There are two Isothermal Processes and two Adiabatic Processes present in the Carnot cycle. The theoretical heat engine that operates on this cycle is the Carnot Engine. It consists of four operations in succession as shown in below figure 12.11; (a) isothermal expansion (heat absorption) at a higher temperature \(T_h\) (b) the adiabatic expansion between temperature \(T_h\) and \(T_l\) (c) isothermal compression (heat rejection) at a constant lower temperature \(T_l\) and (d) adiabatic compression between temperatures \(T_l\) and \(T_h\).
The working substance (gas) is to be placed in the cylinder, which expands and compresses and makes the piston move up and down with respect to the processes.
It consists of 4 reversible processes, as shown in the figure given below. They are:

1. Process 1-2: Reversible Isothermal Expansion.
2. Process 2-3: Reversible Adiabatic Expansion
3. Process 3-4: Reversible Isothermal Compression.
4. Process 4-1: Reversible Adiabatic Compression.

Efficiency of the Carnot Engine

To find the efficiency of the Carnot engine, let us consider all the steps of the Carnot cycle as shown in the above figure, which is given following one by one:

Step 1 -> 2: Isothermal expansion: Isothermal expansion of the gas taking its state from \(\left(P_1, V_1, T_1\right)\) to \(\left(P_2, V_2, T_1\right)\).
The heat absorbed by the gas \(\left(Q_1\right)\) from the reservoir at temperature \(T_1\) is given by Eq. (12.12). This is also the work done \(\left(W_{1 \rightarrow 2}\right)\) by the gas on the environment.
\(
W_{1 \rightarrow 2}=Q_1=\mu R T_1 \ln \left(\frac{V_2}{V_1}\right) \dots(12.23)
\)

Step 2 -> 3: Adiabatic expansion: Adiabatic expansion of the gas from \(\left(P_2, V_2, T_1\right)\) to \(\left(P_3, V_3, T_2\right)\) Work done by the gas, using Eq. (12.16), is
\(
W_{2 \rightarrow 3}=\frac{\mu R\left(T_1-T_2\right)}{\gamma-1} \dots(12.24)
\)

Step 3 -> 4: Isothermal compression: Isothermal compression of the gas from \(\left(P_3, V_3, T_2\right)\) to \(\left(P_4, V_4, T_2\right)\).
Heat released \(\left(Q_2\right)\) by the gas to the reservoir at temperature \(T_2\) is given by Eq. (12.12). This is also the work done \(\left(W_{3 \rightarrow 4}\right)\) on the gas by the environment.
\(
W_{3 \rightarrow 4}=\Theta_2=\mu R T_2 \ln \left(\frac{V_3}{V_4}\right) \dots(12.25)
\)

Steps 4 -> 1: Adiabatic compression: Adiabatic compression of the gas from \(\left(P_4, V_4, T_2\right)\) to \(\left(P_1, V_1, T_1\right)\).
Work done on the gas, [using Eq.(12.16), is
\(
W_{4 \rightarrow 1}=\mu R\left(\frac{T_1-T_2}{\gamma-1}\right) \dots(12.26)
\)
From Eqs. (12.23) to (12.26) total work done by the gas in one complete cycle is
\(
\begin{aligned}
& W=W_{1 \rightarrow 2}+W_{2 \rightarrow 3}-W_{3 \rightarrow 4}-W_{4 \rightarrow 1} \\
& =\mu R T_1 \ln \left(\frac{V_2}{V_1}\right)-\mu R T_2 \ln \left(\frac{V_3}{V_4}\right) \dots(12.27)
\end{aligned}
\)
The efficiency \(\eta\) of the Carnot engine is
\(
\begin{aligned}
& \eta=\frac{W}{Q_1}=1-\frac{Q_2}{Q_1} \\
& =1-\left(\frac{T_2}{T_1}\right) \frac{\operatorname{In}\left(\frac{V_3}{V_4}\right)}{\operatorname{In}\left(\frac{V_2}{V_1}\right)} \dots(12.28)
\end{aligned}
\)
Now since step \(2 \rightarrow 3\) is an adiabatic process,
\(
T_1 V_2^{\gamma-1}=T_2 V_3^{\gamma-1}
\)
\(
\text { i.e. } \frac{V_2}{V_3}=\left(\frac{T_2}{T_1}\right)^{1 /(\gamma-1)} \dots(12.29)
\)
Similarly, since step \(4 \rightarrow 1\) is an adiabatic process
\(
T_2 V_4^{\gamma-1}=T_1 V_1^{\gamma-1}
\)
i.e. \(\frac{V_1}{V_4}=\left(\frac{T_2}{T_1}\right)^{1 / \gamma-1} \dots(12.30)\)
From Eqs. (12.29) and (12.30),
\(
\frac{V_3}{V_4}=\frac{V_2}{V_1} \dots(12.31)
\)
Using Eq. (12.31) in Eq. (12.28), we get
\(
\eta=1-\frac{T_2}{T_1} \text { (Carnot engine) } \dots(12.32)
\)

Thus, the efficiency of the engine depends only on the temperatures of the hot and cold bodies between which the engine works.

