## Chapter 11 Thermodynamics

### Multiple Choice Questions (MCQs)

**1.** An ideal gas undergoes four different processes from the same initial state (figure). Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1,2, 3 and 4 which one is adiabatic?

(a) 4

(b) 3

(c) 2

(d) 1

## Show Answer

**Thinking Process**

The slope of the curve for the adiabatic process will be more that is the curve will be steeper. Slope of $p-V$ curve in adiabatic process $=\gamma(p / V)$ where as slope of is otemal process $=-p / v$

**Answer**
(c) For the curve 4 pressure is constant, so this is an isobaric process.

For the curve 1, volume is constant, so it is isochoric process. Between curves 3 and 2 , curve 2 is steeper, so it is adiabatic and 3 is isothermal.

Note We should be careful while deciding between isothermal and adiabatic curves because these curves look similar.

**2.** If an average person jogs, he produces $14.5 \times 10^{3} cal / min$. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming $1 kg$ requires $580 \times 10^{3}$ cal for evaporation) is

(a) $0.25 kg$

(b) $2.25 kg$

(c) $0.05 kg$

(d) $0.20 kg$

## Show Answer

**Answer**
(a) Amount of sweat evaporated/minute

$$ \begin{aligned} & =\frac{\text { Sweat produced / minute }}{\text { Number of calories required for evaporation } / kg} \\ & =\frac{\text { Amount of heat produced per minute in jogging }}{\text { Latent heat (in cal } / kg \text { ) }} \\ & =\frac{14.5 \times 10^{3}}{580 \times 10^{3}}=\frac{145}{580}=0.25 kg \end{aligned} $$

**3.** Consider $p-V$ diagram for an ideal gas shown in figure.

Out of the following diagrams, which figure represents the $T-p$ diagram?

(a) (iv)

(b) (ii)

(c) (iii)

(d) (i)

## Show Answer

**Answer**
(c) According to the question given that $p V=$ constant Hence, we can say that the gas is going through an isothermal process.

Clearly, from the graph that between process 1 and 2 temperature is constant and the gas expands and pressure decreases i.e., $p_2<p_1$ which corresponds to diagram (iii).

**4.** An ideal gas undergoes cyclic process $A B C D A$ as shown in given $p-V$ diagram. The amount of work done by the gas is

(a) $6 p_0 V_0$

(b) $-2 p_0 V_0$

(c) $+2 p_0 V_0$

(d) $+4 p_0 V_0$

## Show Answer

**Thinking Process**

Work done in a process by which a gas is going through can be calculated by area of the $p$ - $V$ diagram.

**Answer**
(b) Consider the $p$ - $V$ diagram given in the question.

Work done in the process $A B C D=$ area of rectangle $A B C D A$

$$ \begin{aligned} & =(A B) \times B C=(3 V_0-V_0) \times(2 p_0-p_0) \\ & =2 V_0 \times p_0=2 p_0 V_0 \end{aligned} $$

As the process is going anti-clockwise, hence there is a net compression in the gas. So, work done by the gas $=-2 p_0 V_0$.

**5.** Consider two containers $A$ and $B$ containing identical gases at the same pressure, volume and temperature. The gas in container $A$ is compressed to half of its original volume isothermally while the gas in container $B$ is compressed to half of its original value adiabatically. The ratio of final pressure of gas in $B$ to that of gas in $A$ is

(a) $2^{\gamma-1}$

(b) $(\frac{1}{2})^{\gamma-1}$

(c) $(\frac{1}{1-\gamma})^{2}$

(d) $(\frac{1}{\gamma-1})^{2}$

## Show Answer

**Answer**
(a) Consider the $p-V$ diagram shown for the container $A$ (isothermal) and for container $B$ (adiabatic).

Container $A$ (Isothermal)

Container $B$ (Adiabatic)

Both the process involving compression of the gas.

