should change simultaneously while their product (temperature) should be constant.

#### Page No 77:

#### Question 2:

Let *Q* and *W* denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.

(a) *Q* = 0

(b) *W* = 0

(c) *Q* ≠ *W*

(d) *Q* = *W*

#### Answer:

(d) *Q = W*

In an isothermal process, temperature of the system stays constant, i.e. there's no change in internal energy. Thus*, U* = 0, where U denotes the change in internal energy of the system. According to the first law of thermodynamics, heat supplied to the system is equal to the sum of change in internal energy and work done by the system, such that *Q = U + W*. As *U* = 0, *Q = W*.

#### Page No 77:

#### Question 3:

Let *Q* and *W* denote the amount of heat given to an ideal gas and the work done by it in an adiabatic process.

(a) *Q* = 0

(b) *W* = 0

(c) *Q* = *W*

(d) *Q* ≠ *W*

#### Answer:

(a) *Q* = 0

(d) *Q* ≠ *W*

In an adiabatic process, no heat is supplied to the system; so, *Q *= 0.*Q = W+U* and as *Q* = 0, *W* = -*U *and *Q* ≠ *W*.

#### Page No 77:

#### Question 4:

Consider the processes *A* and *B* shown in the figure. It is possible that

Figure

(a) both the processes are isothermal

(b) both the processes are adiabatic

(c) *A* is isothermal and *B* is adiabatic

(d) *A* is adiabatic and *B* is isothermal

#### Answer:

(c) *A* is isothermal and *B* is adiabatic

The slope of an adiabatic process is greater than that of an isothermal process. Since *A* and *B* are initiated from the same initial state, both cannot be isothermal or adiabatic, as they would be overlapping. But the curve of process *B* is steeper than the curve of process *A.* Hence, *A* is isothermal and *B* is adiabatic.

#### Page No 77:

#### Question 5:

Three identical adiabatic containers *A*, *B* and *C* contain helium, neon and oxygen, respectively, at equal pressure. The gases are pushed to half their original volumes.

(a) The final temperatures in the three containers will be the same.

(b) The final pressures in the three containers will be the same.

(c) The pressures of helium and neon will be the same but that of oxygen will be different.

(d) The temperatures of helium and neon will be the same but that of oxygen will be different.

#### Answer:

(c) The pressures of helium and neon will be the same but that of oxygen will be different.

(d) The temperatures of helium and neon will be the same but that of oxygen will be different.

Adiabatic process is expressed as

$p{V}^{\gamma}=\mathrm{constant}\mathrm{or}\phantom{\rule{0ex}{0ex}}T{V}^{\gamma -1}=\mathrm{constant},$

where $\gamma =\frac{{C}_{p}}{{C}_{v}}$ is the ratio of molar heat capacities at constant pressure and volume.

We know that $\gamma $ is equal to 1.67 and 1.40 for a monatomic gas and a diatomic gas, respectively. Helium and neon are monatomic gases and oxygen is a diatomic gas. Therefore, changing the state of the gases, i.e. reducing the volume will lead to identical changes in temperature and pressure for helium and neon and that will be different for oxygen.

#### Page No 77:

#### Question 6:

A rigid container of negligible heat capacity contains one mole of an ideal gas. The temperature of the gas increases by 1° C if 3.0 cal of heat is added to it. The gas may be

(a) helium

(b) argon

(c) oxygen

(d) carbon dioxide

#### Answer:

(a) helium

(b) argon

The temperature of one mole of a gas kept in a container of fixed volume is increased by 1 degree Celsius if 3 calories, i.e. 12.54 J of heat is added to it. So, its molar heat capacity, *C _{v }*= 12.54 J $J{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$, as molar heat capacity at fixed volume is the heat supplied to a mole of gas to increase its temperature by a degree. For a monatomic gas,

*C*= 12.54 JK

_{v $\simeq $$\frac{3}{2}\mathrm{R}=1.5\times 8.314$}^{-1}mol

^{-1}. Among the given gases, only helium and argon are inert and, hence, monoatomic. Therefore, the gas may be helium or argon.

#### Page No 77:

#### Question 7:

Four cylinders contain equal number of moles of argon, hydrogen, nitrogen and carbon dioxide at the same temperature. The energy is minimum in

(a) argon

(b) hydrogen

(c) nitrogen

(d) carbon dioxide

#### Answer:

(a) argon

The energy of a gas is measured as *C _{v}*

*T*

*.*All the four cylinders are at the same temperature but the gases in them have different values of

*C*such that it is least for the monatomic gas and keeps on increasing as we go from monatomic to tri-atomic. Among the above gases, argon is monatomic, hydrogen and nitrogen are diatomic and carbon dioxide is tri-atomic. Therefore, the energy is minimum in argon.

_{v,}#### Page No 77:

#### Question 1:

A vessel containing one mole of a monatomic ideal gas (molecular weight = 20 g mol^{−1}) is moving on a floor at a speed of 50 m s^{−1}. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.

#### Answer:

Number of moles of the ideal gas,* n* = 1 mole

Molecular weight of the gas*, W* = 20 g/mole

Mass of the gas, *m* =20 g

Velocity of the vessel, *V* = 50 m/s

Decrease in K.E. of the vessel = Internal energy gained by the gas

$KE=\frac{1}{2}m({u}^{2}-{v}^{2})\phantom{\rule{0ex}{0ex}}KE=\frac{1}{2}\times 20\times {10}^{-3}\times (0-50\times 50)\phantom{\rule{0ex}{0ex}}KE=-25\mathrm{J}=\mathrm{gain}\mathrm{in}\mathrm{internal}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{gas}\phantom{\rule{0ex}{0ex}}\text{C}\mathrm{hange}\mathrm{in}\mathrm{internal}\mathrm{energy}\mathrm{of}\mathrm{a}\mathrm{gas}=\frac{d}{2}nR(\u2206T),\phantom{\rule{0ex}{0ex}}\mathrm{where}d\mathrm{is}\mathrm{the}\mathrm{degre}\mathrm{of}\mathrm{freedom}\mathrm{of}\mathrm{the}\mathrm{gas}\phantom{\rule{0ex}{0ex}}\text{F}\mathrm{or}\text{a}\mathrm{monoatomic}\mathrm{gas},d=3.\phantom{\rule{0ex}{0ex}}\mathrm{So},25=\frac{3}{2}nR(\u2206T)\phantom{\rule{0ex}{0ex}}\Rightarrow 25=1\times \frac{3}{2}\times 8.31\times \u2206T\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206T=\frac{50}{(3\times 8.3)}=2\mathrm{K}$

#### Page No 77:

#### Question 2:

5 g of a gas is contained in a rigid container and is heated from 15°C to 25°C. Specific heat capacity of the gas at constant volume is 0.172 cal g^{−1} °C^{−1} and the mechanical equivalent of heat is 4.2 J cal^{−1}. Calculate the change in the internal energy of the gas.

#### Answer:

Given:

Mass of the gas, *m* = 5 g

Change in temperature of the system, ∆*T* = 25 − 15°C = 10°C

Specific heat at constant volume, C_{v} = 0.172 cal/g -°C

Mechanical equivalent, *J* = 4.2 J/cal

From the first law of thermodynamics,

*dQ = dU + dW*

Now,

$\u2206$*V *= 0 (Rigid wall of the container keeps the volume constant)

So, *dW = P$\u2206V$=* 0

Therefore,

*dQ = dU *(From the first law)

*Q = m${c}_{v}$dT* = 5 × 0.172 × 10

= 8.6 cal = 8.6 × 4.2 J

= 36.12 J

So, change in internal energy of the system is 36.12 J.

#### Page No 77:

#### Question 3:

The figure shows a cylindrical container containing oxygen (γ = 1.4) and closed by a 50-kg frictionless piston. The area of cross-section is 100 cm^{2}, atmospheric pressure is 100 kPa and *g* is 10 m s^{−2}. The cylinder is slowly heated for some time. Find the amount of heat supplied to the gas if the piston moves out through a distance of 20 cm.

Figure

#### Answer:

Given:

Mass of the piston (*m*) = 50 kg

Adiabatic constant of the gas, *γ *= 1.4

Area of cross-section of the piston (*A*) = 100 cm^{2}

Atmospheric pressure (*P _{0}*) = 100 kPa

*g*= 10 m/s

^{2}

Distance moved by the piston ,

*x*= 20 cm

Work done by the gas,

$d\mathrm{W}=Pdv\phantom{\rule{0ex}{0ex}}\text{Thep}\mathrm{ressure}\left(p\right)\mathrm{is}\mathrm{because}\mathrm{of}\mathrm{two}\mathrm{factors}:\text{the}\mathrm{first}\mathrm{is}\mathrm{the}\mathrm{initial}\mathrm{pressure}\mathrm{and}\text{the}\mathrm{second}\mathrm{is}\mathrm{the}\mathrm{weight}\mathrm{of}\mathrm{the}\mathrm{piston}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}W=\left(\frac{mg}{\mathrm{A}}+{P}_{0}\right)\times Adx\phantom{\rule{0ex}{0ex}}W=\left(\frac{50\times 10}{100\times {10}^{-4}}+{10}^{5}\right)\times 100\times {10}^{-4}\times 20\times {10}^{-4}$

*W*= (5 × 10

^{4}+ 10

^{5}) × 20 × 10

^{−4}

*W*= 1.5 × 10

^{5}× 20 × 10

^{−4}

*W*= 300 J

Hence,

*nRd*T = P$\u2206V$ = 300

$\Rightarrow d\mathrm{T}=\frac{300}{nR}\phantom{\rule{0ex}{0ex}}\mathrm{So},d\mathrm{Q}=n{\mathrm{C}}_{p}dT=n{c}_{p}\times \left(\frac{300}{nR}\right)\phantom{\rule{0ex}{0ex}}\text{U}\mathrm{sing}{C}_{p}-{C}_{v}=Rand\frac{{C}_{p}}{{C}_{v}}=\gamma ,\phantom{\rule{0ex}{0ex}}d\mathrm{Q}=\frac{n\gamma R300}{(\gamma -1)n\mathrm{R}}\phantom{\rule{0ex}{0ex}}d\mathrm{Q}=\left(\frac{300\times 1.4}{0.4}\right)=1050\mathrm{J}$

#### Page No 77:

#### Question 4:

The specific heat capacities of hydrogen at constant volume and at constant pressure are 2.4 cal g^{−1} °C^{−1} and 3.4 cal g^{−1} °C^{−1} respectively. The molecular weight of hydrogen is 2 g mol^{−1} and the gas constant, *R* = 8.3 × 10^{7} erg °C^{−1} mol^{−1}. Calculate the value of *J*.