We have already seen that a Carnot engine is a reversible engine. Indeed it is the only reversible engine possible that works between two reservoirs at different temperatures. Each step of the Carnot cycle given in Fig. \(12.11\) can be reversed. This will amount to taking heat \(Q_2\) from the cold reservoir at \(T_2\), doing work \(W\) on the system, and transferring heat \(Q_1\) to the hot reservoir. This will be a reversible refrigerator.

Carnot’s Theorem

Carnot engine is a reversible engine. It can be proved from the second law of thermodynamics that: All reversible engines operating between the same two temperatures have equal efficiency and no engine operating between the same two temperatures can have an efficiency greater than this.

This theorem is called Carnot’s theorem. It is a consequence of the second law and puts a theoretical limit \(\eta=1-\frac{T_2}{T_1}\) to the maximum efficiency of heat engines.

Refrigerator or Heat Pump

A heat engine takes heat from a hot body, converts part of it into work, and rejects the rest to a cold body. The reverse operation is done by a refrigerator also known as a heat pump. It takes an amount \(Q_2\) of heat from a cold body, an amount \(W\) of work is done on it by the surrounding and the total energy \(Q_1=Q_2+W\) is supplied to a hot body in the form of heat. Thus, heat is passed from the cold body to the hot body. Figure (12.12) shows the process schematically. If the heat is taken at a single low temperature \(T_2\), it is rejected at a single high temperature \(T_1\) and all the parts of the process are carried out reversibly, we get a Carnot refrigerator. If the operating temperatures are fixed, a Carnot refrigerator needs minimum amount of work done to extract a given amount \(Q_2\) of heat from the colder body.

In this case,
\(
\begin{array}{rlrl}
\frac{Q_1}{Q_2} =\frac{T_1}{T_2} \dots(12.33)\\
\text { or, } \quad \frac{Q_2+W}{Q_2} =\frac{T_1}{T_2} \\
\text { or, } W =Q_2\left(\frac{T_1}{T_2}-1\right) .
\end{array}
\)
A minimum of this much work has to be done by the surrounding if we wish to transfer heat \(Q_2\) from the low-temperature body to the high-temperature body. This leads to another statement of the second law of thermodynamics as follows:

It is not possible to design a refrigerator which works in a cyclic process and whose only result is to transfer heat from a body to a hotter body.

This is known as the Claussius statement of the second law.

Example 1: How much work does a Carnot engine perform when it takes \(1000 \mathrm{kcal}\) of heat from a reservoir at \(827^{\circ} \mathrm{C}\) and exhausts it to a sink at \(27^{\circ} \mathrm{C}\)? And also find the efficiency of the engine?

Solution:

Given, \(Q_1=10^6 \mathrm{Cal}\)
\(
T_1=(827+273)=1100 \mathrm{~K}
\)
And,
\(
T_2=(27+273)=300 \mathrm{~K}
\)
as,
\(
\begin{aligned}
& \frac{Q_2}{Q_1}=\frac{T_2}{T_1} \Rightarrow Q_2=\frac{T_2}{T_1} \times Q_1=\left(\frac{300}{1100}\right) \times\left(10^6\right) \\
& =2.72 \times 10^5 \mathrm{cal} \\
& W=Q_1-Q_2=7.28 \times 10^5 \mathrm{cal}
\end{aligned}
\)
The efficiency of the cycle,
\(
\begin{aligned}
& \eta=\left(1-\frac{T_2}{T_1}\right) \times 100 \\
& \eta=\left(1-\frac{300}{1100}\right) \times 100 \\
& \eta=72.72 \%
\end{aligned}
\)

Example 2: The amounts of heat involved when an ideal gas is taken through a cyclic thermodynamic process through four steps are given as \(Q_1=5960 \mathrm{~J}, Q_2=-5585 \mathrm{~J}, Q_3=-2980 \mathrm{~J}\) and \(Q_4=3645 \mathrm{~J}\) respectively. And the corresponding quantities of work involved are given as \(W_1=2200 \mathrm{~J}, W_2=-825 \mathrm{~J}, W_3=-1100 \mathrm{~J}\) and \(W_4\), respectively. Then find the value of \(W_4\).

Solution:

In a cyclic process, \(\Delta U=0\)
Therefore,
\(
Q_{n e t}=W_{n e t}
\)
Or,
\(
Q_1+Q_2+Q_3+Q_4=W_1+W_2+W_3+W_4
\)
Hence,
\(
\begin{aligned}
& W_4=\left(Q_1+Q_2+Q_3+Q_4\right)-\left(W_1+W_2+W_3\right) \\
& =\{(5960-5585-2980+3645)-(2200-825-1100)\}
\end{aligned}
\)
Or,
\(
W_4=765 \mathrm{~J}
\)

Example 3: The figure below shows a cylindrical tube of volume \(V\) with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by \(1.5 \mathrm{nRT}\). The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are \(p_1, T_1\) on the left, and \(p_2, T_2\) on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides. (a) How much work has been done by the gas on the left part? (b) Find the final pressures on the two sides. (c) Find the final equilibrium temperature. (d) How much heat has flown from the gas on the right to the gas on the left?

