Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 12 - Thermodynamics

Answer:

From the graph we can observe that along 1 volume remains constant hence it an isochoric process.
In 4, pressure remains constant hence it is isobaric process
The slope of isothermal curve is given by $-\frac{P}{V}$ where P ,V represent pressure and Volume at that point respectively.
The slope of adiabatic curve is given by $-\gamma \frac{P}{V}$ where $\gamma$ is the ratio of specific heats and its value is always greater than 1.
Thus adiabatic is steeper than isothermal curve which implies 2 is adiabatic process while 3 is isothermal.
Hence the correct answer is option c.

Answer:

Let the amount of sweat evaporated be x kg in 1 minute.
So in one minute:
calories produced = x â¨¯ latent heat of evaporation

$$\begin{gathered} 14.5\times {10}^3=x\times 580\times {10}^3 \\ x=\frac{14.5}{580} \\ x=0.025 \mathrm{kg} \end{gathered}$$

Disclaimer: None of the given options is correct.

 

Answer:

The equation for the process is PV=constant, which means that the process is isothermal in nature i.e. the temperature remains constant.
Also from graph pressure at 1 is greater than pressure at 2. (P₁ > P₂).
The graph that satisfies the above conditions is (iii)
Hence the correct answer is option c

Answer:

In a cyclic process the work done by the gas is given by area under the P-V diagram.
Work done in process ABCDA = Area of rectangle ABCD
                                                   = AB⨯BC
                                                   = (3V_o–V_o)⨯(2P_o–P_o)
                                                   = 2P_oV_o
Since the cyclic process is anti-clockwise ( as shown by the arrow in the diagram), so the work done by the gas is negative i.e., –2P_oV_o.
Hence the correct answer is option (b).

Answer:

Consider two containers A and B containing identical gases at the same pressure P_o initially.
Since both gases are compressed, the Pressure will increase in both.
$$\begin{gathered} \mathrm{Gas} \mathrm{A} \mathrm{under} \mathrm{goes} \mathrm{Isothermal} \mathrm{compression} \\ \mathrm{The} \mathrm{process} \mathrm{equation} \mathrm{for} \mathrm{A} \mathrm{will} \mathrm{be} \mathrm{PV}=\mathrm{constant} \\ {\mathrm{P}}_1{\mathrm{V}}_1={\mathrm{P}}_2{\mathrm{V}}_2 \\ {\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_{\mathrm{o}}={\mathrm{P}}_2\frac{{\mathrm{V}}_{\mathrm{o}}}{2} \\ {\mathrm{P}}_2=2{\mathrm{P}}_{\mathrm{o}} \\ \mathrm{Final} \mathrm{pressure} \mathrm{of} \mathrm{A}, {\mathrm{P}}_{\mathrm{A}}={\mathrm{P}}_2=2{\mathrm{P}}_{\mathrm{o}} \\ \mathrm{Gas} \mathrm{B} \mathrm{under} \mathrm{goes} \mathrm{adiabatic} \mathrm{compression} \\ \mathrm{The} \mathrm{process} \mathrm{equation} \mathrm{for} \mathrm{A} \mathrm{will} \mathrm{be} {\mathrm{PV}}^{\mathrm{\gamma }}=\mathrm{constant} \\ {\mathrm{P}}_1{\mathrm{V}}_1^{{}^{\mathrm{\gamma }}}=\mathrm{P}{\text{'}}_2{\mathrm{V}}_2^{{}^{\mathrm{\gamma }}} \\ {\mathrm{P}}_{\mathrm{o}}{\left({\mathrm{V}}_{\mathrm{o}}\right)}^{{}^{\mathrm{\gamma }}}=\mathrm{P}{\text{'}}_2{\left(\frac{{\mathrm{V}}_{\mathrm{o}}}{2}\right)}^{{}^{\mathrm{\gamma }}} \\ \mathrm{P}{\text{'}}_2={\left(2\right)}^{{}^{\mathrm{\gamma }}}{\mathrm{P}}_{\mathrm{o}} \\ \mathrm{Final} \mathrm{pressure} \mathrm{of} \mathrm{B}, {\mathrm{P}}_{\mathrm{B}}=\mathrm{P}{\text{'}}_2={\left(2\right)}^{{}^{\mathrm{\gamma }}}{\mathrm{P}}_{\mathrm{o}} \\ \mathrm{The} \mathrm{ratio} \mathrm{of} \mathrm{final} \mathrm{pressure} \mathrm{of} \mathrm{gas} \mathrm{B} \mathrm{to} \mathrm{A} \mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by}, \\ \frac{P_B}{P_{\mathrm{A}}}=\frac{{\left(2\right)}^{{}^{\mathrm{\gamma }}}{\mathrm{P}}_{\mathrm{o}}}{2{\mathrm{P}}_{\mathrm{o}}}=2^{\mathrm{\gamma }-1} \end{gathered}$$
Hence the correct answer is option a.