#### Answer:

Specific heat capacity at constant volume, *C _{v}*(H

_{2}) = 2.4 cal/g-°C

Specific heat capacity at constant pressure,

*C*(H

_{p}_{2}) = 3.4 cal/g-°C

Molecular weight,

*M*= 2 g/mol

Gas constant,

*R*= 8.3 × 10

^{7}erg/mol-°C

We know: C

_{p}− C

_{v}

_{ }= 1 cal/g-°C,

where C

_{p}

_{ and}

_{ }C

_{v}

_{ }are molar specific heat capacities.

So, difference of molar specific heat,

*C*= 1 cal/mol-°C

_{p}× M − C_{v}× MNow, 2 ×

*J*=

*R*

⇒ 2 ×

*J*= 8.3 × 10

^{7}erg/mol-°C

⇒

*J*= 4.15 × 10

^{7}erg/cal

#### Page No 77:

#### Question 5:

The ratio of the molar heat capacities of an ideal gas is *C _{p}*/

*C*= 7/6. Calculate the change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K (a) keeping the pressure constant (b) keeping the volume constant and (c) adiabatically.

_{v}#### Answer:

Given:

*$\left(\frac{{\mathrm{C}}_{p}}{{\mathrm{C}}_{v}}\right)=\frac{7}{6}$*

Number of moles of the gas, *n* = 1 mol

Change in temperature of the gas, ∆*T* = 50 K

(a) Keeping the pressure constant:

Using the first law of thermodynamics,

*dQ = dU + dW*

∆*T* = 50 K and $\gamma =\frac{7}{6}$

*dQ = dU + dW*

Work done,* dW* = *PdV*

As pressure is kept constant, work done = *P($\mathit{\u2206}V$)*

Using the ideal gas equation *PV* = *nRT,*

*P($\mathit{\u2206}V$) = **nR**($\u2206$T)*

⇒ *dW = nR($\u2206$T*)

At constant pressure, * dQ = nC _{p}_{ }dT *

Substituting these values in the first law of thermodynamics, we get

*nC*

_{p}*=*

_{ }dT*dU*

*+ RdT*

⇒

*dU*

*= nC*

_{p}dT − RdTUsing $\frac{{C}_{p}}{{C}_{V}}\mathit{=}\gamma \mathit{}\mathit{}\mathit{}\mathit{}\mathit{\text{and}}\mathit{}{C}_{P}\mathit{-}{C}_{V}\mathit{=}R,\text{weget}$

$dU=1\times \frac{R\gamma}{(\gamma -1)}\times d\mathrm{T}-Rd\mathrm{T}$

= 7 R

*d*T − R

*d*T

= 7 R

*d*T − R

*d*T = 6 R

*d*T

= 6 × 8.3 × 50 = 2490 J

(b) Keeping volume constant:

Work done = 0

Using the first law of thermodynamics,

*dU = dQ*

dU = nC

dU = nC

_{v}dT$=1\times \frac{R}{\gamma -1}\times d\mathrm{T}\phantom{\rule{0ex}{0ex}}=1\times \left(\frac{8.3}{{\displaystyle \frac{7}{6}}-1}\right)\times 50$

= 8.3 × 50 × 6 = 2490 J

(c) Adiabatically,

*dQ*= 0

Using the first law of thermodynamics, we get

*dU = − dW*

$=\left[\frac{n\times R}{\gamma -1}({T}_{1}-{T}_{2})\right]\phantom{\rule{0ex}{0ex}}=\frac{1\times 8.3}{7/6-1}=({T}_{2}-{T}_{1})$

= 8.3 × 6 × 50 = 2490 J

#### Page No 78:

#### Question 6:

A sample of air weighing 1.18 g occupies 1.0 × 10^{3} cm^{3} when kept at 300 K and 1.0 × 10^{5} Pa. When 2.0 cal of heat is added to it at constant volume, its temperature increases by 1°C. Calculate the amount of heat needed to increase the temperature of air by 1°C at constant pressure if the mechanical equivalent of heat is 4.2 × 10^{7} erg cal^{−1}. Assume that air behaves as an ideal gas.

#### Answer:

Here,

m = 1.18 g = 1.18×10^{-3} kg

ΔQ = 2.0×4.2 J

P = 1.0×10^{6} Pa

V = 1.0×10^{3} cm^{3} = 1.0×10^{-3} m^{3}

T = 300K

Applying eqn. of state

PV = nRT

=> n = PV/RT

=> n = 1.0×10^{5}×1.0×10^{-3}/(8.314×300)

=> n = 0.04

ΔT = 1^{0}C

(ΔH)_{v} = nC_{v}_{ }ΔT

2.0×4.2 = nC_{v}×1

=> C_{v} = 8.4/n= 8.4/0.04

=> C_{v} = 210

Again we know

C_{p} – C_{v} =R

=> C_{p} = R + C_{v}

=> C_{p} = 8.3 + 210

=> C_{p} = 218.3

Now at constant pressure

(ΔH)_{p} = nC_{p}_{ }ΔT

=> (ΔH)_{p} = 218.3×0.04×1 = 8.732 J

In calories

=> (ΔH)_{p} = 8.732/4.2 = 2.08 cal

#### Page No 78:

#### Question 7:

An ideal gas expands from 100 cm^{3} to 200 cm^{3} at a constant pressure of 2.0 × 10^{5} Pa when 50 J of heat is supplied to it. Calculate (a) the change in internal energy of the gas (b) the number of moles in the gas if the initial temperature is 300 K (c) the molar heat capacity *C _{p}* at constant pressure and (d) the molar heat capacity

*C*at constant volume.

_{v}#### Answer:

Initial volume of the gas, *V*_{1} = 100 cm^{3}

Final volume = *V*_{2} = 200 cm^{3}

Pressure = 2 × 10^{5} Pa

Heat supplied, *dQ** *= 50 J

(a) According to the first law of thermodynamics,

*dQ* = *dU* + *dW*

d*W* = *P*$\u2206V=2\times {10}^{5}\times (200-100)\times {10}^{-6}=20$

⇒ 50 = *dU* + 2 × 10

⇒ *dU* = 30 J

(b) For a monatomic gas,

*U = $\frac{3}{2}n$RT*

30 = *n* × $\frac{3}{2}$ × 8.3 × 300

$\Rightarrow n=\frac{2}{(83\times 3)}=\frac{2}{249}=0.008$

(c) Also,

*dU = nC _{v}dT*

$\Rightarrow {C}_{\mathit{v}}\mathit{=}\frac{\mathit{d}\mathit{U}}{\mathit{n}\mathit{d}\mathit{T}}=\frac{30}{0.008\times 300}=12.5$

*C*R = 12.5 + 8.3 = 20.8 J/mol-K

_{p}= C_{v}+(d) C

_{v}= 12.5 J/mol-K

#### Page No 78:

#### Question 8:

An amount *Q* of heat is added to a monatomic ideal gas in a process in which the gas performs a work *Q*/2 on its surrounding. Find the molar heat capacity for the process.

#### Answer:

Given:

Amount of heat given to the gas* = Q*

So, ∆*Q** = Q*

Work done by the gas, ∆*W* = *$\frac{\mathrm{Q}}{2}$*

From the first law of thermodynamics,

∆*Q* = ∆*W* + ∆*U*

$\Rightarrow \u2206U=\mathrm{Q}-\frac{\mathrm{Q}}{2}=\frac{\mathrm{Q}}{2}$

For a monoatomic gas,

∆*U** = $\frac{3}{2}$ nRdT*

*$\Rightarrow \frac{\mathrm{Q}}{2}=ndT\times \frac{3}{2}R$*

$\Rightarrow $*Q* = 3*nRdT*

Again, for expansion at constant pressure,

*Q = nC _{p}dT, *

where

*C*is the molar heat capacity at constant pressure.

_{p}So, 3

*RndT*

*=*

*nC*

_{p}dT⇒

*C*= 3

_{p}*R*

#### Page No 78:

#### Question 9:

An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation *p* = *kV*. Show that the molar heat capacity of the gas for the process is given by *C *= *C _{v}* + $\frac{R}{2}$.

#### Answer:

Relation between pressure and volume of a gas is *P = kV*.

Ideal gas equation is *PV = nRT*.

*$\Rightarrow \frac{nRT}{V}=kV\phantom{\rule{0ex}{0ex}}\Rightarrow nRT=k{V}^{2}$*

For simplicity, take the number of moles of a gas, *n *= 1.

⇒* RdT = 2 kVdV
$\Rightarrow \frac{RdT}{2kV}=dV$*

From the first law of thermodynamics,

*dQ*

*= dU + dW*

⇒

*n${C}_{p}$dT = C*

_{v}dT + PdV$\Rightarrow n{C}_{p}dT=\mathit{}{C}_{v}\mathit{}dT+\frac{PRdT}{\mathit{2}kV}\phantom{\rule{0ex}{0ex}}\Rightarrow 1\times {C}_{p}=\mathit{}{C}_{v}+\frac{PR}{2kV}\phantom{\rule{0ex}{0ex}}\therefore {C}_{p}={C}_{v}+\frac{R}{2}$

#### Page No 78:

#### Question 10:

An ideal gas (*C _{p}_{ }*/

*C*= γ) is taken through a process in which the pressure and the volume vary as

_{v}_{ }*p*=

*aV*. Find the value of

^{b}*b*for which the specific heat capacity in the process is zero.

#### Answer:

As the process has specific heat capacity zero, the process is essentially an adiabatic process.

#### Page No 78:

#### Question 11:

Two ideal gases have the same value of *C _{p}* /

*C*= γ. What will be the value of this ratio for a mixture of the two gases in the ratio 1 : 2?