Solution:

Let \(n_1, U_1\) and \(n_2, U_2\) be the no. of moles, and internal energy of ideal gas in the left chamber and right chamber respectively.
(a) As the diathermic wall is fixed, so the final volume of the chambers will be the same. Thus, \(\Delta V=0\), hence work done \(\Delta \mathrm{W}=\mathrm{P} \Delta \mathrm{V}=0\)
by eq. of state in the first and second chamber
\(
\begin{aligned}
& P_1 \frac{V}{2}=n_1 R T_1 \\
& \Rightarrow n_1=\frac{P_1 V}{2 R T_1} \\
& P_2 \frac{V}{2}=n_2 R T_2 \\
& \Rightarrow n_2=\frac{P_2 V}{2 R T_2} \\
& \mathrm{n}=\mathrm{n}_1+\mathrm{n}_2 \\
& \Rightarrow n=\frac{P_1 V}{2 R T_1}+\frac{P_2 V}{2 R T_2}=\frac{V}{2 R}\left(\frac{P_1 T_2+P_2 T_1}{T_1 T_2}\right)
\end{aligned}
\)
Again,
\(
\begin{aligned}
& \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \mathrm{T} \\
& \Rightarrow n C_{\mathrm{v}} T=1.5 n R T \\
& \Rightarrow C_{\mathrm{v}}=1.5 R
\end{aligned}
\)
In the first and second chamber internal energy is given by
\(
\begin{aligned}
& \mathrm{U}_1=\mathrm{n}_1 \mathrm{C}_{\mathrm{v}} \mathrm{T}_1=\mathrm{n}_1 1.5 R \mathrm{~T}_1 \\
& \mathrm{U}_2=\mathrm{n}_2 \mathrm{C}_{\mathrm{v}} \mathrm{T}_2=\mathrm{n}_2 1.5 R \mathrm{~T}_2 \\
& \mathrm{U}=\mathrm{U}_1+\mathrm{U}_2
\end{aligned}
\)
\(
\begin{aligned}
& 1.5 \mathrm{nRT}=\mathrm{n}_1 1.5 \mathrm{RT}_1+\mathrm{n}_2 1.5 \mathrm{RT}_2 \\
& \Rightarrow \mathrm{nT}=\mathrm{n}_1 \mathrm{~T}_1+\mathrm{n}_2 \mathrm{~T}_2 \\
& \Rightarrow n T=\frac{P_1 V}{2 R T_1} T_1+\frac{P-2 V}{2 R T_2} T_2=\frac{\left(P_1+P_2\right) V}{2 R} \\
& \Rightarrow T=\frac{\left(P_1+P_2\right) V}{2 n R}=\frac{\left(P_1+P_2\right) V}{2 R \frac{V}{2 R}\left(\frac{P_1 T_2+P_2 T_1}{T_1 T_2}\right)}=\frac{T_1 T_2\left(P_1+P_2\right)}{P_1 T_2+P_2 T_1} . \dots(1)
\end{aligned}
\)
b) Let final pressure in the first and second compartment \(P_1^{\prime}\) and \(P_2^{\prime}\).
By five variable equation of state in the first chamber
\(
\begin{aligned}
& \frac{P_1 \frac{V}{2}}{T_1}=\frac{P_1 \prime \frac{V}{2}}{T} \\
& \Rightarrow P_1 \prime=\frac{P_1}{T_1} T
\end{aligned}
\)
By eq. (1)
\(
\Rightarrow P_1 \prime=\frac{P_1}{T_1} \frac{T_1 T_2\left(P_1+P_2\right)}{P_1 T_2+P_2 T_1}=\frac{P_1 T_2\left(P_1+P_2\right)}{P_1 T_2+P_2 T_1}
\)
Similarly,
\(
\Rightarrow P_2^{\prime}=\frac{P_2}{T_2} T=\frac{P_2 T_1\left(P_1+P_2\right)}{P_1 T_2+P_2 T_1}
\)
c) Final temperature will be
\(
T=\frac{T_1 T_2\left(P_1+P_2\right)}{P_1 T_2+P_2 T_1} \ldots \ldots \ldots \ldots (Eq(1))
\)
d) Heat lost by the right chamber will be
\(
\begin{aligned}
& \mathrm{n}_2 \mathrm{C}_{\mathrm{v}} \mathrm{T}_2-\mathrm{n}_2 \mathrm{C}_{\mathrm{v}} \mathrm{T} \\
& =\frac{P_2 V}{2 R T_2} 1.5 R T_2-\frac{P_2 V}{2 R T_2} 1.5 R \frac{T_1 T_2\left(P_1+P_2\right)}{P_1 T_2+P_2 T_1} \\
& =\frac{3 P_2 V}{4}-\frac{3 P_2 V}{4} \frac{T_1\left(P_1+P_2\right)}{P_1 T_2+P_2 T_1} \\
& =\frac{3 P_2 V}{4}\left[1-\frac{T_1\left(P_1+P_2\right)}{P_1 T_2+P_2 T_1}\right] \\
& =\frac{3 P_2 V}{4}\left[\frac{P_1 T_2+P_2 T_1-T_1\left(P_1+P_2\right)}{P_1 T_2+P_2 T_1}\right] \\
& =\frac{3 P_1 P_2 V}{4}\left[\frac{T_2-T_1}{P_1 T_2+P_2 T_1}\right]
\end{aligned}
\)