For isothermal process (gas $A$ ) (during $1 \rightarrow 2$ )

$ p_1 V_1 =p_2 V_2 $

$\Rightarrow p_0(2 V_0) =p_2(V_0) $

$\Rightarrow p_2 =2 p_0$

For adiabatic process, (gas $B$ ) (during $1 \rightarrow 2$ ) $$ \begin{array}{rlrl} p_1 V _1 ^\gamma & =p _2 V _2 ^\gamma \\ \Rightarrow & p _0\left(2 V_0\right)^\gamma =p_2\left(V_0\right)^\gamma \\ \Rightarrow & p _2 =\left(\frac{2 V _0}{V _0}\right)^\gamma p_0=(2)^\gamma p _0 \end{array} $$

Hence, $\frac{\left(p_2\right) _B}{\left(p _2\right)_A}=$ Ratio of final pressure $=\frac{(2)^\gamma p_0}{2 p _0}=2^{\gamma-1}$ where, $\gamma$ is ratio of specific heat capacities for the gas.

**6.** Three copper blocks of masses $M_1, M_2$ and $M_3 kg$ respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at $T_1, T_2, T_3(T_1>T_2>T_3)$. Assuming there is no heat loss to the surroundings, the equilibrium temperature $T$ is ( $s$ is specific heat of copper)

(a) $T=\frac{T_1+T_2+T_3}{3}$

(b) $T=\frac{M_1 T_1+M_2 T_2+M_3 T_3}{M_1+M_2+M_3}$

(c) $T=\frac{M_1 T_1+M_2 T_2+M_3 T_3}{3(M_1+M_2+M_3)}$

(d) $T=\frac{M_1 T_1 S+M_2 T_2 s+M_3 T_3 S}{M_1+M_2+M_3}$

## Show Answer

**Answer**
(b) Let the equilibrium temperature of the system is $T$.

Let us assume that $T_1, T_2<T<T_3$.

According to question, there is no net loss to the surroundings.

Heat lost by $M_3=$ Heat gained by $M_1+$ Heat gained by $M_2$

$$ \begin{matrix} \Rightarrow & M_3 s(T_3-T)=M_1 s(T-T_1)+M_2 s(T-T_2) \\ \Rightarrow & \quad \text { (where, s is specific heat of the copper material) } \\ \Rightarrow & T[M_1+M_2+M_3]=M_3 T_3+M_1 T_1+M_2 T_2 \\ \Rightarrow & T=\frac{M_1 T_1+M_2 T_2+M_3 T_3}{M_1+M_2+M_3} \end{matrix} $$

### Multiple Choice Questions (More Than One Options)

**7.** Which of the processes described below are irreversible?

(a) The increase in temperature of an iron rod by hammering it

(b) A gas in a small container at a temperature $T_1$ is brought in contact with a big reservoir at a higher temperature $T_2$ which increases the temperature of the gas

(c) A quasi-static isothermal expansion of an ideal gas in cylinder fitted with a frictionless piston

(d) An ideal gas is enclosed in a piston cylinder arrangement with adiabatic walls. A weight $w$ is added to the piston, resulting in compression of gas

## Show Answer

**Thinking Process**

If any process can be returned back such that both, the system and the surroundings return to their original states, with no other change anywhere else in the universe, then this process is called reversible process.

**Answer**
$(a, b, d)$

(a) When the rod is hammered, the external work is done on the rod which increases its temperature. 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 $T_2$.

(d) As the weight is added to the cylinder arrangement in the form of external pressure hence, it cannot be reversed back itself.

**8.** An ideal gas undergoes isothermal process from some initial state $i$ to final state $f$. Choose the correct alternatives.

(a) $d U=0$

(b) $d Q=0$

(c) $d Q=d U$

(d) $d Q=d W$

## Show Answer

**Answer**
$(a, d)$

For an isothermal process change in temperature of the system $d T=0 \Rightarrow T=$ constant.

We know that for an ideal gas $d U=$ change in internal energy $=n C_{V} d T=0$

[where, $n$ is number of moles and $C_{V}$ is specific heat capacity at constant volume] From first law of thermodynamics,

$$ \begin{aligned} d Q & =d U+d W \\ & =0+d W \Rightarrow d Q=d W \end{aligned} $$

**9.** Figure shows the $p-V$ diagram of an ideal gas undergoing a change of state from $A$ to $B$. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.