_{v}#### Answer:

For the first ideal gas,

*C*_{p1} = specific heat at constant pressure

*C*_{v1} = specific heat at constant volume

*n*_{1} = number of moles of the gas

$\frac{{C}_{{\mathrm{p}}_{1}}}{{C}_{{\mathrm{V}}_{1}}}$ = γ and C_{p1} − C_{v1} = *R*

⇒ *γC*_{v1}* − **C*_{v1}* = R*

⇒ *C*_{v1}* (γ − 1)* = *R*

$\Rightarrow {\mathrm{Cv}}_{1}=\frac{R}{(\mathrm{\gamma}-1)}\phantom{\rule{0ex}{0ex}}C{\mathrm{p}}_{1}=\mathrm{\gamma}\frac{R}{(\mathrm{\gamma}-1)}$

For the second ideal gas,

*C*_{p2} = specific heat at constant pressure

*C*_{v2}* _{ }*= specific heat at constant volume

*n*= number of moles of the gas

_{2}$\frac{{\mathrm{C}}_{{\mathrm{p}}_{2}}}{{\mathrm{C}}_{{\mathrm{v}}_{2}}}$ = γ and

*C*

_{p2}

*−*

*C*

_{v2}=

*R*

*⇒*

*γC*

_{v2}

*−*

*C*

_{v2}

*=*

*R*

⇒

*C*

_{v2}

*(γ − 1)*=

*R*

$\Rightarrow {\mathrm{Cv}}_{2}=\frac{R}{(\mathrm{\gamma}-1)}\mathrm{and}\phantom{\rule{0ex}{0ex}}{\mathrm{Cp}}_{2}=\mathrm{\gamma}\frac{R}{(\mathrm{\gamma}-1)}$

Given:

*n*= 1 : 2

_{1}= n_{2}*dU*

_{1}=

*nC*

_{v1}

*dt*

*dU*

_{2}= 2

*nC*

_{v2}

*dT*

When the gases are mixed,

*nC*

_{v1}

*dT*+ 2

*n*C

_{v2}

*dT*= 3

*n*C

_{v}

*dT*

${\mathrm{C}}_{\mathrm{v}}=\frac{{\mathrm{C}}_{{v}_{1}}+2{\mathrm{C}}_{{v}_{2}}}{3}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{R}{(\mathrm{\gamma}-1)}}+{\displaystyle \frac{2R}{(\mathrm{\gamma}-1)}}}{3}\phantom{\rule{0ex}{0ex}}=\frac{3R}{(\mathrm{\gamma}-1)3}=\frac{R}{\mathrm{\gamma}-1}$

Hence, C

*p*/ C

*v*in the mixture is γ.

#### Page No 78:

#### Question 12:

A mixture contains 1 mole of helium (*C _{p}* = 2.5

*R*,

*C*= 1.5

_{v}*R*) and 1 mole of hydrogen (

*C*= 3.5

_{p}*R*,

*C*

_{v}_{ }= 2.5

*R*). Calculate the values of

*C*,

_{p}*C*and γ for the mixture.

_{v}#### Answer:

Specific heat at constant pressure of helium, *C _{p}*' = 2.5

*R*

Specific heat at constant pressure of hydrogen,

*C*= 3.5

_{p}"*R*

Specific heat at constant volume of helium

*,*

*C*

_{v}

*'*= 1.5

*R*

Specific heat at constant volume of hydrogen,

*C*

_{v}

*"*= 2.5

*R*

*n*= 1 mol

_{1}= n_{2}_{$dU\mathit{=}n{C}_{v}dT$}

For the mixture of two gases,

$d{U}_{1}+d{U}_{2}=1\mathrm{mol}$

[

*n*

_{1}

*+*

*n*

_{2}]

*C*

_{v}dT =*n*

_{1}

*C'*

_{v}dT*+*

*n*

_{2}

*C*

*"*,

_{v}dTwhere ${C}_{V}$ is the heat capacity of the mixture.

$\Rightarrow {C}_{v}\mathit{=}\frac{{n}_{1}C{\mathit{\text{\'}}}_{v}\mathit{+}{n}_{2}C{\mathit{"}}_{v}}{{n}_{1}\mathit{+}{n}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}=\frac{1.5R+2.5R}{2}=2R$

*Cp = Cv + R*= 2

*R*+

*R*= 3

*R*

$\mathrm{\gamma}=\frac{\mathrm{C}p}{\mathrm{C}v}=\frac{3R}{2R}=1.5$

#### Page No 78:

#### Question 13:

Half mole of an ideal gas (γ = 5/3) is taken through the cycle *abcda*, as shown in the figure. Take $R=\frac{25}{3}\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$. (a) Find the temperature of the gas in the states *a*, *b*, *c* and *d*. (b) Find the amount of heat supplied in the processes *ab* and *bc*. (c) Find the amount of heat liberated in the processes *cd* and *da*.

Figure

#### Answer:

Given:

Number of moles of the gas, $n=0.5\mathrm{mol}\phantom{\rule{0ex}{0ex}}$

$R=\frac{25}{3}\mathrm{J}/\mathrm{mol}-\mathrm{K}\phantom{\rule{0ex}{0ex}}\gamma =\frac{5}{3}$

(a) Temperature at *a* = ${T}_{a}$

${P}_{\mathit{a}}{V}_{\mathit{a}}\mathit{=}nR{T}_{\mathit{a}}$

$\Rightarrow Ta\mathit{=}\frac{PaVa}{nR}=\frac{100\times {10}^{3}\times 5000\times {10}^{-6}}{0.5\times {\displaystyle \frac{25}{3}}}=120\mathrm{K}$

Similarly, temperature at *b*,

${T}_{b}\mathit{=}\frac{{P}_{b}{V}_{b}}{nR}\phantom{\rule{0ex}{0ex}}{T}_{b}=\frac{100\times {10}^{3}\times 10000\times {10}^{-6}}{0.5\times {\displaystyle \frac{25}{3}}}\phantom{\rule{0ex}{0ex}}{T}_{b}=240K$

Similarly, temperature at *c* is 480 K and at *d* is 240 K.

(b) For process *ab*,

d*Q** = **nc _{p}*d

*T*

[Since

*ab*is isobaric]

$dQ=\frac{1}{2}\times \frac{R\gamma}{\gamma \mathit{-}\mathit{1}}\mathit{(}{T}_{b}\mathit{-}{T}_{a})\phantom{\rule{0ex}{0ex}}dQ=\frac{1}{2}\times \frac{{\displaystyle \frac{25\times 5}{3\times 3}}}{{\displaystyle \frac{5}{3}}-1}\times (240-120)\phantom{\rule{0ex}{0ex}}dQ=\frac{1}{2}\times \frac{125}{9}\times \frac{3}{2}\times \left(120\right)\phantom{\rule{0ex}{0ex}}dQ=1250\mathrm{J}$

For line

*bc,*volume is constant. So, it is an isochoric process.

d

*Q*= d

*U*+

*dW*

[

*d*W = 0, isochoric process]

*dQ = dU = nC*

_{v}dT*dQ*=

*nC*

_{v}(${T}_{c}-{T}_{b})$$dQ=\frac{1}{2}\times \frac{\left({\displaystyle \frac{25}{3}}\right)}{\left[\left({\displaystyle \frac{5}{3}}\right)-1\right]}\times \left(240\right)\phantom{\rule{0ex}{0ex}}dQ=\frac{1}{2}\times \frac{25}{3}\times \frac{3}{2}\times 240=1500\mathrm{J}$

(c) Heat liberated in

*cd*(isobaric process),

*dQ = − nC*

_{p}dT$dQ=-\frac{1}{2}\times \frac{\gamma R}{\mathit{(}\gamma \mathit{-}\mathit{1}\mathit{)}}\mathit{\times}\mathit{(}{T}_{d}\mathit{-}{T}_{c})\phantom{\rule{0ex}{0ex}}dQ=-\frac{1}{2}\times \frac{125}{9}\times \frac{3}{2}\times (240-480)\phantom{\rule{0ex}{0ex}}dQ=-\frac{1}{2}\times \frac{125}{6}\times 240=2500\mathrm{J}$

Heat liberated in

*da*(isochoric process),

$\Rightarrow $

*dQ = dU*

*Q= −nC*

_{v}dT$dQ=-\frac{1}{2}\times \frac{R}{\gamma \mathit{-}\mathit{1}}\left(Ta\mathit{-}Td\right)\phantom{\rule{0ex}{0ex}}dQ=-\frac{1}{2}\times \frac{25}{2}\times (120-240)\phantom{\rule{0ex}{0ex}}dQ=\frac{25}{4}\times 120=750\mathrm{J}$

#### Page No 78:

#### Question 14:

An ideal gas (γ = 1.67) is taken through the process *abc* shown in the figure. The temperature at point *a* is 300 K. Calculate (a) the temperatures at *b* and *c* (b) the work done in the process (c) the amount of heat supplied in the path *ab* and in the path *bc* and (d) the change in the internal energy of the gas in the process.

Figure

#### Answer:

(a) For line *ab,* volume is constant.

So, from the ideal gas equation,

$\frac{{P}_{\mathit{1}}}{{T}_{\mathit{1}}}\mathit{=}\frac{{P}_{\mathit{2}}}{{T}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{100}{300}=\frac{200}{{\mathrm{T}}_{2}}\phantom{\rule{0ex}{0ex}}$

${T}_{\mathit{2}}=\left(\frac{200\times 300}{100}\right)=600\mathrm{K}$

For line *bc*, pressure is constant.

$\mathrm{So},\frac{{\mathit{V}}_{\mathit{1}}}{{\mathit{T}}_{\mathit{1}}}\mathit{=}\frac{{\mathit{V}}_{\mathit{2}}}{{\mathit{T}}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{100}{600}=\frac{150}{{\mathrm{T}}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{T}}_{2}\left(\frac{600\times 150}{100}\right)=900\mathrm{K}$

(b)

As process *ab* is isochoric, ${W}_{ab}=0$.

During process *bc,*

*P* = 200 kPa

The volume is changing from 100 to 150 cm^{3} .

Therefore, work done = 50 × 10^{−6} × 200 × 10^{3} J

= 10 J

(c) For* ab* (isochoric process), work done = 0.

