Hc Verma II Solutions for Class 11 Science Physics Chapter 26 Laws Of Thermodynamics are provided here with simple step-by-step explanations. These solutions for Laws Of Thermodynamics are extremely popular among Class 11 Science students for Physics Laws Of Thermodynamics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma II Book of Class 11 Science Physics Chapter 26 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Hc Verma II Solutions. All Hc Verma II Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

#### Page No 60:

#### Answer:

Change in internal energy of a system, $\u2206U={C}_{v}\u2206T$

Here,

*C _{v}* = Specific heat at constant volume

$\u2206$

*T*= Change in temperature.

If $\u2206$

*T*= 0, then $\u2206$

*U*= 0, i.e. in isothermal processes, where temperature remains constant, the internal energy doesn't change even on adding heat to the system.

Thus, the internal energy of a system should not necessarily increase if heat is added to it.

#### Page No 60:

#### Answer:

Internal energy of a system increases if its temperature increases. This is valid only for the system of ideal gases and not for all the systems.

For example: During meting process, temperature of the system remains constant, but internal energy change increases by *mL*.

$\Rightarrow \u2206U=mL$

Here,

*m *= Mass of the solid

*L* = Latent heat of the solid

#### Page No 60:

#### Answer:

As a cylinder is lifted from the first floor to the second floor, there is decrease in the atmospheric pressure on the gas and it expands. Therefore, some work is done by the gas on its surroundings. Work done on the gas is zero.

Work done by the gas, *W = P$\u2206V$ *(positive)

The increase in the internal energy and temperature of the system will depend on the types of the walls of the system (conducting or insulating).

#### Page No 60:

#### Answer:

If force *F* is applied on a block of mass *M *and displacement of block is *d*, then work done by the force is given by

*W = F.d = Fd *cos(0^{o}) = *Fd*

This work done does not change the internal energy of the block as the internal energy does not include the energy due to motion or location of the system as a whole.

#### Page No 60:

#### Answer:

As the outer surface of a cylinder containing a gas is rubbed vigorously by a polishing machine, no work is done on the cylinder. Volume of the gas remains constant and the heat energy generated due to friction between the machine and the cylinder gets transferred to the gas as heat energy. This heat energy leads to an increase in the temperature of the cylinder and its gas.

#### Page No 60:

#### Answer:

When we rub our hands, they become warm. In this process, heat is supplied to the hands due to the friction between the hands.

#### Page No 60:

#### Answer:

A closed bottle containing some liquid is shaken vigorously such that work is done on the bottle (against the viscous force). Hence, the temperature of the liquid increases. However, no external heat is supplied to the system.

#### Page No 60:

#### Answer:

Work done by the system is neither necessarily zero nor necessarily non-zero.

If in a certain process, the pressure *P* stays constant, then

$\u2206W=P\u2206V\phantom{\rule{0ex}{0ex}}\Rightarrow W=P({V}_{2}-{V}_{1})\phantom{\rule{0ex}{0ex}}\mathrm{As}{V}_{2}={V}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow W=0$

(Initial volume, *V*_{1}_{ }= Final volume, *V*_{2})

Hence, it is an isobaric process.

Even if *P = P*(*V*), net work done will be zero if *V*_{2}_{ }= *V*_{1}. In this case, work done is zero.

If the system goes through a cyclic process, then initial volume gets equal to the final volume after one cycle. But work done by the gas is non-zero.

#### Page No 60:

#### Answer:

If the system goes through a cyclic process, then initial volume gets equal to the final volume after one cycle. But work done by the gas is non-zero. So, work can be done by a system without changing its volume.

#### Page No 60:

#### Answer:

An ideal gas is continuously being pumped into the container. Therefore, the number of moles, n are continuously increasing. In a certain interval,

Pressure, *P*_{2} = 2*P*_{1}

*n*_{2} = 2*n*_{1}

Thus, internal energy, *U = nC _{v}T* will double as the number of moles get doubled.

#### Page No 60:

#### Answer:

When a tyre bursts, adiabatic expansion of air takes place. The pressure inside the tyre is greater than the atmospheric pressure of the surrounding due to which the expansion of air occurs with some work done against the surrounding leading to decrease in the internal energy of the air present inside the tyre. This decrease of internal energy leads to fall in temperature of the inside air. Hence, the air coming out is cooler than that of the surrounding.

#### Page No 60:

#### Answer:

When we heat an object, it expands, i.e. its volume increases.

Work done by the system, $\Delta W=P\Delta V$

Using the first law of thermodynamics, we get

$\u2206$*Q* = $\u2206U+\u2206W$

Since the volume changes,$\u2206$*W* has some non-zero positive value. Thus, heat given to the object is not equal to the increase in the internal energy of the system.