Answer:

Let the final temperature be T, and assume that T₁ > T > T₂ >T₃
Since no heat is lost to the surrounding
Heat lost by M₁ = Heat gained by M₂ + heat gained by M₃ 

$$\begin{gathered} {\mathrm{M}}_1\mathrm{s}({\mathrm{T}}_1-\mathrm{T})={\mathrm{M}}_2\mathrm{s}(\mathrm{T}-{\mathrm{T}}_2)+{\mathrm{M}}_3\mathrm{s}(\mathrm{T}-{\mathrm{T}}_3) \\ \mathrm{T}[{\mathrm{M}}_1+{\mathrm{M}}_2+{\mathrm{M}}_3]={\mathrm{M}}_3{\mathrm{T}}_3+{\mathrm{M}}_2{\mathrm{T}}_2+{\mathrm{M}}_1{\mathrm{T}}_1 \\ \mathrm{T}=\frac{{\mathrm{M}}_1{\mathrm{T}}_1+{\mathrm{M}}_2{\mathrm{T}}_2+{\mathrm{M}}_3{\mathrm{T}}_3}{{\mathrm{M}}_1+{\mathrm{M}}_2+{\mathrm{M}}_3} \end{gathered}$$
Hence the correct answer is option b.

Answer:

A reversible process is one which can be reversed in such a way that all changes occurring in the direct process are exactly repeated in the opposite order and inverse sense and no change is left in any of the bodies taking part in the process or in the surroundings.
The conditions for reversibility are:
• There must be complete absence of dissipative forces such as friction, viscosity, electric resistance etc. 
• The direct and reverse processes must take place infinitely slowly.
• The temperature of the system must not differ appreciably from its surroundings.

Any process which is not reversible exactly is an irreversible process. All natural processes such as conduction, radiation, radioactive decay etc. are irreversible. All practical processes such as free expansion, Joule-Thomson expansion, electrical heating of a wire are also irreversible.
(a) In this case internal energy of the rod is increased from external work done by hammer which in turn increases its temperature. So, the process cannot be retraced itself.
(b) In this process energy in the form of heat is transferred to the gas in the small container by big reservoir at temperature T₂ . Process is irreversible.
(c) In a quasi-static isothermal expansion, the gas is ideal, this process is reversible because the cylinder is fitted with frictionless piston.
(d) As the weight is added to the cylinder arrangement in the form of external pressure hence, it cannot be reversed back itself.

Hence the correct answers are options a, b, d.

Answer:

In an isothermal process dT = 0
Since, dU=nC_vdT
​∴  isothermal process dU = 0.

According to First Law of Thermodynamics, dQ = dU + dW
Since for isothermal process dU = 0,
​∴  for isothemal process dQ = dW

Hence the correct answers are options a, d. 


 

Answer:

Change in internal energy of a gas is state dependent i.e. it only depends only on the initial and final state of the gas.
In all the process the starting state and ending state is same which implies change in internal energy is same in all the processes.

Work done in a P-V diagram is equal to area under the curve, more the area under the curve more is the work done.
It can be concluded from graph that,
$A_I > A_{IV} > A_{II} > A_{III}$ 
∴  $W_I > W_{IV} > W_{II} > W_{III}$

Hence the correct answers are options b, c.

Answer:

For a cyclic process,
$$\begin{gathered} \mathrm{Since}, ∆U_{cyclic} = 0 \\ \mathrm{From} \mathrm{first} \mathrm{law} \mathrm{of} \mathrm{thermodynamics}, ∆Q_{cyclic}=∆W_{cyclic} \end{gathered}$$
In the given cyclic process, all heat given is being converted to work done, which violates the second law of thermodynamics.
The curves 2 to 3 and 3 to 1 represent adiabatic curves that should not intersect each other since it is reversible process.
Hence the correct answers are options a, c.