From the first law,

*dQ = dU = nC _{v}dT*

$\Rightarrow $Heat supplied =

*nC*

_{v}dTNow,

${Q}_{ab}\mathit{=}\left(\frac{PV}{RT}\right)\mathit{\times}\left(\frac{R}{\gamma \mathit{-}\mathit{1}}\right)\mathit{\times}dT\phantom{\rule{0ex}{0ex}}=\frac{200\times {10}^{3}\times 100\times {10}^{-6}\times 300}{600\times 0.67}\phantom{\rule{0ex}{0ex}}=14.925(\therefore \mathrm{\gamma}=1.67)\phantom{\rule{0ex}{0ex}}\text{F}\mathrm{or}\mathrm{bc}(\mathrm{isobaric}\mathrm{process}):\phantom{\rule{0ex}{0ex}}\mathrm{Heat}\mathrm{supplied}\mathrm{in}\mathrm{bc}=n{C}_{p}dT\mathit{}\mathit{}\mathit{}\mathit{}\left({C}_{p}\mathit{=}\frac{\gamma R}{\gamma \mathit{-}\mathit{1}}\right)\phantom{\rule{0ex}{0ex}}\mathit{=}\frac{PV}{RT}\mathit{\times}\frac{\gamma R}{\gamma \mathit{-}\mathit{1}}\mathit{\times}dT\phantom{\rule{0ex}{0ex}}=\frac{200\times {10}^{3}\times 150\times {10}^{-6}}{600\times 0.67}\times 300\phantom{\rule{0ex}{0ex}}=10\times \frac{1.67}{0.67}=\frac{16.7}{0.67}=24.925\phantom{\rule{0ex}{0ex}}$

(d) d

*Q*

*= dU + dW*

Now,

*dU*=

*dQ*−

*dW*

= Heat supplied − Work done

= (24.925 + 14.925) − 10

= 39.850 − 10 = 29.850 J.

#### Page No 78:

#### Question 15:

In Joly's differential steam calorimeter, 3 g of an ideal gas is contained in a rigid closed sphere at 20°C. The sphere is heated by steam at 100°C and it is found that an extra 0.095 g of steam has condensed into water as the temperature of the gas becomes constant. Calculate the specific heat capacity of the gas in J g^{−1} K^{−1}. The latent heat of vaporisation of water = 540 cal g^{−1}

#### Answer:

For Joly's differential steam calorimeter,

${C}_{v}\mathit{=}\frac{{m}_{\mathit{2}}L}{{m}_{\mathit{1}}\mathit{}\mathit{(}{\theta}_{2}\mathit{-}{\theta}_{1}\mathit{)}}$,

where

*m*_{2} = mass of steam condensed

*m*_{2} = 0.095 g

Latent heat of vapourization, *L* = 540 cal/g = 540 × 4.2 J/g

*m*_{1} = mass of gas present

*m*_{1} = 3 g

Initial temperature, *θ _{1}* = 20°C

Final temperature,

*θ*= 100°C

_{2}$\Rightarrow {C}_{v}=\frac{0.095\times 540\times 4.2}{3\times (100-20)}$

= 0.89 = 0.9 J/g-K

#### Page No 78:

#### Question 16:

The volume of an ideal gas (γ = 1.5) is changed adiabatically from 4.00 litres to 3.00 litres. Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.

#### Answer:

Given,

*γ* = 1.5

Since the process is adiabatic, *PV ^{γ}* = constant.

(a)

*P*

_{1}

*V*

_{1}

^{γ}*=*

*P*

_{2}

*V*

_{2}

^{γ}Given,

*V*

_{1}= 4 L

*V*

_{2}= 3 L

We need to find

*$\frac{{P}_{2}}{{P}_{1}}$*.

$\Rightarrow \frac{{\mathit{P}}_{\mathit{2}}}{{\mathit{P}}_{\mathit{1}}}\mathit{=}{\left(\frac{{\mathit{V}}_{\mathit{1}}}{{\mathit{V}}_{\mathit{2}}}\right)}^{\mathit{\gamma}}\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{3}\right)}^{1.5}=1.5396=1.54$

(b) Also, for an adiabatic process,

*TV*= constant

^{γ−1}*T*

_{1}

*V*

_{1}

^{γ−1}=*T*

_{2}

*V*

_{2}

^{γ−1}$\Rightarrow \frac{{T}_{\mathit{2}}}{{T}_{\mathit{1}}}\mathit{=}{\left(\frac{{V}_{\mathit{1}}}{{V}_{\mathit{2}}}\right)}^{\gamma \mathit{-}\mathit{1}}={\left(\frac{4}{3}\right)}^{0.5}=1.154$

#### Page No 78:

#### Question 17:

An ideal gas at pressure 2.5 × 10^{5} Pa and temperature 300 K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure (b) the final temperature and (c) the work done by the gas in the process. Take γ = 1.5

#### Answer:

Initial pressure of the gas, *P*_{1} = 2.5 × 10^{5} Pa

Initial temperature, *T*_{1} = 300 K

Initial volume, *V*_{1} = 100 cc

(a) For an adiabatic process,

*P*_{1}*V*_{1}^{γ}* = **P*_{2}*V*_{2}^{γ}

⇒ 2.5 × 10^{5} × *V*^{1.5} = ${\left(\frac{\mathrm{V}}{2}\right)}^{1.5}$ × *P*_{2}

⇒ *P*_{2} = 7.07 × 10^{5}

= 7.1 × 10^{5} Pa

(b) Also, for an adiabatic process,

*T _{1}V*

_{1}

^{γ−}^{1}

*=*

*T*

_{2}

*V*

_{2}

^{γ}^{−}^{1}

⇒ 300 × (100)

^{1.5−1}= T

_{2}× ${\left(\frac{100}{2}\right)}^{1.5-1}$

=

*T*

_{2}× (50)

^{1.5−1}

⇒ 300 × 10 =

*T*

_{2}

_{ }× 7.07

⇒

*T*

_{ 2}= 424.32 K = 424 K

(c) Work done by the gas in the process,

$W\mathit{=}\frac{nR}{\mathit{(}\gamma \mathit{-}\mathit{1}\mathit{)}}\mathit{}[{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}]\phantom{\rule{0ex}{0ex}}\mathit{=}\frac{{P}_{\mathit{1}}{V}_{\mathit{1}}}{T\mathit{(}\gamma \mathit{-}\mathit{1}\mathit{)}}\mathit{}\mathit{}[{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}]\phantom{\rule{0ex}{0ex}}=\frac{2.5\times 10}{300\times 0.5}\times (-124)\phantom{\rule{0ex}{0ex}}=-20.67\approx -21\mathrm{J}$

#### Page No 78:

#### Question 18:

Air (γ = 1.4) is pumped at 2 atm pressure in a motor tyre at 20°C. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre? Neglect any mixing with the atmospheric air.

#### Answer:

Given:

For air,* γ* = 1.4

Initial temperature of air, *T*_{1} = 20°C = 293 K

Initial pressure, *P*_{1} = 2 atm

Final pressure, *P*_{2} = 1 atm

The bursting of the tyre is an adiabatic process. For an adiabatic process,

*P*_{1}^{1}^{−}^{γ} × *T*_{1}^{1}^{−}^{γ}* = **P*^{1}^{−}^{γ}* × **T*_{2}^{γ}

(2)^{1−1.4} × (293)^{1.4} = (1)^{1−1.4} ×* T*_{2}^{1.4}

⇒ (2)^{−0.4} × (293)^{1.4} = *T*_{2}^{1.4}

⇒ 2153.78 = *T*_{2}^{1.4}

⇒ *T*_{2}* *= (2153.78)^{1/1.4}

= 240.3 K

#### Page No 78:

#### Question 19:

A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure, volume and temperature of the gas are 100 kPa, 400 cm^{3} and 300 K, respectively. The ratio of the specific heat capacities of the gas, *C _{p}* /

*C*= 1.5. Find the pressure and the temperature of the gas if it is (a) suddenly compressed (b) slowly compressed to 100 cm

_{v}^{3}.

#### Answer:

Initial pressure of the gas, *P*_{1} = 100 kPa

Initial volume of the gas,*V*_{1} = 400 cm^{3}

= 400 × 10^{−6} m^{3}

Initial temperature of the gas, *T*_{1} = 300 K

*$\gamma =\frac{\mathrm{C}p}{\mathrm{C}v}=1.5$*

(a) The gas is suddenly compressed to volume, *V*_{2}* =* 100 cm^{3} .

So, this is an adiabatic process.

For an adiabatic process,

*P*_{1}*V*_{1}^{γ}* = **P*_{2}*V*_{2}^{γ}

⇒ 10^{5} × (400)^{1.5} = *P*_{2} (100)^{1.5 }

⇒ *P*_{2}* *= 10^{5}(4)^{1.5} = 800 kPa

Also,

*T*_{1}*V ^{γ−}*

^{1}

*=*

*T*

_{2}

*V*

_{2}

^{γ−}^{1}

⇒ 300 × (400)

^{1.5−1}=

*T*

_{2}(100)

^{1.5−1}

⇒ 300 × (400)

^{0.5}=

*T*

_{2}(100)

^{0.5 }

⇒

*T*

_{2}= 600 K

(b) If the container is slowly compressed, the heat transfer is zero, even thought the walls are adiabatic.

Thus, the values remain same. Thus,

*P*

_{2}= 800 kPa

*T*

_{2}= 600 K

#### Page No 78:

#### Question 20:

The initial pressure and volume of a given mass of a gas (*C _{p}*/

*C*= γ) are

_{v}*p*

_{0}and

*V*

_{0}. The gas can exchange heat with the surrounding. (a) It is slowly compressed to a volume

*V*

_{0}/2 and then suddenly compressed to

*V*

_{0}/4. Find the final pressure. (b) If the gas is suddenly compressed from the volume

*V*

_{0}to

*V*

_{0}/2 and then slowly compressed to

*V*

_{0}/4, what will be the final pressure?

#### Answer:

Given:

For the gas, *$\frac{\mathit{C}\mathit{p}}{\mathit{C}\mathit{v}}\mathit{=}\gamma $*

Initial pressure of the gas = *P*_{0}

Initial volume of the gas = *V*_{0}

(a)

(i) As the gas is slowly compressed, its temperature will remain constant.