Example 4: An adiabatic vessel of total volume \(V\) is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas ( \(U=1.5 \mathrm{nRT})\) and the part on the right contains two moles of the same gas. Initially, the pressure on each side is \(p\). The system is left for sufficient time so that a steady state is reached. Find (a) the work done by the gas in the left part during the process, (b) the temperature on the two sides in the beginning, (c) the final common temperature reached by the gases, (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part.

Solution:

(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 the left side
Let the initial pressure on both sides of the wall be \(p\).
We know,
Volume \(=\frac{V}{2}\)
Number of moles, \(n=1\)
Let the initial temperature be \(T_1\).
Using the ideal gas equation, we get
\(
\begin{aligned}
& \frac{P V}{2}=n R T_1 \\
& \Rightarrow \frac{P V}{2}=(1) R T_1 \\
& \Rightarrow T_1=\frac{P V}{(2 \text { moles }) R}
\end{aligned}
\)
For right side
Number of moles, \(n=2\)
Let the initial temperature be \(T_2\).
We know,
\(
\begin{aligned}
& \text { Volume }=\frac{V}{2} \\
& \frac{P V}{2}=n R T_2 \\
& \Rightarrow T_2=\frac{P V}{(4 \text { moles }) R}
\end{aligned}
\)
(c)
Here,
\(
U=1.5 n R T
\)
\(\mathrm{T}=\) temperature at the equilibrium
\(
P_1=P_2=P
\)
\(
\mathrm{n}_1=1 \mathrm{~mol}
\)
\(
\mathrm{n}_2=2 \mathrm{~mol}
\)
Let \(T_1\) and \(T_2\) be the initial temperatures of the left and right chamber respectively.
Applying eqn. of state
For the left side chamber
\(
\begin{aligned}
& P \frac{V}{2}=n_1 R T_1 \\
& \Rightarrow P \frac{V}{2}=R T_1 \\
& \Rightarrow P V=2 R T_1 \\
& \Rightarrow T_1=\frac{P V}{2 R}
\end{aligned}
\)
Right side chamber
\(
\begin{aligned}
& P \frac{V}{2}=n_2 R T_2 \\
& \Rightarrow P \frac{V}{2}=2 R T_2 \\
& \Rightarrow T_2=\frac{P V}{4 R}
\end{aligned}
\)
We know that total \(n=n_1+n_2=3\)
\(
\begin{aligned}
& \mathrm{U}_1=\mathrm{n}_1 C_{\mathrm{v}} \mathrm{T}_1=\mathrm{C}_{\mathrm{v}} \mathrm{T}_1 \\
& \mathrm{U}_2=\mathrm{n}_2 \mathrm{C}_{\mathrm{v}} \mathrm{T}_2=2 \mathrm{C}_{\mathrm{v}} \mathrm{T}_2 \\
& \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \mathrm{T} \\
& \mathrm{U}=\mathrm{U}_1+\mathrm{U}_2 \\
& \Rightarrow 3 C_{\mathrm{v}} T=C_{\mathrm{v}} T_1+2 C_{\mathrm{v}} T_2 \\
& \Rightarrow 3 T=T_1+2 T_2 \\
& \Rightarrow 3 T=\frac{P V}{2 R}+\frac{2 P V}{4 R} \\
& \Rightarrow T=\frac{P V}{(3 \text { moles }) R}
\end{aligned}
\)

Example 5: Explain why
(a) Two bodies at different temperatures \(T_1\) and \(T_2\) if brought in thermal contact do not necessarily settle to the mean temperature \(\left(T_1+T_2\right) / 2\).
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Solution:

a) When two bodies at different temperatures \(T_1\) and \(T_2\) are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature \(\left(T_1+T_2\right) / 2\) only when the thermal capacities of both the bodies are equal.
b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.
c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, the temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.
d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Example 6: Two cylinders \(A\) and \(B\) of equal capacity are connected to each other via a stopcock. \(A\) contains a gas at standard temperature and pressure. \(B\) is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :
(a) What is the final pressure of the gas in \(A\) and \(B\)?
(b) What is the change in the internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

Solution:

(a) Since the final temperature and initial temperature remain the same,
\(
\therefore \mathrm{P}_2 \mathrm{~V}_2=\mathrm{P}_1 \mathrm{~V}_1
\)
But \(P_1=1 \mathrm{~atm}, \mathrm{V}_1=\mathrm{V}, \mathrm{V}_2=2 \mathrm{~V}\) and \(\mathrm{P}_2=?\)
\(
\begin{aligned}
& \therefore P_2=\frac{P_1 V_1}{V_2}=\frac{1 \times V}{2 V} \\
& =0.5 \mathrm{~atm}
\end{aligned}
\)

(b) Since the temperature of the system remains unchanged, the change in internal energy is zero. 