#### Page No 60:

#### Answer:

When we stir a liquid vigorously, we do work on it due to which its temperature rises and it becomes warm. To bring back the liquid to the initial temperature, heat needs to be extracted from it. But it is an irreversible process that cannot be brought back to the initial state by stirring the liquid again in the opposite direction.

#### Page No 60:

#### Answer:

Efficiency of Carnot engine, $\eta =\frac{W}{{Q}_{1}}=\frac{{Q}_{1}-{Q}_{2}}{{Q}_{1}}$

Here,

*W* = Work done by the engine

*Q*_{1} = Heat abstracted from the source

*Q*_{2} = Heat transferred to the sink

Thus, for $\eta =1,W={Q}_{1}or{Q}_{2}=0$, the total heat which is abstracted from the source gets converted into work.

#### Page No 60:

#### Answer:

When an object cools down, heat is withdrawn from it. Hence, the entropy of the object decreases. But the decrease in entropy leads to the transfer of energy to the surrounding. The second law is not violated here, which states that entropy of the universe always increases as the net entropy increases.

Here,

Net entropy = Entropy of object + Entropy of surrounding

#### Page No 61:

#### Answer:

(d) conservation of energy

Heat is a form of energy. Since the first law of thermodynamics deals with the conservation of heat, it actually refers to the conservation of energy in the broader sense.

The first law of thermodynamics is just the restatement of the law of conservation of energy. We observe that the energy supplied to a system will contribute to change in its internal energy and the amount of work done by the system on its surroundings.

#### Page No 61:

#### Answer:

(b) the gas will do positive work

Internal energy of the gas, $U=\int {C}_{v}dT\phantom{\rule{0ex}{0ex}}$

Here, *Cv* is the specific heat at constant volume and *dT *is the change in temperature.

If the process is isothermal, i.e. *dT* = 0, *dU* = 0.

Using the first law of thermodynamics, we get

$\Delta Q=\Delta U+\Delta W\phantom{\rule{0ex}{0ex}}But\Delta U=0\phantom{\rule{0ex}{0ex}}\Rightarrow \Delta Q=+\Delta W$

Here, $\Delta $*W* is the positive work done by the gas.

#### Page No 61:

#### Answer:

(a) ∆*Q*_{1} > ∆*Q*_{2}

Both the processes A and B have common initial and final points. So, change in internal energy, *∆U* is same in both the cases. Internal energy is a state function that does not depend on the path followed.

In the *P-V *diagram, the area under the curve represents the work done on the system, *∆**W*. Since area under curve A > area under curve B, *∆W*_{1}> *∆**W*_{2}.

Now,

$\Delta {Q}_{1}=\Delta U+\Delta {W}_{1}\phantom{\rule{0ex}{0ex}}\Delta {Q}_{2}=\Delta U+\Delta {W}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{But}\Delta {W}_{1}\Delta {W}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \Delta {Q}_{1}\Delta {Q}_{2}$

Here, *∆ Q*

_{1}and

*∆*

*Q*_{2}denote the heat given to the system in processes

*A*and

*B,*respectively.

#### Page No 61:

#### Answer:

(b) ∆*U*_{1} = ∆*U*_{2}

The internal energy of the system is a state function, i.e. it only depends on the initial and final point of the process and doesn't depend on the path followed. Both processes A and B have common initial and final points. Therefore, change in internal energy in process A is equal to the change in internal energy in process B. Thus,

∆*U*_{1} = ∆*U*_{2}_{ }= 0

#### Page No 61:

#### Answer:

(a) continuously increases

Work done by a system, $W=\int PdV$

Here,

*P* = Pressure on the system

*dV* = change in volume.

Since *dV* is positive, i.e. the volume is continuously increasing, work done by the system also continuously increases.

#### Page No 61:

#### Answer:

(c) B is correct but A is wrong.

If heat is added to a system in an isothermal process, then there'll be no change in the temperature.

$\mathrm{Work}\mathrm{done}\mathrm{by}\mathrm{a}\mathrm{system},\u2206W=P\u2206V\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206W=\mathrm{positive}\Rightarrow \u2206V=\mathrm{positive}$

Here,

*P* = Pressure

*$\u2206$V* = change in volume

#### Page No 61:

#### Answer:

(c) is zero

Work done by the gas during the process, $\u2206W=P\u2206V\phantom{\rule{0ex}{0ex}}$

Here,

*P* = Pressure

$\u2206$*V* = change in volume.

Since the process described in the figure is isochoric, *P = kT*. As volume remains constant (*$\u2206$V* = 0), $\u2206$*W* = 0.