(a) Change in internal energy is same in IV and III cases, but not in I and II

(b) Change in internal energy is same in all the four cases

(c) Work done is maximum in case I

(d) Work done is minimum in case II

## Show Answer

**Thinking Process**

Internal energy is a state function and work done by the gas is a path dependent function. The work done in a thermodynamical process is equal to the area bounded between $p$ - $V$ curve.

**Answer**
(b, c)

Change in internal energy for the process $A$ to $B$

$$ d U_{A arrow B}=n C_{V} d T=n C_{V}(T_{B}-T_{A}) $$

which depends only on temperatures at $A$ and $B$.

Work done for $A$ to $B, d W_{A arrow B}=$ Area under the $p-V$ curve which is maximum for the path $I$.

**10.** Consider a cycle followed by an engine (figure.) 1 to 2 is isothermal

2 to 3 is adiabatic

3 to 1 is adiabatic

Such a process does not exist, because

(a) heat is completely converted to mechanical energy in

such a process, which is not possible

(b) mechanical energy is completely converted to heat in this process, which is not possible

(c) curves representing two adiabatic processes don’t intersect

(d) curves representing an adiabatic process and an isothermal process don’t intersect

## Show Answer

**Answer**
$(a, c)$

(a) The given process is a cyclic process i.e., it returns to the original state 1 .

Hence, change in internal energy $d U=0$

$\Rightarrow \quad d Q=d U+d W=0+d W=d W$

Hence, total heat supplied is converted to work done by the gas (mechanical energy) which is not possible by second law of thermodynamics.

(c) When the gas expands adiabatically from 2 to 3 . It is not possible to return to the same state without being heat supplied, hence the process 3 to 1 cannot be adiabatic.

**11.** Consider a heat engine as shown in figure. $Q_1$ and $Q_2$ are heat added both to $T_1$ and heat taken from $T_2$ in one cycle of engine. $W$ is the mechanical work done on the engine.

If $W>0$, then possibilities are

(a) $Q_1>Q_2>0$

(b) $Q_2>Q_1>0$

(c) $Q_2<Q_1<0$

(d) $Q_1<0, Q_2>0$

## Show Answer

**Answer**
(a, c)

Consider the figure we can write

$$Q_1=W+Q_2$$

$$\Rightarrow W=Q _1-Q _2>0 \text { (By question) }$$

$$\Rightarrow Q _1>Q _2>0 (\text { If both } Q _1 \text { and } Q _2 \text { are positive }) $$

$$\text { We can also, write } Q _2<Q _1<0 (\text { If both } Q _1 \text { and } Q _2 \text { are negative }). $$

### Very Short Answer Type Questions

**12.** Can a system be heated and its temperature remains constant?

## Show Answer

**Answer**Yes, this is possible when the entire heat supplied to the system is utilised in expansion. i.e., its working against the surroundings.

**13.** A system goes from $P$ to $Q$ by two different paths in the $p-V$ diagram as shown in figure. Heat given to the system in path 1 is $1000 J$. The work done by the system along path 1 is more than path 2 by $100 J$. What is the heat exchanged by the system in path 2 ?

## Show Answer

**Thinking Process**

We have to apply first law of thermodynamics for each path.

**Answer**
For path 1, Heat given $Q_1=+1000 J$ Work done $=W_1$ (let)

For path 2,

$$ \begin{aligned} \text { Work done }(W_2) & =(W_1-100) J \\ \text { Heat given } Q_2 & =? \end{aligned} $$

As change in internal energy between two states for different path is same.

$$ \begin{matrix} \therefore & \Delta U & =Q_1-W_1=Q_2-W_2 \\ & \Rightarrow & 1000-W_1 & =Q_2-(W_1-100) \\ & Q_2 & =1000-100=900 J \end{matrix} $$

**14.** If a refrigerator’s door is kept open, will the room become cool or hot? Explain.

## Show Answer

**Answer**If a refrigerator’s door is kept open, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.

**15.** Is it possible to increase the temperature of a gas without adding heat to it? Explain.

## Show Answer

**Answer**
Yes, during adiabatic compression the temperature of a gas increases while no heat is given to it.

In adiabatic compression, $\quad d Q=0$

$\therefore$ From first law of thermodynamics, $\quad d U=d Q-d W$

$d U=-d W$

In compression work is done on the gas i.e., work done is negative.