Answer:

Using conservation of energy we can write,

$$\begin{gathered} Q_1=Q_2+W \\ \mathrm{Then}, \\ W=Q_1-Q_2 \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} W >0 \\ \mathrm{So}, Q_1 \mathrm{should} \mathrm{be} \mathrm{greater} \mathrm{than} Q_2. \end{gathered}$$

There are two possibilities:
(i) If both Q₁ and Q₂ are positive, then:
Q₁ > Q₂ > 0
(ii) If both Q₁ and Q₂ are negative, then:
​Q₂ < Q₁ < 0

Hence the correct answers are options a, c.

Answer:

Yes. This is what happens in an isothermal expansion. Heat absorbed by the gas is completely utilised in doing work, thus the change in internal energy is zero, which implies that temperature remains constant.

$$\begin{gathered} \mathrm{From} \mathrm{First} \mathrm{Law} \mathrm{of} \mathrm{thermodynamics}, ∆Q=∆U+∆W \\ \mathrm{If} ∆T=0, \mathrm{implies} \mathrm{that} ∆U=0 \\ \therefore ∆Q=∆W \end{gathered}$$
 

Answer:

Let,
Heat given in path 1 = ΔQ₁
Heat given in path 2 = ΔQ₂
Work done in path 1= ΔW₁
Work done in path 2 = ΔW₂
It is given that, ΔQ₁ = 1000 J and  ΔW₁ - ΔW₂ = 100 J

The change in internal energy between two states for different paths is same.

$$\begin{gathered} \mathrm{Using} \mathrm{first} \mathrm{law} \mathrm{of} \mathrm{thermodynamics}, ∆Q=∆U+∆W \\ ∆U=∆Q-∆W \\ ∆U_1=∆U_2 \\ ∆Q_1-∆W_1=∆Q_2-∆W_2 \\ \mathrm{we} \mathrm{get}, \\ ∆Q_2=∆Q_1-(∆W_1-∆W_2) \\ =1000 - 100 \\ = 900 \mathrm{J} \end{gathered}$$

Answer:

Room will become hotter, because the heat removed is less than the heat supplied and hence the room, including the refrigerator (which is not insulated from the room) becomes hotter.
 A refrigerator as a device that transfers heat from inside a box to its surroundings. The room around a refrigerator is warmed as it receives the heat removed from inside the refrigerator. If a refrigerator’s door is kept open, heat is merely recycled from the room into the refrigerator, then back into the room. A net room temperature increase would result from the heat of the motor that would be constantly running to move energy around in a circle.

Answer:

According to the first law of thermodynamics,
$$\begin{gathered} ∆Q=∆U+∆W \\ \mathrm{No} \mathrm{added} \mathrm{heat} \mathrm{implies} ∆Q=0 (\mathrm{Adiabatic} \mathrm{Process}) \\ \mathrm{we} \mathrm{get}, \\ ∆U=-∆W \\ \mathrm{If} ∆W \mathrm{is} \mathrm{negative} (\mathrm{gas} \mathrm{under} \mathrm{goes} \mathrm{compression}), \\ ∆U \mathrm{is} \mathrm{positive}, \mathrm{which} \mathrm{implies} \mathrm{temperature} \mathrm{of} \mathrm{gas} \mathrm{rises}. \end{gathered}$$ 
Hence we can increase the temperature of the gas by Adiabatic compression.
 

Answer:

Answer:

Temperature of Source (T₁) = 500 K
Temperature of Sink (T₂) = 300 K
Work done by engine = 1 kJ
Heat Absorbed from Source = Q₁
Heat released to Sink = Q₂
Efficiency of Carnot engine is given by,
$$\begin{gathered} \eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1} \\ 1 - \frac{300}{500}=\frac{1}{Q_1} \\ {Q}_1 = 2.5 \mathrm{kJ} \\ \mathrm{Using} \mathrm{Conservation} \mathrm{of} \mathrm{energy} \mathrm{we} \mathrm{can} \mathrm{write}, \\ {Q}_1 = W + Q_2 \\ 2.5 = 1 + Q_2 \\ {Q}_2 = 1.5 \mathrm{kJ} \end{gathered}$$

 

Answer:

height of stairs, h = 10 m
The energy lost by person in form of fat is utilised to increase the potential energy of the person when he climbs up. It is also given that he burns twice as much fat while going up than coming down.