#### Page No 61:

#### Answer:

(c) ∆*W*_{1} < ∆*W*_{2}

Work done by the system, $\u2206W=P\u2206V$

Here,

*P* = Pressure in the process

$\u2206V$ = Change in volume during the process

Let *V _{i}* and

*V* be the volumes in the initial states and final states for processes A and B, respectively. Then,

_{f}$\Delta {W}_{1}={P}_{1}\Delta {V}_{1}\phantom{\rule{0ex}{0ex}}\Delta {W}_{2}={P}_{2}\Delta {V}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{But}\Delta {V}_{2}=\Delta {V}_{1},\left[\left({V}_{{f}_{1}}-{V}_{{i}_{1}}\right)=\left({V}_{{f}_{2}}-{V}_{{i}_{2}}\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\Delta {W}_{1}}{\Delta {W}_{2}}=\frac{{P}_{1}}{{P}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \Delta {W}_{1}\Delta {W}_{2}\left[\because {P}_{2}{P}_{1}\right]$

#### Page No 61:

#### Answer:

(b) decreases

As the piston of a metallic cylinder containing gas is moved to compress the gas, the volume in which the gas is contained reduces, leading to increase in pressure and temperature. When the time elapses, the heat generated radiates through the metallic cylinder as metals are good conductors of heat. Consequently, the pressure of the gas in the cylinder decreases because of decrease in the temperature.

#### Page No 61:

#### Answer:

(b) The work done by the system is positive.

(c) The temperature of the system must increase.

Work done by the system depends on the change in volume during the process $(\Delta W=p\Delta V)$, where *p* be the pressure and *$\u2206$V* be the change in volume of the ideal gas. Since in this process the volume is continuously increasing, the work done by the system will be positive.

According to the ideal gas equation, $pV=nRT$.

Since both *p* and *V* are increasing, temperature (*T* ) must also increase.

#### Page No 61:

#### Answer:

(a) The initial temperature must be equal to the final temperature.

(b) The initial internal energy must be equal to the final internal energy.

a) Let initial pressure, volume and temperature be *P*_{1}, *V*_{1} and *T*_{1} and final pressure, volume and temperature be *P*_{2}, *V*_{2} and *T*_{2}. Then,

$\frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}_{1}={T}_{2}\left[{P}_{1}={P}_{2}\mathrm{and}{V}_{1}={V}_{2}\right]$

b) Internal energy is given by Δ*U* = *nC*_{v}Δ*T*.

Since Δ*T* = 0, Δ*U* = 0.

c) This may not be true because the system may be isothermal and in between, heat may have been given to the system. Also, the system may have done mechanical expansion work and returned back to its original state.

d) Not necessary because an isothermal system may have done work absorbing heat from outside and coming back to the same state losing heat.

#### Page No 62:

#### Answer:

(d) ∆*Q* − ∆*W*

A system is taken from an initial state to the final state by two different methods. So, work done and heat supplied in both the cases will be different as they depend on the path followed. On the other hand, internal energy of the system (*U*) is a state function, i.e. it only depends on the final and initial state of the process. They are the same in the above two methods.

Using the first law of thermodynamics, we get

$\Delta Q=\Delta U+\Delta W\phantom{\rule{0ex}{0ex}}\Rightarrow \Delta U=\Delta Q-\Delta W$

Here, $\Delta $*U* is the change in internal energy, $\Delta $*Q* is the heat given to the system and $\Delta $*W* is the work done by the system.

#### Page No 62:

#### Answer:

(a) ∆*U*_{1} + ∆*U*_{2} = 0

(c) ∆*Q* − ∆*W* = 0

The process that takes place through A and returns back to the same state through B is cyclic. Being a state function, net change in internal energy, *∆**U* will be zero, i.e.

*∆U*_{1} (Change in internal energy in process A) = *∆U*_{2} (Change in internal energy in process B)

$\Rightarrow \Delta U=\Delta {U}_{1}+\Delta {U}_{2}=0$

Here, *∆U* is the total change in internal energy in the cyclic process.

Using the first law of thermodynamics, we get

$\Delta Q-\Delta W=\Delta U$

Here, *∆ Q* is the net heat given to the system in process

*A*+

*B*and

*∆*is the net work done by the system in the process

*W**A*+

*B*.

Thus,

*∆Q - ∆W*= 0

#### Page No 62:

#### Answer:

(a) The process must be adiabatic

(d) The temperature must decrease.

Using the first law of thermodynamics, we get

$\Delta Q=\Delta W+\Delta U\phantom{\rule{0ex}{0ex}}\Rightarrow \Delta Q=0\left[\because \Delta W=-\Delta U\right]$

Thus, no heat is exchanged in the process, i.e. the process is adiabatic and since the internal energy is decreasing, the temperature must decrease because the gas is an ideal gas.

On the other hand, volume and pressure of the gas are varying, leading to positive work done. So, the process cannot be isochoric and isobaric.

#### Page No 62:

#### Answer:

Given:

The system comprises of an insulated copper vessel that contains water.