Therefore, $\quad d U=$ Positive

Hence, internal energy of the gas increases due to which its temperature increases.

**16.** Air pressure in a car tyre increases during driving. Explain.

## Show Answer

**Answer**
During, driving, temperature of the gas increases while its volume remains constant.

So, according to Charle’s law, at constant volume $(V)$,

Pressure $(p) \propto$ Temperature $(T)$

Therefore, pressure of gas increases.

### Short Answer Type Questions

**17.** Consider a Carnot’s cycle operating between $T_1=500 K$ and $T_2=300 K$ producing $1 kJ$ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

## Show Answer

**Thinking Process**

The efficiency of a Carnot’s engine is $\eta=1-\frac{T_2}{T_1^{\prime}}$

where, $T_2$ is temperature of the sink and $T_1$ is temperature of the source.

**Answer**
Given, temperature of the source $T_1=500 K$

Temperature of the $sink T_2=300 K$

Work done per cycle $W=1 k J=1000 J$

Heat transferred to the engine per cycle $Q_1=$ ?

Efficiency of a Carnot engine $(\eta)=1-\frac{T_2}{T_1}=1-\frac{300}{500}=\frac{200}{500}=\frac{2}{5}$

$$ \begin{matrix} \text { and } & \eta=\frac{W}{Q_1} \\ \Rightarrow & Q_1=\frac{W}{\eta}=\frac{1000}{(2 / 5)}=2500 J \end{matrix} $$

**18.** A person of mass $60 kg$ wants to lose $5 kg$ by going up and down a $10 m$ high stairs. Assume he burns twice as much fat while going up than coming down. If $1 kg$ of fat is burnt on expending $7000 kcal$, how many times must he go up and down to reduce his weight by $5 kg$ ?

## Show Answer

**Thinking Process**

Potential energy(PE) of an object at height $(h)$ is $m g h$. The energy in the form offat will be utilised to increase PE of the person. Thus, the calorie consumed by the person in going up is mgh, then according to problem calorie consumed by the person in coming down is $\frac{1}{2} m g h$.

**Answer**
Given,

height of the stairs $=h=10 m$

Energy produced by burning $1 kg$ of fat $=7000 kcal$

$\therefore$ Energy produced by burning $5 kg$ of fat $=5 \times 7000=35000 kcal$

$$ =35 \times 10^{6} cal $$

Energy utilised in going up and down one time

$$ \begin{aligned} & =m g h+\frac{1}{2} m g h=\frac{3}{2} m g h \\ & =\frac{3}{2} \times 60 \times 10 \times 10 \\ & =9000 J=\frac{9000}{4.2}=\frac{3000}{1.4} cal \end{aligned} $$

$\therefore$ Number of times, the person has to go up and down the stairs

$$ \begin{aligned} & =\frac{35 \times 10^{6}}{(3000 / 1.4)}=\frac{35 \times 1.4 \times 10^{6}}{3000} \\ & =16.3 \times 10^{3} \text { times } \end{aligned} $$

**19.** Consider a cycle tyre being filled with air by a pump. Let $V$ be the volume of the tyre (fixed) and at each stroke of the pump $\Delta V(«V)$ of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from $p_1$ to $p_2$ ?

## Show Answer

**Thinking Process**

There is no exchange of heat in the process, hence this can be considered as an adiabatic process.

**Answer**
Let, volume is increased by $\Delta V$ and pressure is increased by $\Delta p$ by an stroke.

For just before and after an stroke, we can write

$$ \begin{matrix} p_1 V_1^{\gamma} & =p_2 V_2^{\gamma} & \\ \Rightarrow \quad p(V+\Delta V)^{\gamma} & =(p+\Delta p) V^{\gamma} \\ \Rightarrow \quad p V^{\gamma}(1+\frac{\Delta V}{V})^{\gamma} & =p(1+\frac{\Delta p}{p}) V^{\gamma} \\ \Rightarrow \quad p V^{\gamma}(1+\gamma \frac{\Delta V}{V}) & \approx p V^{\gamma}(1+\frac{\Delta p}{p}) \quad(\because \text { volume is fixed }) \\ \Rightarrow \quad \gamma \frac{\Delta V}{V} & =\frac{\Delta p}{p} \Rightarrow \Delta V=\frac{1}{\gamma} \frac{V}{p} \Delta p \\ d V & =\frac{1}{\gamma} \frac{V}{p} d p \end{matrix} $$