For isothermal compression,

*P*_{1}*V*_{1}* = **P*_{2}*V*_{2}

*So, **P*_{0}*V*_{0}* = **P*_{2}* $\frac{{\mathrm{V}}_{0}}{2}$* ⇒ *P*_{2} = 2*P*_{0}

(ii) Sudden compression means that the gas could not get sufficient time to exchange heat with its surroundings. So, it is an adiabatiac compression.

So, for adiabatic compression,

*P*_{1}*V*_{1}^{γ}* = **P*_{2}*V*_{2}^{γ}

$\text{Or}2{P}_{0}{\left(\frac{{V}_{0}}{2}\right)}^{\gamma}={\mathrm{P}}_{2}{\left(\frac{{V}_{0}}{4}\right)}^{\mathrm{\gamma}}\phantom{\rule{0ex}{0ex}}\Rightarrow {P}_{2}=\frac{{{V}_{\mathit{0}}}^{\gamma}}{{2}^{\mathrm{\gamma}}}\times 2{P}_{0}\times \frac{{4}^{\mathrm{\gamma}}}{{{V}_{\mathit{0}}}^{\gamma}}$

= 2^{γ} × 2*P*_{0} = *P*_{0}2^{γ}^{+1}

(b)

(i) Adiabatic compression:

*P*_{1}*V*_{1}^{γ}* = **P*_{2}*V*_{2}^{γ}

* **P*_{0}*V*_{0}^{γ}* = P'${\left(\frac{{V}_{0}}{2}\right)}^{\gamma}$*

⇒ *P' = **P*_{0}*2 ^{γ}*

(ii) Isothermal compression:

*P*

_{1}

*V*

_{1}

*=*

*P*

_{2}

*V*

_{2}

${2}^{\gamma}{P}_{0}\times \frac{{V}_{0}}{2}=P"\left(\frac{{V}_{0}}{2}\right)$

*P" = P*2

_{0}2^{γ}×⇒

*P" =*

*P*

_{0}2

^{γ}

^{+1 }

#### Page No 78:

#### Question 21:

Consider a given sample of an ideal gas (*C _{p}*/

*C*= γ) having initial pressure

_{v}*p*

_{0}and volume

*V*

_{0}. (a) The gas is isothermally taken to a pressure

*p*

_{0}/2 and from there, adiabatically to a pressure

*p*

_{0}/4. Find the final volume. (b) The gas is brought back to its initial state. It is adiabatically taken to a pressure

*p*

_{0}/2 and from there, isothermally to a pressure

*p*

_{0}/4. Find the final volume.

#### Answer:

(a) Given,

Initial pressure of the gas = *p*_{0}

Initial volume of the gas = *${V}_{0}$*

For an isothermal process,

*PV* = constant

So, *P*_{1}*V*_{1} = *P*_{2}*V*_{2}

*${P}_{\mathit{2}}\mathit{=}\frac{{\mathit{P}}_{\mathit{0}}{\mathit{V}}_{\mathit{0}}}{\mathit{\left(}{\displaystyle \frac{{P}_{0}}{2}}\mathit{\right)}}=2{\mathrm{V}}_{0}$*

For an adiabatic process,* P _{3} = $\frac{{P}_{0}}{4}$, V_{3} *= ?

*P*

_{2}

*V*

_{2}

^{γ}*=*

*P*

_{3}

*V*

_{3}

^{γ}$\Rightarrow {\left(\frac{{V}_{3}}{{V}_{2}}\right)}^{\gamma}\mathit{=}\left(\frac{{P}_{2}}{{P}_{3}}\right)\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}{\left(\frac{{V}_{3}}{{V}_{2}}\right)}^{\gamma}\mathit{=}\left(\frac{{\displaystyle \frac{{P}_{0}}{2}}}{{\displaystyle \frac{{P}_{0}}{4}}}\right)=2\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{{V}_{3}}{{V}_{2}}={2}^{\frac{1}{\gamma}}\phantom{\rule{0ex}{0ex}}\therefore {V}_{3}\mathit{=}{V}_{2\mathit{}}{\mathit{2}}^{\frac{\mathit{1}}{\gamma}}=2{V}_{0}{2}^{\frac{\mathit{1}}{\gamma}}\phantom{\rule{0ex}{0ex}}\mathit{=}{2}^{\frac{\gamma \mathit{+}1}{\gamma}}{V}_{0}$

(b)

*P*

_{1}

*V*

_{1}

^{γ}*=*

*P*

_{2}

*V*

_{2}

^{γ}*$\text{Or}\left(\frac{{V}_{2}}{{V}_{1}}\right)={\left(\frac{{P}_{1}}{{P}_{2}}\right)}^{\frac{1}{\gamma}}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{2}={V}_{0}{2}^{\frac{1}{\gamma}}$*

Again, for an isothermal process,

*P*

_{2}

*V*

_{2}

*=*

*P*

_{3}

*V*

_{3}

*$\therefore \mathit{}{V}_{\mathit{3}}\mathit{=}\frac{{P}_{\mathit{2}}{V}_{\mathit{2}}}{{P}_{\mathit{3}}}={22}^{\frac{1}{\gamma}}{V}_{0}\phantom{\rule{0ex}{0ex}}={2}^{\frac{\mathrm{\gamma}+1}{\mathrm{\gamma}}}{V}_{0}$*

#### Page No 78:

#### Question 22:

A sample of an ideal gas (γ = 1.5) is compressed adiabatically from a volume of 150 cm^{3} to 50 cm^{3}. The initial pressure and the initial temperature are 150 kPa and 300 K. Find (a) the number of moles of the gas in the sample (b) the molar heat capacity at constant volume (c) the final pressure and temperature (d) the work done by the gas in the process and (e) the change in internal energy of the gas.

#### Answer:

The ideal gas equation is

*PV = nRT*

Given, *${P}_{1}$ *= 150 kPa = 150 × 10^{3} Pa

*${V}_{1}$* = 150 cm^{3}^{ }= 150 × 10^{−6} m^{3}

*${T}_{1}$*= 300 K

(a)

$n\mathit{=}\frac{PV}{RT}=9.036\times {10}^{-3}\phantom{\rule{0ex}{0ex}}n=0.009$

(b)

$\frac{{C}_{p}}{{C}_{v}}\mathit{=}\gamma \mathit{,}\mathit{}{C}_{p}\mathit{-}{C}_{v}\mathit{=}R\phantom{\rule{0ex}{0ex}}\text{So}\mathit{,}\mathit{}{C}_{v}\mathit{=}\frac{R}{\gamma \mathit{-}\mathit{1}}=\mathit{2}R=\frac{8.3}{0.5}=16.6\mathrm{J}/\mathrm{mole}-\mathrm{K}$

(c) Given,

*P*_{1} = 150 kPa = 150 × 10^{3} Pa

*P*_{2} = ? *V*_{1} = 150 cm^{3}

= 150 × 10^{−6} m^{3}

*γ *= 1.5

*V*_{2} = 50 cm^{3} = 50 × 10^{−6} m^{3},

*T*_{1}* *= 300 K

*T*_{2} = ?

Since the process is adiabatic, using the equation of an adiabatic process,we get

*P*_{1}*V*_{1}^{γ}* = **P*_{2}*V*_{2}^{γ}

⇒ 150 × 10^{3} × (150 × 10^{−6})^{γ} = *P*_{2} × (50 × 10^{−6})^{γ}

$\Rightarrow {P}_{2}=150\times {10}^{3}\times \frac{(150\times {10}^{-6}{)}^{1.5}}{(50\times {10}^{-6}{)}^{1.5}}$

*P*_{2}_{ }= 150000 × (3)^{1.5}

*P*_{2} = 779.422 × 10^{3} Pa

*P*_{2}_{ }= 780 kPa

Again,

*P*_{1}^{1}^{−γ}* **T*_{1}^{γ}* = **P*_{1}^{1}^{−γ}* **T*_{2}^{γ}

⇒ (150 × 10^{3})^{1−1.5 }× (330)^{1.5} = (780 × 10^{3})^{1−1.5} × T_{2}^{1.5}

⇒ *T*_{2}^{1.5} = (150 × 10^{3})^{1−1.5} × (300)^{1.5} × 300^{1.5}

*T*_{2}^{1.5 }= 11849.050

⇒ *T*_{2} = (11849.050)^{1/1.5}

* T*_{2} = 519.74 = 520 K

(d) d*Q* = d*W* + d*U*

Or *dW* = −d*U* [ Since d*Q* = 0 in an adiabatic process]

*dW *= −*nC _{v}dT*

*dW*= −0.009 × 16.6 × (520 − 300)

*dW*= −0.009 × 16.6 × 220

*dW*= −32.87 J $\approx $ −33 J

(e)

*dU*

*= nC*

_{v}dT*dU*= 0.009 × 16.6 × 220 $\approx $ 33 J

#### Page No 79:

#### Question 23:

Three samples *A*, *B* and *C* of the same gas (γ = 1.5) have equal volumes and temperatures. The volume of each sample is doubled, the process being isothermal for *A*, adiabatic for *B* and isobaric for *C*. If the final pressures are equal for the three samples, find the ratio of the initial pressures.

#### Answer:

There are three gases A, B and C.

It is given that initially,

*V*_{A}* = **V*_{B}* = V*_{C}* *and* **T*_{A}* = T*_{B}* = **T*_{C} .

For A, the process is isothermal and for an isothermal process, *PV *= constant.

*P*_{A}*V*_{A} = *P'*_{A}2*V*_{A}

*$\Rightarrow P{\mathit{\text{'}}}_{\mathrm{A}}\mathit{=}{P}_{\mathrm{A}}\mathit{\times}\frac{1}{2}$*

For B, the process is adiabatic. So,

*P*_{B}*V*_{B}^{γ}* = **P*_{B}*'(*2*V*_{B}*) ^{γ}*

$\Rightarrow {P}_{\mathrm{B}}\text{'}\mathit{=}\frac{{P}_{\mathrm{B}}}{{2}^{1.5}}$

For C, the process is isobaric, which implies that the pressure will remain constant.