(c) The system being thermally insulated, there is no change in temperature (because of free expansion). Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.

(d) The expansion is a free expansion. Therefore, the intermediate states are non-equilibrium states and the gas equation is not satisfied in these states. As a result, the gas cannot return to an equilibrium state which lies on the P-V-T surface.

Another solution for (d): No. The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non-equilibrium states, they do not lie on the P-V-T surface of the system.

Example 7: The initial state of a certain gas is \(\left(P_{\mathrm{i}}, V_{\mathrm{i}}, T_{\mathrm{i}}\right)\). It undergoes expansion till its volume becomes \(V_{\mathrm{f}}\). Consider the following two cases:
(a) the expansion takes place at a constant temperature.
(b) the expansion takes place at constant pressure.
Plot the \(P\) – \(V\) diagram for each case. In which of the two cases, is the work done by the gas more?

Solution:

The situation is shown in the given \(\mathrm{P}-\mathrm{V}\) graph, where variation is shown for each process. It is clear from the graph that Process 1 is isobaric and Process 2 is isothermal.
Since work done is equal to the area under the \(\mathrm{P}-\mathrm{V}\) curve. Here, area under the \(\mathrm{P}-\mathrm{V}\) curve 1 is more. So, work done is more when the gas expands in the isobaric process as in comparison of gas expands in isothermal.

Example 8: Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in the figure below.
(a) Find the work done when the gas is taken from state 1 to state 2.
(b) What is the ratio of temperature \(T_1 / T_2\), if \(V_2=2 V_1\)?
(c) Given the internal energy for one mole of gas at a temperature \(T\) is \((3 / 2) R T\), find the heat supplied to the gas when it is taken from state 1 to 2, with \(V_2=2 V_1\).

Solution:

Let \(P V^{1 / 2}=K=\) constant, \(P=\frac{K}{\sqrt{V}}\) Hence we can write, \(P_1 V_1^{1 / 2}=P_2 V_2^{1 / 2}=K\)
(a) Work done for the process 1 to 2 ,
\(
\begin{aligned}
W_{1 \rightarrow 2} & =\int_{V_1}^{V_2} P d V=\int_{V_1}^{V_2} \frac{K}{\sqrt{V}} d V \\
& =K\left[\frac{\sqrt{V}}{1 / 2}\right]_{V_1}^{V_2}=2 K\left(\sqrt{V_2}-\sqrt{V_1}\right) \\
& =2 P_1 V_1^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right)=2 P_2 V_2^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right)
\end{aligned}
\)
(b) From the ideal gas equation,
\(
\begin{gathered}
\quad P V=n R T \Rightarrow T=\frac{P V}{n R}=\frac{P \sqrt{V} \sqrt{V}}{n R} \\
\Rightarrow \quad T=\frac{K \sqrt{V}}{n R} \quad(\text { As, } P \sqrt{V}=K) \\
\text { i.e., } T \propto \sqrt{V} \\
\text { Thus, } \frac{T_2}{T_1}=\sqrt{\frac{V^2}{V_1}}=\sqrt{\frac{2 V_1}{V_1}}=\sqrt{2} \quad \text { (as } V_2=2 V_1 \text { ) }
\end{gathered}
\)
(c) Given, the internal energy of the gas, \(U=\left(\frac{3}{2}\right) R T\)
\(
\begin{aligned}
\Delta U & =U_2-U_1=\frac{3}{2} R\left(\mathrm{~T}_2-\mathrm{T}_1\right) \\
& =\frac{3}{2} R\left(\sqrt{2} T_1-T_1\right)=\frac{3}{2} R T_1(\sqrt{2}-1) \quad\left[\because T_2=\sqrt{2} T_1 \text { from (b) }\right] \\
\Delta W & =W_{1 \rightarrow 2}=2 P_1 V_1^{1 / 2}\left(V_2^{1 / 2}-V_1^{1 / 2}\right) \\
& =2 P_1 V_1\left(\frac{V_1^{1 / 2}}{V_1^{1 / 2}}-1\right)=2 R T_1(\sqrt{2}-1)
\end{aligned}
\)
\(
\left(\text { as } P_1 V_1=R T_1 \text { and } \frac{V_2^{1 / 2}}{V_1^{1 / 2}}=\sqrt{2}\right)
\)
\(
\begin{aligned}
\Delta Q=\Delta U+\Delta W & =\frac{3}{2} R T_1(\sqrt{2}-1)+2 R T_1(\sqrt{2}-1) \\
& =\frac{7}{2} R T_1(\sqrt{2}-1)
\end{aligned}
\)
This is the amount of heat supplied.