Work done in one trip (up and down stairs)
$$\begin{gathered} = mgh + \frac{1}{2}mgh = \frac{3}{2}mgh \\ = \frac{3}{2}\times 60\times 10\times 10 = 9\times {10}^3 \mathrm{J} \end{gathered}$$

Work done to burn 5 kg of fat 
$$\begin{gathered} =5\times 7000\times {10}^3\times 4.2 \\ = 147\times {10}^6 \mathrm{J} \end{gathered}$$

Number of times the person has to go up and downstairs 
$N=\frac{147\times {10}^6}{9\times {10}^3}=16.3\times {10}^3$

Answer:

$$\begin{gathered} \mathrm{The} \mathrm{process} \mathrm{is} \mathrm{adiabatic}, \mathrm{it}\text{'}\mathrm{s} \mathrm{process} \mathrm{equation} \mathrm{is} \\ P{V}^{\gamma }=\mathrm{constant} \\ P{\left(V+∆V\right)}^{\gamma } =(P+∆p)V^{\gamma } \\ \mathrm{Taking} V^{\gamma } \mathrm{common} \mathrm{on} \mathrm{LHS} \mathrm{and} P \mathrm{from} \mathrm{RHS} \\ P{\left(1+\frac{∆V}{V}\right)}^{\gamma }{V}^{\gamma } =P(1+\frac{∆p}{P})V^{\gamma } \\ \mathrm{Since} ∆V <<V \\ \mathrm{we} \mathrm{can} \mathrm{use} \mathrm{binomial} \mathrm{expansion} \\ P\left(1+\gamma \frac{∆V}{V}\right) =P\left(1+\frac{∆p}{P}\right) \\ \gamma \frac{∆V}{V}=\frac{∆p}{P} \end{gathered}$$

If $∆V << V$, we can write it as dV.
$$\begin{gathered} dV=\frac{V}{\gamma P}dP \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{a} \mathrm{gas} \mathrm{is} \mathrm{given} \mathrm{by}, W={\int }_{V_1}^{V_2}PdV \\ \mathrm{Substituting} \mathrm{the} \mathrm{value} \mathrm{of}\mathit{ }dV \mathrm{we} \mathrm{get}, \\ W={\int }_{P_1}^{P_2}P\frac{V}{\gamma P}dP \\ \mathrm{Since} V \mathrm{is} \mathrm{constant} \mathrm{on} \mathrm{solving} \mathrm{we} \mathrm{get} \\ W=\frac{\left(P_2-P_1\right)V}{\gamma } \end{gathered}$$

Answer:

Q₁ = Heat rejected to sink at higher temperature (T₁ = 300 K)
Q₂ = Heat absorbed from source at lower temperature (T₂ = 270 K)

$$\begin{gathered} \mathrm{For} \mathrm{a} \mathrm{refrigerator}, \\ {Q}_1=W+Q_2 \\ \eta =1 - \frac{Q_2}{Q_1} \\ \mathrm{For} \mathrm{perfect} \mathrm{engine}, \\ \eta =1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1} \\ \eta =1-\frac{270}{300}=0.1 \\ \mathrm{For} \mathrm{the} \mathrm{given} \mathrm{refigerator} \mathrm{efficiency} \mathrm{will} \mathrm{be} 50\% \mathrm{of} \mathrm{perfect} \mathrm{engine} \\ {\eta }_r=0.1\times 0.5=0.05 \\ {\eta }_r=\frac{W}{Q_1}=0.05 \\ {Q}_1 = \frac{W}{0.05}=\frac{1}{.05}=20 \mathrm{kJ} \\ \mathrm{Since}, Q_1=W+Q_2 \\ {Q}_2 = Q_1 - W \\ {Q}_2=20 - 1 \\ =19 \mathrm{kJ} \end{gathered}$$

Answer:

Q₁ = Heat rejected to sink at higher temperature (T₁ = 300 K)
Q₂ = Heat absorbed from source at lower temperature (T₂ = ? )

$$\begin{gathered} \mathrm{Coefficeint} \mathrm{of} \mathrm{performance} \mathrm{for} \mathrm{refrigerator} \mathrm{is} \mathrm{given} \mathrm{by}, \\ \beta =\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}=5 \\ \frac{T_2}{300-T_2}=5 \\ {T}_2=250 \mathrm{K} \\ \mathrm{In} \mathrm{Celcius} \mathrm{scale}, T_2=250-273 = -23 {}^{\mathrm{o}}\mathrm{C} \end{gathered}$$

Answer:

Below is the P-V diagram for the two cases.