*t*_{1} = 15°C, *t*_{2} = 17°C

*t*_{1}* *is the initial temperature of the system

*t*_{2}* *is the final temperature of the system

*∆t* = Change in the temperature of the system = *t*_{2} − *t*_{1}

= 17°C − 15°C = 2°C = 275 K

Mass of the vessel, *m _{v}_{ }*= 100 g = 0.1 kg

Mass of water,

*m*= 200 g = 0.2 kg

_{w}Specific heat capacity of copper,

*c*= 420 J/kg-K

_{u}Specific heat capacity of water, c

_{w}= 4200 J/kg-K

(a) Since the system is insulated from the surroundings, no heat is transferred between the system and the surroundings. This implies that the heat transferred to the liquid vessel system is zero. The internal heat is shared between the vessel and water.

(b) Work done on the system$={m}_{w}{c}_{w}\u2206t+{m}_{v}{c}_{u}\u2206t$

⇒

*dW*= 100 × 10

^{−3}× 420 × 2 + 200 × 10

^{−3}× 4200 × 2

⇒

*dW*= 84 + 84 × 20 = 84 × 21

⇒

*dW*= 1764 J

(c) Using the first law of thermodynamics, we get

$dQ=dW+dU\phantom{\rule{0ex}{0ex}}\mathrm{Here},dW=pdV$

Work is done by the system. Thus, work done is negative.

⇒

*dQ*= 0 (given)

*dU*

*=*−

*dW*

= $-$($-$1764) = 1764 J

#### Page No 62:

#### Answer:

(a) Heat is not given to the liquid; instead, the mechanical work done is converted to heat. Also the container is adiabatic. So, no heat can enter or exit the system. This implies that the heat given to the liquid is zero.

(b) Since the 12 kg mass falls through a distance of 70 cm under gravity, energy is lost by this mass. As this mass is connected to the paddle wheel, energy lost by this mass is gained by the paddle wheel.

$\Rightarrow $Work done on the liquid = PE lost by the 12 kg mass

Now,

PE lost by the 12 kg mass = *mgh
= *12 × 10 × 0.70

= 84 J

(c) Suppose

*∆t*is the rise in temperature of the paddle wheel when the system gains energy.

$\Rightarrow $ 84 =

*ms∆t*

If

*s*is the specific heat of the system, then

84 = 1 × 4200 × ∆t (for '

*m*' = 1 kg)

⇒ $\u2206t=\frac{84}{4200}=\frac{1}{50}=0.02\mathrm{K}$

#### Page No 62:

#### Answer:

Here,

*m* = 100 kg

*u* = 2.0 m/s

*v* = 0

*μ*_{k} = 0.2

a) Internal energy of the belt-block system will decrease when the block will lose its KE in heat due to friction. Thus,

KE lost =

b) Velocity of the frame is given by

*u _{f}* = 2.0 m/s

*u’*=

*u*-

*u*= 2 – 2= 0

_{f}*v*’ = 0 – 2 = -2 m/s

KE lost =

c) Force of friction is given by

#### Page No 62:

#### Answer:

Given: Heat supplied to the system, *$\u2206$Q =* 100 J

Using the first law of thermodynamics, we get

*$\u2206U=\u2206Q-\u2206W$*

Since the container is rigid, initial volume of the system is equal to the final volume of the system. Thus,

*$\u2206$V = ${V}_{f}-{V}_{i}$=* 0

*$\u2206$W = P$\u2206$V = *0

*∆U = ∆Q = *100 J

We see that heat supplied to the system is used up in raising the internal energy of the system.

#### Page No 62:

#### Answer:

Initial pressure of the system, *P _{1}* = 10 kPa

*=*10 × 10

^{3}Pa

Final pressure of the system,

*P*= 50 kPa = 50 × 10

_{2}^{3}Pa

Initial volume of the system,

*V*= 200 cc

_{1}Final volume of the system,

*V*= 50 cc

_{2}(i) Work done on the gas = Pressure $\times $ Change in volume of the system

Since pressure is also changing, we take the average of the given two pressures.

Now,

*P*= $\left(\frac{1}{2}\right)\left(10+50\right)\times {10}^{3}$

= $30\times {10}^{3}$ Pa

Work done by the system of gas can be given by

$30\times {10}^{3}\times \left(50-200\right)\times {10}^{-6}\phantom{\rule{0ex}{0ex}}=-4.5\mathrm{J}$

(ii) Since no heat is supplied to the system, ∆

*Q*= 0.