So, using the ideal gas equation

*P$=\frac{\mathit{n}\mathit{R}\mathit{T}}{\mathit{V}}$*, we get

$\frac{{V}_{\mathrm{C}}}{{T}_{\mathrm{C}}}\mathit{=}\frac{V{\mathit{\text{'}}}_{\mathrm{C}}}{T{\mathit{\text{'}}}_{\mathrm{C}}}\mathit{\Rightarrow}\frac{{V}_{\mathrm{C}}}{{T}_{\mathrm{C}}}\mathit{=}\frac{2{V}_{\mathrm{C}}}{T{\mathit{\text{'}}}_{\mathit{C}}}$

⇒

*T*

_{C}

*' =*2

*T*

_{C}

As the final pressures are equal,

$\frac{{P}_{\mathit{A}}}{2}=\frac{{P}_{\mathit{B}}}{{2}^{1.5}}={P}_{\mathit{C}}\phantom{\rule{0ex}{0ex}}\Rightarrow {P}_{\mathit{A}}\mathit{}\mathit{:}\mathit{}{P}_{\mathit{B}}\mathit{}\mathit{:}\mathit{}{P}_{\mathit{C}}=2:{2}^{1.5}:1$

#### Page No 79:

#### Question 24:

Two samples *A* and *B,* of the same gas have equal volumes and pressures. The gas in sample *A* is expanded isothermally to double its volume and the gas in *B* is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 2^{1−γ}^{ }= (γ − 1) ln2.

#### Answer:

Let,

Initial pressure of the gas =* P _{1}*

Initial volume of the gas =

*V*

_{1}_{ }Final pressure of the gas=

*P*

_{2}Final volume of the gas =

*V*

_{2}Given,

*V*

_{2}

*=*2

*V*

_{1}, for each case.

In an isothermal expansion process,

work done$=nR{T}_{1}\mathit{}\mathrm{ln}\frac{{V}_{2}}{{V}_{1}}$

Adiabatic work done,

$W=\frac{{P}_{1}{V}_{1}-{P}_{2}{V}_{2}}{\gamma -1}$

It is given that same work is done in both cases.

So,

$nR{T}_{1}\mathit{}\mathrm{ln}\mathit{\left(}\frac{{V}_{2}}{{V}_{1}}\mathit{\right)}\mathit{=}\frac{{P}_{1}{V}_{1}\mathit{-}{P}_{2}{V}_{2}}{\gamma \mathit{-}1}...\left(1\right)$

In an adiabatic process,

${P}_{2}={P}_{1}{\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma}={P}_{1}{\left(\frac{1}{2}\right)}^{\gamma}$

From eq (1),

$nR{T}_{1}\mathit{}\mathrm{ln}\mathit{}2\mathit{=}\frac{{P}_{1}{V}_{1}\left(1-{\displaystyle \frac{1}{{2}^{\mathrm{\gamma}}}}\times 2\right)}{\gamma \mathit{-}1}$

and

*nRT*

_{1}*=*

*P*

_{1}V_{1}

$\mathrm{So},\mathrm{ln}2=\frac{1-{\displaystyle \frac{1}{{2}^{\mathrm{\gamma}}}}.2}{\mathrm{\gamma}-1}$

Or (γ − 1)

*l*n 2 = 1 − 2

^{1−γ}

#### Page No 79:

#### Question 25:

1 litre of an ideal gas (γ = 1.5) at 300 K is suddenly compressed to half its original volume. (a) Find the ratio of the final pressure to the initial pressure. (b) If the original pressure is 100 kPa, find the work done by the gas in the process. (c) What is the change in internal energy? (d) What is the final temperature? (e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (f) The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.

#### Answer:

Given:

*γ =* 1.5

*T* = 300 K

Initial volume* *of the gas*, **V*_{1}* =* 1 L

Final volume, *V*_{2} = $\frac{1}{2}$ L

(a) The process is adiabatic because volume is suddenly changed; so, no heat exchange is allowed.

*P*_{1}*V*_{1}^{γ}* = **P*_{2}*V*_{2}^{γ}

$\text{Or}{P}_{2}={P}_{1}{\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma}={P}_{1}(2{)}^{\gamma}\phantom{\rule{0ex}{0ex}}\frac{{P}_{2}}{{P}_{1}}={2}^{1.5}=2\sqrt{2}$

(b) *P*_{1} = 100 kPa = 10^{5} Pa

and *P*_{2} = $2\sqrt{2}$ × 10^{5} Pa

Work done by an adiabatic process,

$W=\frac{{P}_{1}{V}_{1}-{P}_{2}{V}_{2}}{\gamma -1}\phantom{\rule{0ex}{0ex}}W=\frac{{10}^{5}\times {10}^{-3}-2\sqrt{2}\times {10}^{5}\times {\displaystyle \frac{1}{2}}\times {10}^{-3}}{1.5-1}\phantom{\rule{0ex}{0ex}}W=-82\mathrm{J}$

(c) Internal energy,

*dQ* = 0, as it is an adiabatic process.

⇒ d*U* = − d*W* = − (− 82 J) = 82 J

(d)

Also, for an adiabatic process,

*T*_{1}*V*_{1}^{γ−}^{1}* = **T*_{2}*V*_{2}^{γ−}^{1}

*T*_{2}* = **T*_{1}*${\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma -1}$
= *300 $\times $ (2)

^{0.5}

= 300 × $\sqrt{2}$ × = 300 × 1.4142

*T*

_{2}= 424 K

(e) The pressure is kept constant.

The process is isobaric; so, work done =

*P*$\u2206V$=

*nRdT*.

Here, $n=\frac{PV}{RT}=\frac{{10}^{5}\times {10}^{-3}}{R\times 300}=\frac{1}{3R}$

So, work

*d*one = $\frac{1}{3R}\times R\mathit{\times}(300-424)=-41.4\mathrm{J}$

As pressure is constant,

$\frac{{V}_{1}}{{T}_{1}}=\frac{{V}_{2}}{{T}_{2}}...\left(1\right)\phantom{\rule{0ex}{0ex}}{V}_{1}={V}_{2}\frac{{T}_{1}}{{T}_{2}}$

(f)Work done in an isothermal process,

$W=nRT\mathrm{ln}\frac{{V}_{2}}{{V}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3R}\times R\times T\times \mathrm{In}\left(2\right)$

= 100 × ln 2 = 100 × 1.039

= 103 J

(g) Net work done (using first law of thermodynamics)

= − 82 − 41.4 + 103

= − 20.4 J

#### Page No 79:

#### Question 26:

Figure shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas (γ = 1.5) is injected in the two sides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1 : 3. Find the ratio of the temperatures in the two parts of the vessel.

Figure

#### Answer:

Given:

*γ* = 1.5

For an adiabatic process, *TV ^{γ}^{−1}* = constant .

So,

*T*

_{1}

*V*

_{1}

^{γ−}^{1}

*=*

*T*

_{2}

*V*

_{2}

^{γ−}^{1}

As it is an adiabatic process and all the other conditions are same, the above equation can be applied.

In the new position, the slid is dividing the tube in the ratio 3:1.

So, if the total volume is

*V,*then one side will occupy a volume of$\frac{3}{4}V$ and the other side will occupy $\frac{V}{4}$.

So, ${T}_{1}\times {\left(\frac{3v}{4}\right)}^{\gamma -1}={T}_{2}\times {\left(\frac{v}{4}\right)}^{\gamma -1}$

(i) (ii)

$\Rightarrow {T}_{1}\times {\left(\frac{3v}{4}\right)}^{1.5-1}={T}_{2}\times {\left(\frac{v}{4}\right)}^{1.5-1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{T}_{1}}{{T}_{2}}=\frac{\sqrt{3}}{1}$

#### Page No 79:

#### Question 27:

Figure shows two rigid vessels *A* and *B*, each of volume 200 cm^{3}, containing an ideal gas (*C _{v}* = 12.5 J K

^{−1}mol

^{−1}). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75 cm of mercury and the temperature is 300 K. (a) Find the number of moles of the gas in each vessel. (b) 5.0 J of heat is supplied to the gas in vessel

*A*and 10 J to the gas in vessel

*B*. Assuming there's no appreciable transfer of heat from

*A*to

*B*, calculate the difference in the heights of mercury in the two sides of the manometer. Gas constant,

*R*= 8.3 J K

^{−1}mol

^{−1}.

Figure

#### Answer:

Given:

Volume of gas in each vessel,* V* = 200 cm^{3}

Specific heat at constant volume of the gas,* C _{v}* = 12.5 J/mol-K

Initial temperature of the gas,

*T*= 300 K

Initial pressure of the gas,

*P*= 75 cm of Hg

(a) Using the ideal gas equation, number of moles of gases in each vessel,

$75\mathrm{cm}\mathrm{of}\mathrm{Hg}=99991.5\mathrm{N}/{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}n\mathit{=}\frac{PV}{RT}\phantom{\rule{0ex}{0ex}}=\frac{99991.5\times 200\times {10}^{-6}}{8.3\times 300}\phantom{\rule{0ex}{0ex}}=8031.4\times {10}^{-6}\phantom{\rule{0ex}{0ex}}=0.008$

(b) Heat is supplied to the gas, but

*dV*is zero as the container has rigid walls.

So, d

*W*

*= $P\mathit{\u2206}V$ = 0*

From first law of thermodynamics,

*dQ*=

*dU*

⇒ 5 =

*nC*

_{v}dT⇒ 5 = 0.008 × 12.5 ×

*dT*

⇒

*dT*= 50 for A

*$\frac{\mathit{P}}{\mathit{T}}=\frac{{P}_{\mathrm{A}}}{{T}_{\mathrm{A}}}$*because volume is kept constant.