The process in case (a) is isothermal, i.e., at constant temperature and the process in case (b) is isobaric, i.e., at constant pressure.
Area under the P-V diagram gives the work done by the gas in the two process. It can be observed in the diagram that the area under the P-V graph in case (b) is more than that case (a). 
Hence the work done is more when expansion takes place at constant pressure.

Answer:

As given in the question,
$$\begin{gathered} P{V}^{\frac{1}{2}}=k=cons\tan t, \\ \Rightarrow P=\frac{k}{V^{{\displaystyle \frac{1}{2}}}} \\ \Rightarrow P_1{V_1}^{\frac{1}{2}}=P_2{V_2}^{\frac{1}{2}}=k \end{gathered}$$
(a) Now, work done for the process 1 to 2 will be,
$$\begin{gathered} W={\int }_{V_1}^{V_2}PdV \\ ={\int }_{V_1}^{V_2}\frac{k}{V^{{\displaystyle \frac{1}{2}}}}dV \\ =2k\left({V_2}^{\frac{1}{2}}-{V_1}^{{}^{\frac{1}{2}}}\right) \\ =2P_1{V_1}^{{}^{\frac{1}{2}}}\left({V_2}^{\frac{1}{2}}-{V_1}^{{}^{\frac{1}{2}}}\right) \end{gathered}$$
(b) From ideal gas equation:

$$\begin{gathered} PV=nRT \\ \Rightarrow P=\frac{nRT}{V} \\ \mathrm{Substituting} \mathrm{value} \mathrm{of} P \mathrm{in} P{V}^{\frac{1}{2}}=k, \mathrm{we} \mathrm{get} : \\ \frac{nRT}{V}{V}^{{}^{\frac{1}{2}}}=k \\ \Rightarrow \frac{T}{V^{{}^{\frac{1}{2}}}}=k\text{'}\left(\mathrm{constant}\right) \\ \Rightarrow \frac{T_1}{{V_1}^{{}^{\frac{1}{2}}}}=\frac{T_2}{{V_2}^{{}^{\frac{1}{2}}}} \\ \frac{T_2}{T_1}=\frac{{V_2}^{{}^{\frac{1}{2}}}}{{V_1}^{{}^{\frac{1}{2}}}}={\left[\frac{2V_1}{V_1}\right]}^{{}^{\frac{1}{2}}}=\sqrt{2} \left(\mathrm{Given} \mathrm{that} V_2=2V_1\right) \\ \frac{T_1}{T_2}=\frac{1}{\sqrt{2}} \end{gathered}$$

(c) Internal energy of the gas, $U=\frac{3}{2}RT$. 

$$\begin{gathered} \mathrm{So}, \mathrm{change} \mathrm{in} \mathrm{internal} \mathrm{energy}, \\ ∆U = U_2-U_1=\frac{3}{2}R\left(T_2-T_1\right) \\ =\frac{3}{2}R\left(\sqrt{2}{T}_1-T_1\right) \left(\mathrm{as}, T_{2 }=\sqrt{2}{T}_1\right) \\ =\frac{3}{2}R{T}_1\left(\sqrt{2}-1\right) \\ \mathrm{So}, \mathrm{work} \mathrm{done}, \\ W = 2P_1{V}_1^{\frac{1}{2}}\left(V_2^{\frac{1}{2}}-V_1^{\frac{1}{2}}\right) \\ =2P_1{V}_1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\sqrt{2}{V}_1^{\frac{1}{2}}-V_1^{\frac{1}{2}}\right) \left(\mathrm{As}, V_2=2V_1\right) \\ =2P_1{V}_1\left(\sqrt{2}-1\right) \\ =2R{T}_1\left(\sqrt{2}-1\right) \left(\mathrm{using} P_1{V}_1=nR{T}_1\right) \\ \mathrm{Heat} \mathrm{supplied} \mathrm{is} \mathrm{given} \mathrm{by}, \\ Q=∆U+W \\ =\frac{3}{2}R{T}_1\left(\sqrt{2}-1\right) +2R{T}_1\left(\sqrt{2}-1\right) \\ =\frac{7}{2}R{T}_1\left(\sqrt{2}-1\right) \end{gathered}$$