Using the first law of thermodynamics, we get

∆

*U*

*= − ∆W*= 4.5 J

#### Page No 62:

#### Answer:

Let:

*P _{1}* = Initial pressure

*P*= Final pressure

_{2}*T*= Absolute initial temperature

_{1}*T*

_{2}_{ }= Absolute final temperature

Given :

$\frac{{P}_{1}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}}$

Using the ideal gas equation, we get

*PV = nRT*

If

*n*is the number of moles of the gas and

*R*is the universal gas constant, then

$\frac{{P}_{1}}{{T}_{1}}\mathit{=}\frac{\mathit{n}\mathit{R}}{{\mathit{V}}_{\mathit{1}}}\mathrm{and}\frac{{P}_{2}}{{T}_{2}}\mathit{=}\frac{\mathit{n}\mathit{R}}{{\mathit{V}}_{\mathit{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {V}_{1}\mathit{=}{V}_{2}\left[\because \frac{{P}_{\mathit{1}}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\mathit{\u2206}V={V}_{2}\mathit{-}{V}_{1}=0\phantom{\rule{0ex}{0ex}}$

Thus, Work done by gas =

*P∆V*= 0

#### Page No 62:

#### Answer:

Work done during any process, *W* = $P\u2206V$

If both pressure and volume are changing during a process, then work done can be found out by finding the area under the *PV* diagram.

In path ACB, for line AC:

Since initial volume is equal to final volume, $\u2206V=0$.

*$\Rightarrow $**W*_{AC} = $P\u2206V$ = 0

For line BC:

*P* = 30$\times {10}^{3}$ Pa

*${W}_{ACB}$ = W _{AC} + W_{BC} = *0

*+ P∆V*

= 30 × 10

^{3}× (25 − 10) × 10

^{−6}

= 0.45 J

For path AB:

Since both pressure and volume are changing, we use the mean pressure to find the work done.

Mean pressure,

*P*= $\frac{1}{2}\times (30+10)\times {10}^{3}$ = 20 kPa

*W*= $\frac{1}{2}$ × (10 + 30) × 10

_{AB}^{3}× 15 × 10

^{−6}

= $\frac{1}{2}$ × 40 × 15 × 10

^{−3}= 0.30 J

Initial volume in path ADB, along line DB is the same as final volume. Thus, work done along this line is zero.

Along line AD,

*P*= 10 kPa

*W = W*

_{AD}+ W_{DB}= 10 × 10

^{3}(25 − 10) × 10

^{−6}+ 0

= 10 × 15 × 10

^{−3}= 0.15 J

#### Page No 62:

#### Answer:

Initial point is *a* and the final point is *c*.

As internal energy is a state function so it depends only on the initial and final points and not on the path followed by the system. This implies that change in internal energy for path *abc* and path *adc *is the same.

Using the first law of thermodynamics, we get

∆*Q** = ∆U + ∆W*

Here, ∆Q is the amount of heat absorbed and ∆U is the change in internal energy of the system. Also, ∆W is the work done by the system.

For path *abc*:

∆*Q* = 80 J, ∆W = 30 J

∆*U* = (80 − 30) J = 50 J

For path *abc*:

∆*U* =50 J (same as for path abc)

∆*W* = 10 J

∴ ∆*Q* = 10 J + 50 J = 60 J (∆U = 50 J)

#### Page No 63:

#### Answer:

Given:

In path ACB,

*∆Q *= 50 cal = (50 × 4.2) J

*∆Q *= 210 J

∆*W** = **W*_{AC}* + **W*_{CB}

Since initial and final volumes are the same along the line BC, change in volume of the system along BC is zero.

Hence, work done along this line will be zero.

For line AC:

*P = *50 kPa

Volume changes from 200 cc to 400 cc.

$\Rightarrow \u2206V=400-200cc=200cc$

*∆W = W _{AC} + W_{CB}*

= 50 × 10

^{−3}× 200 × 10

^{−6}+ 0

= 10 J

Using the first law of thermodynamics, we get

∆

*Q*

*= ∆U + ∆W*

*⇒ ∆U = ∆Q − ∆W*= (210 − 10) J

*∆U*= 200 J

In path ADB, ∆

*Q*

*= ?*

∆

*U*= 200 J (Internal energy depends only on the initial and final points and not on the path followed.)

*⇒*∆

*W*

*= W*

_{AD}

*+*

*W*

_{DB}

_{ }= ∆WWork done for line AD will also be zero.

For line DB:

*P*= 155 kPa

$\u2206V=400-200=200cc$

*W*= 0 + 155 × 10

^{3}× 200 × 10

^{−6}

*W*= 31 J

∆

*Q*

*= ∆U + ∆W*

∆

*Q*= (200 + 31) J = 231 J

∆

*Q*

*= 55 cal*

#### Page No 63:

#### Answer:

Using the first law of thermodynamics, we get

$\u2206Q\mathit{=}\mathit{\u2206}U\mathit{+}\mathit{\u2206}W$

Since internal energy depends only on the initial and final points and for a cyclic process, initial and final points are the same, change in internal energy of the system during this process will be zero.