$Q=n{C}_{v}dT\phantom{\rule{0ex}{0ex}}T=\frac{Q}{n{C}_{v}}$

$\Rightarrow \frac{75}{300}=\frac{{\mathrm{P}}_{\mathrm{A}}\times 0.008\times 12.5}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow {P}_{\mathrm{A}}=\frac{75\times 5}{300\times 0.008\times 12.5}\phantom{\rule{0ex}{0ex}}=12.5\mathrm{cm}\mathrm{of}\mathrm{Hg}\phantom{\rule{0ex}{0ex}}\text{A}\mathrm{gain},\frac{P}{T}\mathit{=}\frac{{P}_{B}}{{T}_{B}}[\mathrm{For}\mathrm{container}\mathrm{B}]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{75}{300}=\frac{{P}_{\mathrm{B}}\times 0.008\times 12.5}{10}$

*P*= 25 cm of Hg

_{B}The distance moved by the mercury,

*P*= 25 − 12.5 = 12.5 cm

_{B}− P_{A}#### Page No 79:

#### Question 28:

The figure shows two vessels with adiabatic walls, one containing 0.1 g of helium (γ = 1.67, *M* = 4 g mol^{−1}) and the other containing some amount of hydrogen (γ = 1.4, *M* = 2 g mol^{−1}). Initially, the temperatures of the two gases are equal. The gases are electrically heated for some time during which equal amounts of heat are given to the two gases. It is found that the temperatures rise through the same amount in the two vessels. Calculate the mass of hydrogen.

Figure

#### Answer:

Given:

Mass of He, *m*_{He} = 0.1 g

*γ _{1}* = 1.67

Molecular weight of He,

*M*

_{He}= 4 g/mol

*M*

_{H2}= ?

*M*= 2 g/mol

_{H2}*γ*1.4

_{2}=Since it is an adiabatic environment and the system is not dong any external work, the amount of heat given will be used up entirely to raise its internal energy.

For He,

*dQ*= d

*U*=

*nC*...(i)

_{v}dT$=\frac{m}{4}\times \frac{\mathrm{R}}{\gamma -1}\times d\mathrm{T}\phantom{\rule{0ex}{0ex}}=\frac{0.1}{4}\times \frac{\mathrm{R}}{(1.67-1)}\times d\mathrm{T}$

For H

_{2},

*dQ*=d

*U*=

*n*C

*vdT*...(ii)

$=\frac{m}{2}\times \frac{\mathrm{R}}{\gamma -1}\times d\mathrm{T}\phantom{\rule{0ex}{0ex}}=\frac{m}{2}\times \frac{\mathrm{R}}{1.4-1}\times d\mathrm{T},$

where

*m*is the required mass of H

_{2}.

Since equal amount of heat is given to both gases; so

*d*Q is same in both eq (i) and (ii), we get

$\frac{0.1}{4}\times \frac{R}{0.67}dT\phantom{\rule{0ex}{0ex}}=\frac{m}{2}\times \frac{R}{0.4}\times dT\phantom{\rule{0ex}{0ex}}\Rightarrow m=\frac{0.1}{2}\times \frac{0.4}{0.67}\phantom{\rule{0ex}{0ex}}\Rightarrow m=0.0298\approx 0.03\mathrm{g}$

#### Page No 79:

#### Question 29:

Two vessels *A* and *B* of equal volume *V*_{0} are connected by a narrow tube that can be closed by a valve. The vessels are fitted with pistons that can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas (*C _{p}/C_{v}* = γ) at atmospheric pressure

*p*

_{0}

_{ }and atmospheric temperature

*T*

_{0}. The walls of vessel

*A*are diathermic and those of

*B*are adiabatic. The valve is now closed and the pistons are slowly pulled out to increase the volumes of the vessels to double the original value. (a) Find the temperatures and pressures in the two vessels. (b) The valve is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and pressure.

#### Answer:

Initial pressure of the gas in both the vessels =* P _{0}*

Initial temperature of the gas in both the vessels =

*T*

_{0}Initial volume =

*V*

$\frac{\mathrm{C}p}{\mathrm{C}v}=\gamma $

_{0}$\frac{\mathrm{C}p}{\mathrm{C}v}=\gamma $

(a) The temperature inside the diathermic vessel remains constant. Thus,

*P*

_{1}

*V*

_{1}

*=*

*P*

_{2}

*V*

_{2}

*⇒*

*P*

_{0}

*V*

_{0}

*=*

*P*

_{2}

*×*2

*V*

_{0}

*$\Rightarrow {P}_{2}=\frac{{P}_{0}}{2}$*

Temperature =

*T*

_{0}

The temperature inside the adiabatic vessel does not remain constant.

So, for an adiabatic process,

*T*

_{1}

*V*

_{1}

^{γ−}^{1}

*=*

*T*

_{2}

*V*

_{2}

^{γ−}^{1}

*⇒*

*T*

_{0}

*V*

_{0}

^{γ−}^{1}

*= T ×*(2

*V*

_{0})

^{γ−}^{1}

*⇒*

*T*

_{2}

*=*

*T*

_{0}

*×*2

^{1}

^{−γ}*⇒*

*P*

_{1}

*V*

_{1}

^{γ}*=*

*P*

_{2}

*V*

_{2}

^{γ}*P*

_{0}

*V*

_{0}

^{γ}*=*

*P*

_{2}

*×*(2

*V*

_{0})

^{γ}*$\Rightarrow {P}_{2}=\left(\frac{{P}_{0}}{{2}^{\gamma}}\right)$*

(b) When the valves are open, the temperature remains

*T*

_{0}throughout, i.e.

*T*

_{1}

*=*

*T*

_{2}

*=*

*T*

_{0}.

Also, pressure will be same throughout. Thus,

*P*

_{1}

*=*

*P*

_{2}

_{.}As, temperature has not changed on side 1, so pressure on this side will also not change (volume is also fixed due to fixed piston) and will be equal to ${P}_{0}$ (pressure is an intrinsic variable).

On side 2, pressure will change to accommodate the changes in temperature on this side.

So

*,*

*P*

_{0}

*=*

*P*

_{1}

*+*

*P*

_{2}

*⇒*2

*P*

_{1}

*= 2*

*P*

_{2}

So,

*P*

_{1}

*=*

*P*

_{2}

*= $\frac{{\mathrm{P}}_{0}}{2}$*

#### Page No 79:

#### Question 30:

The figure shows an adiabatic cylindrical tube of volume *V*_{0} divided in two parts by a frictionless adiabatic separator. Initially, the separator is kept in the middle, an ideal gas at pressure *p*_{1} and temperature *T*_{1} is injected into the left part and another ideal gas at pressure *p*_{2} and temperature *T*_{2} is injected into the right part. *C _{p}*/

*C*= γ is the same for both the gases. The separator is slid slowly and is released at a position where it can stay in equilibrium. Find (a) the volumes of the two parts (b) the heat given to the gas in the left part and (c) the final common pressure of the gases.

_{v}Figure

#### Answer:

For an adiabatic process,* PV ^{γ}* = Constant

So,

*P*

_{1}

*V*

_{1}

^{γ}*=*

*P*

_{2}

*V*

_{2}

^{γ}*...*(i)

According to the problem,

*V*

_{1}

*+*

*V*

_{2}

*=*

*V*

_{0}...(ii)

Using the relation in eq (ii) in eq (i), we get

*P*

_{1}

*V*

_{1}

^{γ}*=*

*P*

_{2}

*(*

*V*

_{0}

*−*

*V*

_{1}

*)*

$\text{Or}{\left(\frac{{P}_{1}}{{P}_{2}}\right)}^{1/\gamma}=\frac{{V}_{0}-{V}_{1}}{{V}_{1}}\phantom{\rule{0ex}{0ex}}$

${V}_{1}{{P}_{\mathit{1}}}^{\frac{\mathit{1}}{\gamma}}\mathit{=}{V}_{0}{{P}_{\mathit{2}}}^{\frac{\mathit{1}}{\gamma}}\mathit{-}{V}_{1}{{P}_{2}}^{\frac{\mathit{1}}{\gamma}}\phantom{\rule{0ex}{0ex}}{V}_{1}\left({{P}_{1}}^{\frac{\mathit{1}}{\gamma}}+{{P}_{2}}^{\frac{\mathit{1}}{\gamma}}\right)\mathit{=}{V}_{0}{{P}_{2}}^{\frac{\mathit{1}}{\gamma}}\phantom{\rule{0ex}{0ex}}\mathit{}{V}_{\mathit{1}}\mathit{=}\frac{{{P}_{2}}^{{\displaystyle \frac{1}{\gamma}}}{V}_{\mathit{0}}}{{{P}_{1}}^{{\displaystyle \frac{1}{\gamma}}}+{{P}_{2}}^{{\displaystyle \frac{1}{\gamma}}}}$

^{γ}$\text{Or}{\left(\frac{{P}_{1}}{{P}_{2}}\right)}^{1/\gamma}=\frac{{V}_{0}-{V}_{1}}{{V}_{1}}\phantom{\rule{0ex}{0ex}}$

${V}_{1}{{P}_{\mathit{1}}}^{\frac{\mathit{1}}{\gamma}}\mathit{=}{V}_{0}{{P}_{\mathit{2}}}^{\frac{\mathit{1}}{\gamma}}\mathit{-}{V}_{1}{{P}_{2}}^{\frac{\mathit{1}}{\gamma}}\phantom{\rule{0ex}{0ex}}{V}_{1}\left({{P}_{1}}^{\frac{\mathit{1}}{\gamma}}+{{P}_{2}}^{\frac{\mathit{1}}{\gamma}}\right)\mathit{=}{V}_{0}{{P}_{2}}^{\frac{\mathit{1}}{\gamma}}\phantom{\rule{0ex}{0ex}}\mathit{}{V}_{\mathit{1}}\mathit{=}\frac{{{P}_{2}}^{{\displaystyle \frac{1}{\gamma}}}{V}_{\mathit{0}}}{{{P}_{1}}^{{\displaystyle \frac{1}{\gamma}}}+{{P}_{2}}^{{\displaystyle \frac{1}{\gamma}}}}$

Using equation (ii), we get

*${V}_{2}\mathit{=}\frac{{{P}_{1}}^{{\displaystyle \frac{1}{\gamma}}}{V}_{\mathit{0}}}{{{P}_{1}}^{{\displaystyle \frac{1}{\gamma}}}+{{P}_{2}}^{{\displaystyle \frac{1}{\gamma}}}}$*

(b) Since the whole process takes place in adiabatic surroundings, the separator is adiabatic.

Hence, heat given to the gas in the left part is 0.

(c) There will be a common pressure '

*P*' when equilibrium is reached. The slid will move until the pressure on the two sides becomes equal.