Answer:

It is given that,
$\gamma =\frac{5}{3} \mathrm{for} \mathrm{the} \mathrm{gas} \mathrm{and} C_v=\frac{3}{2}R \mathrm{for} \mathrm{one} \mathrm{mole}$
(a) For the process, A to B:
Volume remains constant.
$$\begin{gathered} P_A<P_B \left( V = \mathrm{constant}\right) \\ \mathrm{As}, PV=nRT \left(\mathrm{ideal} \mathrm{gas} \mathrm{equation}\right) \\ \Rightarrow T_B > T_A \\ Q_{A to B} = ∆U+∆W=∆U+0=n{C}_v∆T=n{C}_v(T_B-T_A) > 0 \\ \mathrm{It} \mathrm{means}, \mathrm{heat} \mathrm{is} \mathrm{supplied} \mathrm{to} \mathrm{the} \mathrm{engine}. \end{gathered}$$
(b) For the process C to D:
Volume remains constant.
$$\begin{gathered} P_C>P_D \left(\mathrm{V}= \mathrm{constant}\right) \\ \Rightarrow T_C > T_D \\ Q_{C to D} = ∆U+∆W=n{C}_v∆T+0=n{C}_v(T_D-T_C) < 0 \\ \mathrm{Hence}, \mathrm{in} \mathrm{this} \mathrm{part} \mathrm{heat} \mathrm{is} \mathrm{given} \mathrm{to} \mathrm{the} \mathrm{surrounding} \mathrm{by} \mathrm{the} \mathrm{engine}. \end{gathered}$$

(c) For the complete cyclic process:
$$\begin{gathered} ∆Q_{cyclic} = W_{cyclic} \left(\mathrm{since}, ∆U_{cyclic} = 0\right) \\ Q_{A to B} = n{C}_v∆T=\frac{nR\left(T_B-T_A\right)}{\gamma -1} \\ =\frac{P_B{V}_B\hspace{0.17em}-P_A{V}_A}{\gamma -1} \\ = \frac{\left(P_B\hspace{0.17em}-P_A\right)V_A}{\gamma -1} \left(\mathrm{as}, V_A=V_B\right) ...\left(1\right) \\ Q_{B to C}=0 \left(\mathrm{Adiabatic} \mathrm{process}\right) ...\left(2\right) \\ Q_{C to D} = n{C}_v∆T=\frac{nR(T_D-T_C)}{\gamma -1} \\ =\frac{P_D{V}_D\hspace{0.17em}-P_C{V}_C}{\gamma -1} \\ = \frac{\left(P_D\hspace{0.17em}-P_C\right)V_C}{\gamma -1} \left(\mathrm{as}, V_D=V_C\right) \\ \mathrm{We} \mathrm{know} \mathrm{that}, P_B{V}_B^{\gamma }=P_C{V}_C^{\gamma } \\ \Rightarrow P_B{V}_B^{\gamma }=P_C{\left(2V_B\right)}^{\gamma } \\ \Rightarrow P_C= 2^{-\gamma }{P}_B \\ \mathrm{also}, P_A{V}_A^{\gamma }=P_D{V}_D^{\gamma } \\ \Rightarrow P_A{V}_A^{\gamma }=P_D{\left(2V_A\right)}^{\gamma } \Rightarrow P_D= 2^{-\gamma }{P}_A \\ Q_{C to D}= \frac{\left(P_D\hspace{0.17em}-P_C\right)V_C}{\gamma -1} \\ =\frac{\left(2^{-\gamma }{P}_A\hspace{0.17em}-2^{-\gamma }{P}_B\right)2V_A}{\gamma -1}=\frac{2^{-\gamma +1}\left(P_A\hspace{0.17em}-P_B\right)V_A}{\gamma -1} \\ =-\frac{2^{-\gamma +1}\left(P_B\hspace{0.17em}-P_A\right)V_A}{\gamma -1} ...\left(3\right) \\ Q_{D to A}=0 \left(\mathrm{Adiabatic} \mathrm{process}\right) ...\left(4\right) \\ W_{cyclic}=∆Q_{cyclic}= Q_{A to B}+ Q_{B to C}+ Q_{C to D}+Q_{D to A} \\ W_{cyclic} =\frac{\left(P_B\hspace{0.17em}-P_A\right)V_A}{\gamma -1}+0-\frac{2^{-\gamma +1}\left(P_B\hspace{0.17em}-P_A\right)V_A}{\gamma -1}+0 \\ =\frac{\left(1-2^{-\gamma +1}\right)\left(P_B\hspace{0.17em}-P_A\right)V_A}{\gamma -1}=\frac{3}{2}\left[1-{\left(\frac{1}{2}\right)}^{\frac{2}{3}}\right]\left(P_B\hspace{0.17em}-P_A\right)V_A \\ \left(\mathrm{d}\right) \mathrm{Heat} \mathrm{supplied} =Q_{A to B}=\frac{\left(P_B\hspace{0.17em}-P_A\right)V_A}{\gamma -1} \\ efficiency =\frac{Work done}{Heat Supplied} \\ =\frac{\frac{\left(1-2^{-\gamma +1}\right)\times \left(P_B\hspace{0.17em}-P_A\right)V_A}{\gamma -1}}{\frac{\left(P_B\hspace{0.17em}-P_A\right)V_A}{\gamma -1}}=1-2^{-\gamma +1}=1-2^{-\frac{5}{3}+1}=1-2^{-\frac{2}{3}} \\ =1-{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \end{gathered}$$
 