$\Rightarrow \u2206U=0\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206Q=\u2206W$

Heat absorbed = Work done

Work done = Area under the graph in the given case

Thus,

Heat absorbed = Area of the circle

Diameter of the circle = 300 $-$ 100 = 200

Heat absorbed = $\pi $× 10^{4} × 10^{−6} × 10^{3} J

= 3.14 × 10 = 31.4 J

#### Page No 63:

#### Answer:

Heat given in the process, ∆*Q* = 2.4 cal

∆*W** = **W*_{AB}* + **W*_{BC}* + **W*_{CA}

For line AB:

Change in volume, $\u2206V=0$

$\therefore {W}_{\mathrm{AB}}=0$

For line BC:

$\mathrm{Mean}\mathrm{pressure},P=\frac{1}{2}\times (100+200)\mathrm{kPa}=150\times {10}^{3}\mathrm{Pa}\phantom{\rule{0ex}{0ex}}\u2206V=(700-500)\mathrm{cc}=200\mathrm{cc}\phantom{\rule{0ex}{0ex}}{W}_{\mathrm{BC}}=150\times {10}^{3}\times 200\times {10}^{-6}\phantom{\rule{0ex}{0ex}}{W}_{\mathrm{BC}}=30\mathrm{J}$

For line AC:

*P* = 100 kPa

$\u2206V=200cc$

*W*_{CA}_{ }= 100 × 10^{3} × 200 × 10^{−6}

Total work done in the one cycle is given by

*∆W *= 0 + 150× 10^{3} × 200 × 10^{−6} − 100 × 10^{3} × 200 × 10^{−6}

*∆W *= $\frac{1}{2}$ × 300 × 10^{3} × 200 × 10^{−6} − 20

*∆W *= 30 J − 20 J = 10 J

∆*U* = 0 (in a cyclic process)

∆*Q** = ∆U + ∆W*

⇒ 2.4*J *= 10

$\Rightarrow J=\frac{10}{2.4}=\frac{100}{24}=\frac{25}{6}=4.17\mathrm{J}/\mathrm{cal}$

#### Page No 63:

#### Answer:

Given:

Heat given to the system, ∆*Q* = 2625 cal

Increase in the internal energy of the system, ∆*U* = 5000 J

From the graph, we get

*W* = Area of the rectangle formed under line *ab* + Area under line *bc*

For line BC:

Change in volume = 0

*${W}_{BC}=P\u2206V=0$*

∆*W** *= Area of the rectangle

∆*W* = 200 × 10^{3} × 0.03

= 6000 J

We know,

∆*Q** = ∆W + ∆U*

⇒ 2625 cal = 6000 J + 5000 J

$\Rightarrow J=\frac{11000}{2625}=4.19\mathrm{J}/\mathrm{cal}$

#### Page No 63:

#### Answer:

Given: 70 cal of heat is extracted from the system.

Here,

∆*Q* = $-$70 cal = $-$(70 × 4.2) J = $-$294 J

From the first law of thermodynamics, we get

∆*W** = P$\u2206V$*

If *P* is the average pressure between points A and B and $\u2206V$ is the change in volume of the system while going from point A to B, then

∆*W** *= $-$$\frac{1}{2}$ × (200 + 500) × 10^{3} × (150 × 10^{−6})

∆*W** *= $-$$\frac{1}{2}$ × 700 × 150 × 10^{−3}

∆*W** *= $-$525 × 10^{−1} = $-$52.5 J

Here, negative sign is taken because the final volume is less than the initial volume.

*∆U = *?

*∆Q = ∆U + ∆W
∆Q = $-$*294 J

Here, negative sign indicates that heat is extracted out from the system.

⇒ − 294 = ∆

*U*$-$ 52.5

⇒

*∆U*= − 294 + 52.5 = $-$ 241.5 J

#### Page No 63:

#### Answer:

Let change in volume of the gas be ∆*V**. *

∆*V* = (200 − 100) cm^{3} = 100 cm^{3}

= 10^{−4} m^{3}

*p *= 1 × 10^{5} Pa

Change in internal energy of the system, ∆*U** *= 1.5* pV*

∆*U* = 1.5 × 10^{5} × 10^{−4} = 15 J

∆*W** = p∆V*

= 10^{5} × 10^{−4} = 10 J

Using the first law of thermodynamics, we get

∆*Q** = ∆U + ∆W* = 10 + 15 = 25 J

Thus, heat absorbed by the system is 25 J.

#### Page No 63:

#### Answer:

Given: Heat supplied to the system, *∆Q* = 10 J

Change in volume of the system, ∆*V* = Area of cross section × Displacement of the piston

= *A* × 10 cm

= (4 × 10) cm^{3} = 40 × 10^{−6} m^{3}

*P *= 100 kPa

∆*W** = P∆V* = 100 × 10^{3} × 40 × 10^{−6} m^{3}

= 4 J

∆*U* = ?