*P*

_{1}V_{1}^{γ}+ P_{2}V_{2}^{γ}= PV_{0}^{γ}For equilibrium,

*V*

_{1}= V_{2}= $\frac{{\mathrm{V}}_{0}}{2}$Hence,

${P}_{1}{\left(\frac{{V}_{\mathit{0}}}{2}\right)}^{\gamma}+{P}_{2}{\left(\frac{{V}_{\mathit{0}}}{2}\right)}^{\gamma}\mathit{=}P\mathit{(}{V}_{0}{\mathit{)}}^{\gamma}\phantom{\rule{0ex}{0ex}}\text{Or}\mathit{}P\mathit{=}{\left(\frac{{{P}_{1}}^{{\displaystyle \frac{1}{\gamma}}}+{{P}_{2}}^{{\displaystyle \frac{1}{\gamma}}}}{2}\right)}^{\mathrm{\gamma}}$

#### Page No 79:

#### Question 31:

An adiabatic cylindrical tube of cross-sectional area 1 cm^{2} is closed at one end and fitted with a piston at the other end. The tube contains 0.03 g of an ideal gas. At 1 atm pressure and at the temperature of the surrounding, the length of the gas column is 40 cm. The piston is suddenly pulled out to double the length of the column. The pressure of the gas falls to 0.355 atm. Find the speed of sound in the gas at atmospheric temperature.

#### Answer:

Given:

Area of the tube, *A* = 1 cm^{2} = 1 × 10^{−4} m^{2}

Mass of the gas,* M *= 0.03 g = 0.03 × 10^{−3} kg

Initial pressure, *P* = 1 atm = 10^{5} Pascal

Initial length of the mercury column, *L* = 40 cm = 0.4 m

Final length of the mercury column, *L _{1}* = 80 cm = 0.8 m

Final pressure,

*P'*= 0.355 atm

The process is adiabatic. So,

*P*(

*V*)

*(*

^{γ}= P'*V'*)

^{γ}

⇒ 1 × (A × 0.4)

^{γ}= 0.355 × (A × 0.8)

^{γ}

⇒ 1 × 1 = 0.355 × 2

^{γ}

$\Rightarrow \frac{1}{0.355}={2}^{\gamma}\phantom{\rule{0ex}{0ex}}\gamma \mathrm{log}2=\mathrm{log}\left(\frac{1}{0.355}\right)$

⇒

*γ*= 1.4941

Speed of sound in the gas,

$v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{\gamma P}{m\mathit{/}v}}\phantom{\rule{0ex}{0ex}}v=\sqrt{\frac{1.4941\times {10}^{5}}{m/v}}=\sqrt{\frac{1.4941\times {10}^{5}}{\left({\displaystyle \frac{0.03\times {10}^{-3}}{{10}^{-4}\times 1\times 0.4}}\right)}}\phantom{\rule{0ex}{0ex}}\Rightarrow v=446.33\approx 447\mathrm{m}/\mathrm{s}$

#### Page No 80:

#### Question 32:

The speed of sound in hydrogen at 0°C is 1280 m s^{−1}. The density of hydrogen at STP is 0.089 kg m^{−3}. Calculate the molar heat capacities *C _{p}* and

*C*of hydrogen.

_{v}#### Answer:

Given:

Velocity of sound in hydrogen, *V* = 1280 m/s

Temperature, *T *= 0°C = 273 K

Density of H_{2} = 0.089 kg/m^{3}

*R* = 8.3 J/mol-K

At STP,

*P* = 10^{5} Pa

We know:

${V}_{sound}\mathit{=}\sqrt{\frac{\gamma p}{\rho}}\phantom{\rule{0ex}{0ex}}1280=\sqrt{\frac{\gamma \times {10}^{5}}{0.089}}\phantom{\rule{0ex}{0ex}}\text{Or}\gamma =\frac{1280\times 1280\times 0.089}{{10}^{5}}\phantom{\rule{0ex}{0ex}}=1.46\phantom{\rule{0ex}{0ex}}\frac{{C}_{p}}{{C}_{v}}\mathit{=}\gamma \mathrm{or}\mathit{}{C}_{p}\mathit{-}{C}_{v}\mathit{=}R\phantom{\rule{0ex}{0ex}}{C}_{v}\mathit{=}\frac{R}{\gamma \mathit{-}\mathit{1}}=\frac{8.3}{1.46-1}\phantom{\rule{0ex}{0ex}}=18.0\mathrm{J}/\mathrm{mol}-\mathrm{K}\phantom{\rule{0ex}{0ex}}{C}_{p}\mathit{=}\gamma {C}_{v}=1.46\times 18.0\phantom{\rule{0ex}{0ex}}=26.28\approx 26.3/\mathrm{mol}-\mathrm{K}$

#### Page No 80:

#### Question 33:

4.0 g of helium occupies 22400 cm^{3} at STP. The specific heat capacity of helium at constant pressure is 5.0 cal K^{−1} mol^{−1}. Calculate the speed of sound in helium at STP.

#### Answer:

Given:

Specific heat capacity at constant pressure, *C _{p}_{ }=* 5.0 cal/mol-K

*C*

_{p}= 5.0 × 4.2 J/mol-K

*C*

_{p}

_{ }= 21 J/mol-K

Volume of helium,

*V*= 22400 ${\mathrm{cm}}^{3}$ = 0.0224 ${\mathrm{m}}^{3}$

At STP,

*P*= 1 atm = ${10}^{5}\mathrm{Pa}$

The speed of sound in gas,

$v\mathit{=}\sqrt{\frac{\mathit{\gamma}\mathit{p}}{\mathit{\rho}}}\mathit{=}\sqrt{\frac{\mathit{\gamma}\mathit{R}\mathit{T}}{\mathit{M}}}\mathit{=}\sqrt{\frac{\mathit{\gamma}\mathit{P}\mathit{V}}{\mathit{M}}}\phantom{\rule{0ex}{0ex}}{C}_{\mathit{p}}\mathit{=}\frac{\mathit{R}\mathit{\gamma}}{\mathit{\gamma}\mathit{-}\mathit{1}}$

Or 21(

*γ*− 1) = 8.3γ

⇒ 21

*γ*

*− 8.3γ = 21*

⇒ 12.7

*γ*= 21

$\therefore \gamma =\frac{21}{12.7}=1.653\phantom{\rule{0ex}{0ex}}v=\sqrt{\frac{1.653\times 1.0\times {10}^{5}\times 0.0224}{4\times {10}^{-3}}}\phantom{\rule{0ex}{0ex}}v=960\mathrm{m}/\mathrm{s}$

#### Page No 80:

#### Question 34:

An ideal gas of density 1.7 × 10^{−3} g cm^{−3} at a pressure of 1.5 × 10^{5} Pa is filled in a Kundt's tube. When the gas is resonated at a frequency of 3.0 kHz, nodes are formed at a separation of 6.0 cm. Calculate the molar heat capacities *C _{p}* and

*C*of the gas.

_{v}#### Answer:

Given:

Density of the ideal gas, *ρ* = 1.7 × 10^{−3} g/cm^{3}

= 1.7 k/gm^{3}

Pressure of the gas,* P* = 1.5 × 10^{5} Pa

*R* = 8.3 J/mol-K

Resonance frequency of the gas = 3.0 kHz

Node separation in the Kundt's tube

= $\frac{l}{2}$ = 6 cm

So, *l *= 2$\times $6 = 12 cm = 12 × 10^{−2} m

So*, V = fl *= 3 × 10^{3} × 12 × 10^{−2}

= 360 m/s

Speed of sound, *V* = $\sqrt{\frac{\gamma p}{\rho}}$

Or ${V}^{2}=\frac{\gamma p}{\mathrm{\rho}}$

$\therefore \mathit{}\gamma \mathit{=}\frac{{v}^{\mathit{2}}\rho}{P}\mathit{=}\frac{(360{)}^{2}\times 1.7\times {10}^{-3}}{1.5\times {10}^{5}}\phantom{\rule{0ex}{0ex}}=1.4688\phantom{\rule{0ex}{0ex}}\mathrm{Using}{C}_{p}-{C}_{v}=R\mathit{}and\mathit{}\frac{{C}_{p}}{{C}_{v}}\mathit{=}\gamma \phantom{\rule{0ex}{0ex}}\mathrm{W}\text{e}\mathrm{know}\mathrm{that}\phantom{\rule{0ex}{0ex}}{C}_{v}\mathit{=}\frac{R}{\gamma \mathit{-}\mathit{1}}=\frac{8.3}{0.4688}\phantom{\rule{0ex}{0ex}}=17.7\mathrm{J}/\mathrm{mol}-\mathrm{K}\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{p}=R\mathit{}+{\mathrm{C}}_{v}=8.3+17.7=26\mathrm{J}/\mathrm{mol}-\mathrm{K}$

#### Page No 80:

#### Question 35:

Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K. The separation between the consecutive nodes is 3.3 cm. Calculate the specific heat capacities *C _{p}* and

*C*

_{v}_{ }of the gas.

#### Answer:

Frequency of standing waves, *f* = 5 × 10^{3} Hz

Temperature of oxygen, *T* = 300 K

From Kundt's tube theory, we know that

$\frac{l}{2}$ = node separation = 3.3 cm

*∴ l *= 6.6 × 10^{−2} m

Also,

*v *= *fl* = 5 × 10^{3} × 6.6 × 10^{−2}

= (66 × 5) = 330 m/s

$v\mathit{=}\sqrt{\frac{\gamma RT}{M}}\mathit{}\phantom{\rule{0ex}{0ex}}{v}^{\mathit{2}}\mathit{=}\frac{\gamma RT}{M}\phantom{\rule{0ex}{0ex}}\therefore \gamma =\frac{330\times 330\times 32}{8.3\times 300\times 100}=1.3995\phantom{\rule{0ex}{0ex}}\text{S}\mathrm{pecific}\mathrm{heat}\mathrm{at}\mathrm{constant}\mathrm{volume},{C}_{v}\mathit{=}\frac{R}{\gamma \mathit{-}\mathit{1}}=\frac{8.3}{0.3995}\phantom{\rule{0ex}{0ex}}$

= 20.7 J mol-K

Specific heat at constant pressure, *C _{p} = C_{v} + R*

*C*= 20.7 + 8.3

_{p}_{ }*C*= 29.0 J/mol-K

_{p}_{ }View NCERT Solutions for all chapters of Class 11