Hence, work done is increasing the pressure from $p_1$ to $p_2$

W = $\int_{p_1}^{p_2} p d V= \int_{p_1}^{p_2} p \times \frac{1}{\gamma} \frac{V}{p} d p $

= $\frac{V}{\gamma} \int _{p _ 1}^{p _ 2} d p= \frac{V}{\gamma}(p_2-p_1) $

$W = \frac{(p_2-p_1)}{\gamma} V$

**20.** In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of $1 kW$ power and heat transferred from $-3^{\circ} C$ to $27^{\circ} C$, find the heat taken out of the refrigerator per second assuming its efficiency is $50 %$ of a perfect engine.

## Show Answer

**Thinking Process**

The Carnot engine is the most efficient heat engine operating between two given temperature. This is why it is known as perfect engine. The efficiency of Carnot engine is

$$ \eta=1-\frac{T_2}{T_1} $$

**Answer**
Given, temperature of the source is $27^{\circ} C$

$$ \begin{matrix} \Rightarrow & T_1=(27+273) K=300 K \\ \text { Temperature of sink } & T_2=(-3+273) K=270 K \end{matrix} $$

Efficiency of a perfect heat engine is given by

$$ \eta=1-\frac{T_2}{T_1}=1-\frac{270}{300}=\frac{1}{10} $$

Efficiency of refrigerator is $50 %$ of a perfect engine

$\therefore \quad \eta^{\prime}=0.5 \times \eta=\frac{1}{2} \eta=\frac{1}{20}$

$\therefore$ Coefficient of performance of the refrigerator

$$ \begin{aligned} \beta & =\frac{Q_2}{W}=\frac{1-\eta^{\prime}}{\eta^{\prime}} \\ & =\frac{1-(1 / 20)}{(1 / 20)}=\frac{19 / 20}{1 / 20}=19 \\ \Rightarrow \quad Q_2 & =\beta W=19 W \\ & =19 \times(1 kW)=19 kW=19 kJ / s . \end{aligned} \quad(\because \beta=\frac{Q_2}{W}) $$

Therefore, heat is taken out of the refrigerator at a rate of $19 kJ$ per second.

**21.** If the coefficient of performance of a refrigerator is 5 and operates at the room temperature $(27^{\circ} C)$, find the temperature inside the refrigerator.

## Show Answer

**Thinking Process**

Coefficient of performance $(\beta)$ of a refrigerator is ratio of quantity of heat removed per cycle $(Q_2)$ to the amount of work done on the refrigerator.

**Answer**
Given, coefficient of performace $(\beta)=5$

$$ T_1=(27+273) K=300 K, T_2=? $$

Coefficient of performance $(\beta)=\frac{T_2}{T_1-T_2}$

$$ \begin{aligned} & & 5 & =\frac{T_2}{300-T_2} \Rightarrow 1500-5 T_2=T_2 \\ \Rightarrow & & 6 T_2 & =1500 \Rightarrow T_2=250 K \\ \Rightarrow & & T_2 & =(250-273)^{\circ} C=-23^{\circ} C \end{aligned} $$

**22.** The initial state of a certain gas is $(p_{i}, V_{i}, T_{i})$. It undergoes expansion till its volume becomes $V_{f}$. Consider the following two cases

(a) the expansion takes place at 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?

## Show Answer

**Answer**
Consider the diagram $p-V$, where variation is shown for each process.

Process 1 is isobaric and process 2 is isothermal.

Since, work done $=$ area under the $p-V$ curve. Here, area under the $p-V$ curve 1 is more. So, work done is more when the gas expands in isobaric process.

### Long Answer Type Questions

**23.** Consider a $p$ - $V$ diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in figure.

(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 temperature $T$ is $(3 / 2) R T$, find the heat supplied to the gas when it is taken from state 1 to 2 , with $V_2=2 V_1$.