Using the first law of thermodynamics, we get

10 *= ∆U + ∆W*

⇒ 10 = ∆*U* + 4

⇒ ∆*U** *= 6 J

Here, positive sign indicates that the internal energy of the system has increased.

#### Page No 63:

#### Answer:

(a) Given:

*P*_{1} = 100 kPa,

*V*_{1} = 2 m^{3}

*${V}_{2}$ *= 2.5 m^{3}

∆*V* = 0.5 m^{3}

Work done,* W = P∆V*

$W=100\times {10}^{3}\times 0.5\phantom{\rule{0ex}{0ex}}W=5\times {10}^{4}\mathrm{J}$

W_{AB} = Area under line AB = 5$\times $10^{4} J

If volume is kept constant for line BC, then ∆*V** *= 0.

*W*_{BC}* = P∆V* = 0

Work done while going from point B to C = 0

When the system comes back to the initial point A from C, work done is equal to area under line AC.

W_{CA} = Area of triangle ABC + Area of rectangle under line AB

Total work done, W = Area enclosed by the ABCA

W = W_{AC} $-$W_{AB}_{ }

From the graph, we see that the area under AC is greater than the area under AB. We also see that heat is extracted from the system as change in the internal energy is zero.

(b) Amount of heat extracted = Area enclosed under ABCA

$=\frac{1}{2}\times 0.5\times 100\times {10}^{3}=25000\mathrm{J}$

#### Page No 63:

#### Answer:

Given:

Number of moles of the gas, *n* = 2

∆*Q* = − 1200 J (Negative sign shows that heat is extracted out from the system)

∆*U* = 0 (During cyclic process)

Using the first law of thermodynamics, we get

∆*Q** = ∆U + ∆W*

⇒ −1200 = 0 + (*W*_{AB}* + **W*_{BC} *+ **W*_{CA})

Since the change in volume of the system applies on line CA, work done during CA will be zero.

From the ideal gas equation, we get

*PV = nRT
P∆V = nR∆T
W = P∆V = nR∆T*

*⇒ ∆Q = ∆U + ∆W*

⇒

⇒

*−*1200

*= nR∆T +*

*W*

_{BC}

*+*0

⇒ −1200 = 2 × 8.3 × 200 +

*W*

_{BC}

*W*

_{BC}= − 400 × 8.3 − 1200

= − 4520 J

#### Page No 63:

#### Answer:

Given: Number of moles of the gas, *n* = 2 moles

The system's volume is constant for lines* bc* and* da*. Therefore,

∆*V* = 0

Thus, work done for paths *da *and *bc *is zero.

*$\Rightarrow {W}_{da}={W}_{bc}=0$*

Since the process is cyclic, ∆*U* is equal to zero.

Using the first law, we get

∆*W** = ∆Q*

*∆W = ∆W _{AB} + ∆W_{CD}*

Since the temperature is kept constant during lines

*ab*and

*cd,*these are isothermal expansions.

Work done during an isothermal process is given by

*W = nRT$\mathrm{ln}\frac{{V}_{f}}{{V}_{i}}$*

If ${V}_{f}\mathrm{and}{V}_{i}$ are the initial and final volumes during the isothermal process, then

*$W=n{\mathrm{RT}}_{1}ln\left(\frac{2\mathit{}{\mathrm{V}}_{0}}{{\mathrm{V}}_{0}}\right)+n{\mathrm{RT}}_{2}ln\left(\frac{{\mathrm{V}}_{0}}{2{\mathrm{V}}_{0}}\right)$*

W =

*n*R × 2.303 × log 2 × (500 − 300)

W = 2 × 8.314 × 2.303 × 0.301 × 200

W = 2305.31 J

#### Page No 63:

#### Answer:

Given:

Mass of water, *M** *= 2 kg

Change in temperature of the system, ∆*θ* = 4°C = 277 K

Specific heat of water, *s _{w}* = 4200 J/kg-°C

Initial density,

*p*= 999.9 kg/m

_{0}^{3}

Final density,

*p*= 1000 kg/m

_{f}^{3}

*P*= 10

^{5}Pa

Let change in internal energy be ∆

*U*.