Example 9: A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in the figure below.
A to \(B\) : volume constant
\(\mathrm{B}\) to \(\mathrm{C}\) : adiabatic
C to D : volume constant
D to A: adiabatic
\(V_C=V_D=2 V_A=2 V_B\)
(a) In which part of the cycle heat is supplied to the engine from outside?
(b) In which part of the cycle heat is being given to the surrounding by the engine?
(c) What is the work done by the engine in one cycle? Write your answer in terms of \(P_A, P_B, V_A\).
(d) What is the efficiency of the engine?
\([\gamma=5 / 3 \text { for the gas], }\) \((C_v=\frac{3}{2} R. \text { for one mole) }\)

Solution:

(a) For the process \(A B\) (which is an isochoric process), volume is constant. So,
\(
\begin{aligned}
& d V=0 \Rightarrow d W=0 \\
& d Q=d U+d W=d U \\
& \Rightarrow d Q=d U=\text { Change in internal energy }
\end{aligned}
\)
Hence, in this process heat supplied is utilized to increase, the internal energy of the system.
(b) For the process \(C D\) (which is also an isochoric process), volume is constant but the pressure decreases. Hence, temperature also decreases (because \(\mathrm{Pa} \mathrm{T}\) ) so heat is given to the surroundings.
(c) To calculate work done by the engine in one cycle, we calculate work done in each part separately.
(i) \(W_{A B}=\int_A^B P d V=0\) and (ii) \(W_{C D}=\int_C^D P d V=0\) (as \(V\) is constant, \(d V=0\) )
(iii) \(\begin{aligned} W_{B C} & =\int_B^C P d V=K \int_B^C \frac{d V}{V^\gamma}=K \int_{V_B}^{V_C} V^{-\gamma} d V \\ & =\frac{K}{1-\gamma}\left[V^{1-\gamma}\right]_{V_B}^{V_C}=\frac{K\left[V_C^{1-\gamma}-V_B\right]}{1-\gamma}\end{aligned}\)
\(\left(P V^\gamma=K\right.\) for an adiabatic change \()\)
\(
=\frac{\left[\left(P_C V_C^\gamma\right)\left(V_C^{1-\gamma}\right)-\left(P_B V_B^\gamma\right)\left(V_B^{1-\gamma}\right)\right]}{(1-\gamma)}
\)
Similarly, \(W_{D A}=\int_{V_D}^{V_A} P d V=\frac{1}{(1-\gamma)}\left(P_A V_A-P_D V_D\right)\)
\([\because B C\) is adiabatic process]
Since \(B\) and \(C\) lies on adiabatic curve \(B C\),
\(
\begin{aligned}
& \therefore \quad P_B V_B^\gamma=P_C V_C^\gamma \\
& P_C=P_B\left(\frac{V_B}{V_C}\right)^\gamma=P_B\left(\frac{1}{2}\right)^\gamma=2^{-\gamma} P_B \\
& \text { Similarly, } P_D=2^{-\gamma} P_A
\end{aligned}
\)
Total work done by the engine in one cycle \(A B C D A\)
\(
\begin{aligned}
W & =W_{A B}+W_{B C}+W_{C D}+W_{D A}=W_{B C}+W_{D A} \\
& =\frac{\left(P_C V_C-P_B V_B\right)}{1-\gamma}+\frac{\left(P_A V_A-P_D V_D\right)}{1-\gamma} \\
W & =\frac{1}{1-\gamma}\left[2^{-\gamma} P_B\left(2 V_B\right)-P_B V_B+P_A V_A-2^{-\gamma} P_B\left(2 V_B\right)\right] \\
& =\frac{1}{1-\gamma}\left[P_B V_B\left(2^{-\gamma+1}-1\right)-P_A V_A\left(2^{-\gamma+1}-1\right)\right. \\
& =\frac{1}{1-\gamma}\left(2^{1-\gamma}-1\right)\left(P_B-P_A\right) V_A \\
& =\frac{3}{2}\left[1-\left(\frac{1}{2}\right)^{2 / 3}\right]\left(P_B-P_A\right) V_A
\end{aligned}
\)
(d) Heat \((Q)\) is supplied to the engine only during \(A\) to \(B\). Thus
\(
\begin{aligned}
& Q=\Delta Q_{A B}=\Delta U_{A B}=n C_v \Delta T \\
& \quad=\frac{3}{2} R\left(T_B-T_A\right) \quad\left[\text { as } C_V=(3 / 2) R \text { and } \Delta T=T_B-T_A\right] \\
& P_C V_C^\gamma=P_B V_B^\gamma, P_C=P_B\left(V_B / V_C\right)^\gamma=P_B\left(\frac{1}{2}\right)^\gamma=P_B 2^{-\gamma}\left(\text { as } V_B / V_C=1 / 2\right)
\end{aligned}
\)
Thus, \(P_C V_C^\gamma=\left(P_B 2^{-\gamma}\right)\left(2 V_B\right)=2^{1-\gamma} P_B V_B\)
\(
\begin{aligned}
\quad & \frac{3}{2}\left(P_B V_B-P_A V_A\right)=\frac{3}{2}\left(P_B-P_A\right) V_A \\
(P V= & \left.R T \text { and } V_B=V_A\right)
\end{aligned}
\)
The efficiency of the engine,
\(
\begin{aligned}
\eta & =\frac{W}{Q}=\frac{\frac{3}{2}\left[1-\left(\frac{1}{2}\right)^{2 / 3}\right]\left(P_B-P_A\right) V_A}{\frac{3}{2}\left(P_B-P_A\right) V_A} \\
& =\left[1-(1 / 2)^{2 / 3}\right]
\end{aligned}
\)