Answer:

$$\begin{gathered} \mathrm{AB} \mathrm{is} \mathrm{isochoric} \mathrm{process} \\ {Q}_{AB}=n{C}_v∆T=n\frac{3}{2}R(T_B-T_A) [\mathrm{for} \mathrm{monoatomic} \mathrm{gas} C_v=\frac{3}{2}R] \\ \mathrm{Using} PV=nRT, \mathrm{we} \mathrm{get} \\ {Q}_{AB}=\frac{3}{2}{V}_A(P_B-P_A) \\ \mathrm{BC} \mathrm{is} \mathrm{an} \mathrm{isobaric} \mathrm{process} \\ {Q}_{BC}=n{C}_p∆T=n\frac{5}{2}R(T_C-T_B) [\mathrm{for} \mathrm{monoatomic} \mathrm{gas} C_p=\frac{5}{2}R] \\ \mathrm{Using} PV=nRT, \mathrm{we} \mathrm{get} \\ {Q}_{BC}=\frac{5}{2}{P}_B(V_C-V_B) \\ \mathrm{CD} \mathrm{is} \mathrm{an} \mathrm{adiabatic} \mathrm{process} \\ \mathrm{For} \mathrm{an} \mathrm{adiabatic} \mathrm{process} Q=0 \\ {Q}_{CD}=0 \\ \mathrm{DA} \mathrm{is} \mathrm{an} \mathrm{isobaric} \mathrm{process} \\ {Q}_{DA}=n{C}_p∆T=n\frac{5}{2}R(T_A-T_D) [\mathrm{for} \mathrm{monoatomic} \mathrm{gas} C_p=\frac{3}{2}R] \\ \mathrm{Using} PV=nRT, \mathrm{we} \mathrm{get} \\ {Q}_{DA}=\frac{5}{2}{P}_A(V_A-V_D) \end{gathered}$$

Answer:

Given that,
$$\begin{gathered} P=f\left(V\right), \\ \mathrm{at} \left(P_o,V_o\right), P_o = f\text{'}\left(V_o\right) \\ \mathrm{Spole} \mathrm{of} \mathrm{adiabatic} \mathrm{curve} \mathrm{at} (P_o,V_o) \\ = -\gamma \frac{P_o}{V_o} \\ \mathrm{So}, \text{Heat absorbed in a process is,} \\ dQ=dU+PdV \\ =n{C}_{v}dT+PdV \\ =\frac{nR}{\gamma -1}dT + f\left(V\right)dV \end{gathered}$$
From ideal gas equation,
$$\begin{gathered} PV=nRT \\ T=\frac{PV}{nR}=\frac{f\left(V\right)V}{nR} \\ dT=\frac{1}{nR}\left[V.f\text{'}\left(V\right)dV +\hspace{0.17em}f\left(V\right)dV\right] \\ \mathrm{Now} \mathrm{putting} \mathrm{value} \mathrm{of} dT \mathrm{in} dQ \\ dQ=\frac{nR}{\gamma -1}\times \frac{1}{nR}\left[V.f\text{'}\left(V\right)dV +\hspace{0.17em}f\left(V\right)dV\right]+ f\left(V\right)dV \\ dQ=\frac{\gamma f\left(V\right)dV}{\gamma -1}+\frac{V.f\text{'}\left(V\right)dV}{\gamma -1} \\ =\frac{1}{\gamma -1}\left[\gamma f\left(V\right)+V.f\text{'}\left(V\right)\right]dV \\ \mathrm{Now}, \frac{dQ}{dV}=\frac{1}{\gamma -1}\left[\gamma f\left(V\right)+V.f\text{'}\left(V\right)\right] \\ \mathrm{Heat} \mathrm{is} \mathrm{being} \mathrm{absorbed} \mathrm{only} \mathrm{if} \frac{dQ}{dV}>0 \mathrm{during} \mathrm{expansion} \mathrm{of} \mathrm{gas}: \\ \gamma f\left(V\right)+V.f\text{'}\left(V\right) > 0 \\ \mathrm{Now} \mathrm{at} \left(P_o,V_o \right) \\ \gamma f\left(V_o\right)+V_o.f\text{'}\left(V_o\right) > 0 \\ \gamma P_o+V_o.f\text{'}\left(V_o\right) > 0 \left[ \mathrm{at} \left(P_o,V_o \right), f\left(V_o\right)=P_o\right] \\ f\text{'}\left(V_o\right)>-\frac{\gamma P_o}{V_o} \left(\mathrm{slope}\right) \end{gathered}$$
Hence, it is clear that the gas is absorbing heat at (P₀, V₀) if the slope of the curve P = f (V) is larger than the slope of the adiabat passing through (P₀, V₀ ).

Answer:

Initially the spring is in unstretched condition, so the initial pressure is equal to atmospheric pressure:
$$\begin{gathered} P_o =P_a \\ \left(\mathrm{b}\right) \mathrm{Change} \mathrm{in} \mathrm{length} \mathrm{of} \mathrm{spring} \mathrm{when} \mathrm{volume} \mathrm{increases} \mathrm{from} V_{o }\mathrm{to} V_1 \mathrm{is} \mathrm{given} \mathrm{by}, \\ ∆L=\frac{V_1-V_{o }}{A} \left(\mathrm{where}, A=\mathrm{crossectional} \mathrm{area}=1\right) \\ \mathrm{Hence}, \mathrm{force} \mathrm{due} \mathrm{to} \mathrm{spring} \mathrm{will} \mathrm{be}, \\ F =k\times ∆L= \frac{k\left(V_1-V_{o }\right)}{A} \\ \mathrm{Final} \mathrm{pressure} P_f \mathrm{is} \mathrm{given} \mathrm{by}, \\ P_f = P_a + \frac{k\left(V_1-V_{o }\right)}{A^2}= P_a +k\left(V_1-V_{o }\right) \left(\mathrm{given}, A=1\right) \end{gathered}$$
(c) From first law of thermodynamics:

$$\begin{gathered} Q=∆U+∆W \\ \mathrm{So} \mathrm{change} \mathrm{in} \mathrm{internal} \mathrm{energy} \mathrm{is} \mathrm{given} \mathrm{by} , \\ ∆U = n{C}_v∆T=n{C}_v\left(T_1-T_o\right)=C_v\left(T_1-T_o\right) \left(\mathrm{as}, n=1 \mathrm{mole}\right) \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{the} \mathrm{gas} = \mathrm{potential} \mathrm{energy} \mathrm{stored} \mathrm{in} \mathrm{spring} + \mathrm{work} \mathrm{done} \mathrm{against} \mathrm{atmospheric} \mathrm{pressure} \\ ∆W\hspace{0.17em}= \frac{1}{2}k{\left(\frac{V_1-V_o}{A}\right)}^2 + P_a(V_1-V_o) \\ \mathrm{Total} \mathrm{heat} \mathrm{absorbed} \mathrm{by} \mathrm{gas}: \\ Q = C_v\left(T_1-T_o\right)+\frac{1}{2}k{\left(\frac{V_1-V_o}{A}\right)}^2 + P_a(V_1-V_o) \end{gathered}$$