Using the first law of thermodynamics, we get

∆

*Q*

*= ∆U + ∆W*

Also,

*∆Q = ms∆θ*

*W = P∆V = P(${V}_{f}-{V}_{i}$)*

⇒

*ms∆θ = ∆U + P (V*

_{0}_{ }− V_{4})⇒ 2 × 4200 × 4= ∆U + 10

^{5}(∆V)

⇒ 33600 = ∆

*U*+ 10

^{5}

*$\left(\frac{m}{{p}_{0}}-\frac{m}{{p}_{f}}\right)$*

⇒ 33600 = ∆

*U*+ 10

^{5}× ($-$0.0000002)

⇒ 33600 = ∆

*U*$-$ 0.02

∆

*U*

*= (33600 $-$ 0.02) J*

#### Page No 63:

#### Answer:

Given:

Mass of water, *m* = 10 g = 0.01 kg

Pressure, *P* = 10^{5} Pa

Specific heat capacity of water, *c* = 1000 J/Kg $\xb0C$

Latent heat, *L* = 2.25$\times {10}^{6}\mathrm{J}/\mathrm{Kg}$

$\u2206t=\mathrm{Change}\mathrm{in}\mathrm{temperature}\mathrm{of}\mathrm{the}\mathrm{system}=100\xb0\mathrm{C}=373\mathrm{K}$

∆*Q* = Heat absorbed to raise the temperature of water from 0$\xb0\mathrm{C}\mathrm{to}100\xb0\mathrm{C}$ + Latent heat for conversion of water to steam

∆*Q** *= $mc\u2206t+mL$

= 0.01 × 4200 × 100 + 0.01 × 2.5 × 10^{6}

= 4200 + 25000 = 29200 J

∆*W** = P∆V*

∆*V*$=\mathrm{mass}\left(\frac{1}{\mathrm{final}\mathrm{density}}-\frac{1}{\mathrm{initial}\mathrm{density}}\right)$

$\u2206\mathrm{V}=\left(\frac{0.01}{0.6}\right)-\left(\frac{0.01}{1000}\right)=0.01699$

∆*W** = P∆V* = 0.01699 × 10^{5} = 1699 J

Using the first law, we get

∆*Q** = ∆W + ∆U
∆U = ∆Q − ∆W* = 29200 − 1699

= 27501 = 2.75 × 10

^{4}J

#### Page No 64:

#### Answer:

Let* **n*_{1} , *U*_{1} and *n*_{2} ,*U*_{2} be the no. of moles , internal energy of ideal gas in the left chamber and right chamber respectively.

(a) As the diathermic wall is fixed, so final volume of the chambers will be same. Thus, Δ*V* = 0, hence work done Δ*W*=* **P*Δ*V* = 0

b) Let final pressure in the first and second compartment P_{1}’ and P_{2}’.

#### Page No 64:

#### Answer:

(a) Since the conducting wall is fixed, the work done by the gas on the left part during the process is zero because the change in volume will be zero due to the fixed position of the wall.

(b) For left side:

Let the initial pressure on both sides of the wall be *p*.

We know,

Volume =*$\frac{\mathit{V}}{\mathit{2}}$*

Number of moles, *n* = 1

Let initial temperature be *T _{1}*.

Using the ideal gas equation, we get

$\frac{\mathit{P}\mathit{V}}{\mathit{2}}\mathit{=}nR{T}_{\mathit{1}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{\mathit{P}\mathit{V}}{\mathit{2}}\mathit{=}\mathit{\left(}\mathit{1}\mathit{\right)}\mathit{}RT\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}{T}_{\mathit{1}}\mathit{=}\frac{\mathit{P}\mathit{V}}{\mathit{(}\mathit{2}\mathit{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{)}\mathit{}\mathit{R}}$

For right side:

Number of moles,

*n*= 2

Let the initial temperature be T

_{2}.

We know,

Volume =

*$\frac{\mathit{V}}{\mathit{2}}$*

$\frac{\mathit{P}\mathit{V}}{\mathit{2}}\mathit{=}nR{T}_{\mathit{2}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}{T}_{\mathit{2}}\mathit{=}\frac{\mathit{P}\mathit{V}}{\mathit{(}\mathit{4}\mathit{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{)}\mathit{}\mathit{R}}$

(c)

Here,

*U* = 1.5nRT

*T* = temperature at the equilibrium

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

*n*_{1} = 1 mol

*n*_{2} = 2 mol

Let *T*_{1} and *T*_{2} be the initial temperatures of the left and right chamber respectively.

Applying eqn. of state

We know that total *n* = *n*_{1} + *n*_{2} = 3

(d) For RHS:

∆

*Q*= ∆

*U*as ∆

*W*= 0

∆

*U*= 1.5

*n*

_{2}

*R*

*(T − T*

_{2})When

*T*is the final temperature and

*T*is the initial temperature of side 1, we get

_{2}= 1.5 × 2 ×

*R (T − T*)

_{2}= 1.5 × 2 × $\frac{4\mathrm{PV}-3\mathrm{PV}}{4\times 3\mathrm{mole}}\phantom{\rule{0ex}{0ex}}$

$=\frac{3\times \mathrm{PV}}{4\times 3\mathrm{mole}}=\frac{\mathrm{PV}}{4}$

(e) If d

*W*

*= 0, then using the first law, we get*

*dQ*= −

*dU*

⇒

*dU*= −

*dQ*=

*$\frac{-\mathrm{PV}}{4}$*

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