## Show Answer

**Answer**
Let $p V^{1 / 2}=$ Constant $=K, p=\frac{K}{\sqrt{V}}$

(a) Work done for the process 1 to 2 ,

W = $\int_{V_1}^{V_2} p d V=K \int_{V_1}^{V_2} \frac{d V}{\sqrt{V}}=K[\frac{\sqrt{V}}{1 / 2}]_{V_1}^{V_2}=2 K(\sqrt{V_2}-\sqrt{V_1}) $

$ =2 p_1 V_1^{1 / 2}(\sqrt{V_2}-\sqrt{V_1})=2 p_2 V_2^{1 / 2}(\sqrt{V_2}-\sqrt{V_1})$

(b) From ideal gas equation,

$p V=n R T \Rightarrow T=\frac{p V}{n R}=\frac{p \sqrt{V} \sqrt{V}}{n R} $

$\text{ Hence, } T=\frac{K \sqrt{V}}{n R} (\text{ As, } p \sqrt{V}=K) $

$\Rightarrow T_1=\frac{K \sqrt{V_1}}{n R} \Rightarrow T_2=\frac{K \sqrt{V_2}}{n R} $

$\frac{T_1}{T_2}=\frac{\frac{K \sqrt{V_1}}{n R}}{\frac{K \sqrt{V_2}}{n R}}=\sqrt{\frac{V_1}{V_2}}=\sqrt{\frac{V_1}{2 V_1}}=\frac{1}{\sqrt{2}} \quad(\because V_2=2 V_1)$

(c) Given, internal energy of the gas $=U=(\frac{3}{2}) R T$

$$ \begin{aligned} \Delta U & =U_2-U_1=\frac{3}{2} R(T_2-T_1) \\ & =\frac{3}{2} R T_1(\sqrt{V}-1) \quad[\because T_2=\sqrt{2} T_1 \text { from (b) }] \\ \Delta W & =2 p_1 V_1^{1 / 2}(\sqrt{V_2}-\sqrt{V_1}) \\ & =2 p_1 V_1^{1 / 2}(\sqrt{2} \times \sqrt{V_1}-\sqrt{V_1}) \\ & =2 p_1 V_1(\sqrt{2}-1)=2 R T_1(\sqrt{2}-1) \\ & =\frac{3}{2} R T_1(\sqrt{2}-1)+2 R T_1(\sqrt{2}-1) \\ & =(\sqrt{2}-1) R T_1(2+3 / 2) \\ & =(\frac{7}{2}) R T_1(\sqrt{2}-1) \end{aligned} $$

This is the amount of heat supplied.

**24.** A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in figure.

$A$ to $B$ volume constant, $B$ to $C$ adiabatic, $C$ to $D$ volume constant and $D$ 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 term of $p_{A}, p_{B}, V_{A}$ ?

(d) What is the efficiency of the engine?

$$ (\gamma=\frac{5}{3} \text { for the gas }),(C_{V}=\frac{3}{2} R \text { for one mole }) $$

## Show Answer

**Answer**
(a) For the process $A B$,

$$ \begin{aligned} d V & =0 \Rightarrow d W=0 \quad(\because \text { volume is constant }) \\ \Rightarrow \quad & d Q=d U+d W=d U \\ d Q & =d U=\text { Change in internal energy. } \end{aligned} $$

Hence, in this process heat supplied is utilised to increase, internal energy of the system. Since, $p=(\frac{n R}{V}) T$, in isochoric process, $T \propto p$. So temperature increases with increases of pressure in process $A B$ which inturn increases internal energy of the system i.e., $d U>0$. This imply that $d Q>0$. So heat is supplied to the system in process $A B$.

(b) For the process $C D$, volume is constant but pressure decreases.

Hence, temperature also decreases so heat is given to surroundings.

(c) To calculate work done by the engine in one cycle, we calculate work done in each part separately.

Similarly, $\quad W_{D A}=\frac{p_{A} V_{A}-p_{D} V_{D}}{1-\gamma} \quad[\because B C$ is adiabatic process $]$

$\because B$ and $C$ lies on adiabatic curve $B C$.