Example 10: A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in the figure below. Find heat exchanged by the engine, with the surroundings for each section of the cycle. \(\left(C_v=(3 / 2) R\right)\)
\(\mathrm{AB}:\) constant volume
\(\mathrm{BC}:\) constant pressure
\(\mathrm{CD}\) : adiabatic
\(\mathrm{DA}\) : constant pressure

Solution:

(a) By using first law of thermodynamics, we can find amount of heat associated with each process
For process \(A B\)
Volume is constant, hence work done \(\mathrm{dW}=0\)
According to first law of thermodynamics,
\(
\begin{aligned}
\Delta Q & =\Delta U+\Delta W=\Delta U+0=\Delta U \\
& =n C v \Delta T=n C v\left(T_B-T_A\right) \\
& =\frac{3}{2} R\left(T_B-T_A\right) \quad(\because n=1) \\
& =\frac{3}{2}\left(R T_B-R T_A\right)=\frac{3}{2}\left(P_B V_B-P_A V_A\right) \\
& =\frac{3}{2}\left(P_B-P_A\right) V_A \quad\left[\because V_B=V_A\right]
\end{aligned}
\)
(b) For process \(B C, P=\) constant
\(
\begin{aligned}
\Delta Q & =\Delta U+\Delta W \\
& =\frac{3}{2} R\left(T_C-T_B\right)+P_B\left(V_C-V_B\right) \\
& =\frac{3}{2}\left(P_C V_C-P_B V_B\right)+P_B\left(V_C-V_B\right) \\
& =\frac{5}{2} P_B\left(V_C-V_B\right)
\end{aligned}
\)
Heat exchanged \(=\frac{5}{2} P_B\left(V_C-V_A\right)\)
\(
\left(\because P_B=P_C \text { and } P_B=V_A\right)
\)
(c) For process \(C D, Q_{C D}=0\) (as the change is adiabatic.)
(d) In process \(D A\) involves compression of gas from \(V_D\) to \(V_A\) at constant pressure \(P_A\).
\(\therefore\) Heat exchanged can be calculated in a similar way as process \(B C\).
Hence, \(\Delta Q=\frac{5}{2} P_A\left(V_A-V_D\right)\).

Example 11: Consider that an ideal gas ( \(n\) moles) is expanding in a process given by \(P=f(V)\), which passes through a point \(\left(V_o, P_o\right)\). Show that the gas is absorbing heat at \(\left(P_o, V_o\right)\) If the slope of the curve \(P=f(V)\) is larger than the slope of the adiabat passing through \(\left(P_o, V_o\right)\).

Solution:

Slope of the graph at \(\left(V_0, P_0\right)=\left(\frac{d P}{d V}\right)_{\left(V_0, P_0\right)}\)
We know that for the adiabatic process \(P V^\gamma=K\)
\(
\begin{aligned}
& \Rightarrow P=\frac{K}{V^\gamma} \\
& \Rightarrow \frac{d P}{d V}=K(-\gamma) V^{-\gamma-1} \\
& \Rightarrow \frac{d P}{d V}=-\gamma P V^\gamma V^{-\gamma} V^{-1} \\
& \Rightarrow \frac{d P}{d V}=\frac{-\gamma P}{V} \\
& \Rightarrow\left(\frac{d P}{d V}\right)_{\left(V_0, P_0\right)}=\frac{-\gamma P_0}{V_0}
\end{aligned}
\)
Now, heat absorbed in the process \(P=f(V)\),
\(
\begin{aligned}
& \therefore d Q=d U+d W \\
& \Rightarrow d Q=n C_V d T+P d V \dots(i)
\end{aligned}
\)
We know that \(P V=n R T\). Therefore, we get
\(
\begin{aligned}
& \Rightarrow T=\frac{P V}{n R} \\
& \Rightarrow T=\frac{V}{n R} f(V) \ldots . .(\text { Given, } P=f(V)) \\
& \Rightarrow \frac{d T}{d V}=\frac{1}{n R} \ldots . .\left[f(V)+V f^{\prime}(V)\right]
\end{aligned}
\)
Now, continue solving equation (i)
\(
\begin{aligned}
& \Rightarrow \frac{d Q}{d V}=n C v \frac{d T}{d V}+P \frac{d V}{d V} \\
& \Rightarrow \frac{d Q}{d V}=\frac{n C v}{n R}\left[f(V)+V f^{\prime}(V)\right]+P \\
& \Rightarrow\left(\frac{d Q}{d V}\right)_{V=V_0}=\frac{C v}{R}\left[f\left(V_0\right)+V_0 f^{\prime}\left(V_0\right)\right]+f\left(V_0\right) \\
& \Rightarrow\left(\frac{d Q}{d V}\right)_{V=V_0}=f\left(V_0\right)\left[\frac{C v}{R}+1\right]+V_0 f^{\prime}\left(V_0\right) \frac{C v}{R}
\end{aligned}
\)
We know that \(C_P-C_V=R\)
\(
\begin{aligned}
& \Rightarrow \frac{C_P}{C_V}-1=\frac{R}{C_V} \\
& \Rightarrow \gamma-1=\frac{R}{C_V} \\
& \Rightarrow C_V=\frac{R}{\gamma-1} \\
& \Rightarrow \frac{C_V}{R}=\frac{1}{\gamma-1} \\
& \Rightarrow\left(\frac{d Q}{d V}\right)_{V=V_0}=f\left(V_0\right)\left[\frac{1}{\gamma-1}+1\right]+V_0 f^{\prime}\left(V_0\right) \frac{1}{\gamma-1} \\
& \Rightarrow\left(\frac{d Q}{d V}\right)_{V=V_0}=f\left(V_0\right)\left[\frac{1+\gamma-1}{\gamma-1}\right]+\frac{V_0 f^{\prime}\left(V_0\right)}{\gamma-1} \\
& \Rightarrow\left(\frac{d Q}{d V}\right)_{V=V_0}=\frac{\gamma}{(\gamma-1)} f\left(V_0\right)+V_0 \frac{f^{\prime}\left(V_0\right)}{\gamma-1} \\
& \Rightarrow\left(\frac{d Q}{d V}\right)_{V=V_0}=\frac{1}{(\gamma-1)}\left[\gamma f\left(V_0\right)+V_0 f^{\prime}\left(V_0\right)\right] \\
& \Rightarrow\left(\frac{d Q}{d V}\right)_{V=V_0}=\frac{1}{(\gamma-1)}\left[\gamma P_0+V_0 f^{\prime}\left(V_0\right)\right] \ldots . .\left(\because f\left(V_0\right)=P_0\right)
\end{aligned}
\)
If \(\gamma>1\), so \(\left(\frac{1}{\gamma-1}\right)\) is positive.
Then, heat is absorbed where \(\left(\frac{d Q}{d V}\right)_{V=V_0}>0\) when gas expands.
Hence, \(\gamma P_0+V_0 f^{\prime}\left(V_0\right)>0\)
\(\Rightarrow V_0 f^{\prime}\left(V_0\right)>\left(-\gamma P_0\right)\)
\(\Rightarrow f^{\prime}\left(V_0\right)>\left(\frac{-\gamma P_0}{V_0}\right)\)

Example 12: Consider one mole of a perfect gas in a cylinder of the unit cross-section with a piston attached (Fig. below2). A spring (spring constant \(k\) ) is attached (unstretched length \(L\) ) to the piston and to the bottom of the cylinder. Initially, the spring is unstretched and the gas is in equilibrium. A certain amount of heat \(Q\) is supplied to the gas causing an increase of volume from \(\mathrm{V}_{\mathrm{o}}\) to \(\mathrm{V}_1\).
(a) What is the initial pressure of the system?
(b) What is the final pressure of the system?
(c) Using the first law of thermodynamics, write down a relation between \(Q, P_a, V, V_o\) and \(k\).

Solution:

(a) Initially the piston is in equilibrium hence, \(P_f=P_a\)
(b) On supplying heat, the gas expands from \(V_0\) to \(V_1\)
\(\therefore\) Increase in volume of the gas \(=V_1-V_0\)
As the piston is of the unit cross-sectional area hence, extension in the spring
\(
x=\frac{V_1-V_0}{\text { Area }}=V_1-V_0
\)
\(\therefore\) Force exerted by the spring on the piston
\(
=F=k x=k\left(V_1-V_0\right)
\)
Hence, Final pressure \(=P_f=P_a+k x\)
\(
=P_a+k \times\left(V_1-V_0\right)
\)
(c) From the first law of thermodynamics \(d Q=d u+d W\)
If \(\mathrm{T}\) is the final temperature of the gas. then increases in internal energy \(d U=C_v\left(T-T_0\right)=C_v\left(T-T_0\right)\)
We can write, \(T=\frac{P_f V_1}{R}-\left[\frac{P_a+k\left(V_1-V_0\right)}{R}\right] \frac{V_1}{R}\)
Work done by the gas = PdV + increase in PE of the spring
\(
=P_a\left(V_1-V_0\right)+\frac{1}{2} k x^2
\)
Now, we can write \(d Q=d U+d W\)
\(
\begin{aligned}
& =C_V\left(T-T_0\right)+P_a\left(V-V_0\right)+\frac{1}{2} k x^2 \\
& =C_V\left(T-T_0\right)+P_a\left(V_{V-0}\right)+\frac{1}{2}\left(V_1-V_0\right)^2
\end{aligned}
\)
This is the required relation.