$ \therefore \quad p_{B} V_B^{\gamma}=p_{C} V_C^{\gamma} $

$ p_{C}=p_{B}(\frac{V_{B}}{V_{C}})^{\gamma}=p_{B}(\frac{1}{2})^{\gamma}=2^{-\gamma} p_{B} $ $ p_{D}=2^{-\gamma} p_{A} $

$ W=W_{A B}+W_{B C}+W_{C D}+W_{D A}=W_{B C}+W_{D A} $

$ =\frac{(p_{C} V_{C}-p_{B} V_{B})}{1-\gamma}+\frac{(p_{A} V_{A}-p_{D} V_{D})}{1-\gamma} $

$ W=\frac{1}{1-\gamma}[2^{-\gamma} p_{B}(2 V_{B})-p_{B} V_{B}+p_{A} V_{A}-2^{-\gamma} p_{B}(2 V_{B})] $

$ =\frac{1}{1-\gamma}[p_{B} V_{B}(2^{-\gamma+1}-1)-p_{A} V_{A}(2^{-\gamma+1}-1). $

$=\frac{1}{1-\gamma}(2^{1-\gamma}-1)(p_{B}-p_{A}) V_{A} $

$ ={\frac{3}{2}}[ 1-(\frac{1}{2})^(2/3](p _ B - p _ A) V _ A$

**25.** A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle. $[C_{V}=(3 / 2) R]$

(a) $A B$ : constant volume

(b) $B C$ : constant pressure

(c) $C D$ : adiabatic

(d) $D A$ : constant pressure

## Show Answer

**Thinking Process**

Find amount of heat associated with each process by using first law of thermodynamics.

**Answer**
(a) For process $A B$,

Volume is constant, hence work done $d W=0$

Now, by first law of thermodynamics,

$$ \begin{aligned} d Q & =d U+d W=d U+0=d U \\ & =n C v d T=n C v(T_{B}-T_{A}) \\ & =\frac{3}{2} R(T_{B}-T_{A}) \\ & =\frac{3}{2}(R T_{B}-R T_{A})=\frac{3}{2}(p_{B} V_{B}-p_{A} V_{A}) \end{aligned} $$

Heat exchanged $=\frac{3}{2}(p_{B} V_{B}-p_{A} V_{A})$ (b) For process $B C$,

$$ \begin{aligned} p & =\text { constant } \\ d Q & =d U+d W=\frac{3}{2} R(T_{C}-T_{B})+p_{B}(V_{C}-V_{B}) \\ & =\frac{3}{2}(p_{C} V_{C}-p_{B} V_{B})+p_{B}(V_{C}-V_{B})=\frac{5}{2} p_{B}(V_{C}-V_{B}) \end{aligned} $$

Heat exchanged $=\frac{5}{2} p_{B}(V_{C}-V_{A})$

$(\because p_{B}=p_{C}.$ and $.p_{B}=V_{A})$

(c) For process $C D$, Because $C D$ is adiabatic, $d Q=$ Heat exchanged $=0$

(d) $D A$ involves compression of gas from $V_{D}$ to $V_{A}$ at constant pressure $p_{A}$.

$\therefore$ Heat exchanged can be calculated by similar way as $B C_1$,

Hence,

$$ d Q=\frac{5}{2} p_{A}(V_{A}-V_{D}) $$

**26.** Consider that an ideal gas ( $n$ moles) is expanding in a process given by $p=f(V)$, which passes through a point $(V_0, p_0)$. Show that the gas is absorbing heat at $(p_0, V_0)$ if the slope of the curve $p=f(V)$ is larger than the slope of the adiabatic passing through $(p_0, V_0)$.

## Show Answer

**Answer**
According to question, slope of the curve $=f(V)$, where $V$ is volume.

$\therefore \quad$ Slope of $p=f(V)$, curve at $(V_0, p_0)=f(V_0)$

Slope of adiabatic at $(V_0, p_0)=k(-\gamma) V_0^{-1-\gamma}=-\gamma p_0 / V_0$

Now heat absorbed in the process $p=f(V)$

$ d Q=d U+d W=n C_{V} d T+p d V $

$\because p V=n R T \Rightarrow T=(\frac{1}{n R}) p V $

$\Rightarrow T=(\frac{1}{n R}) V f(V) $

$\Rightarrow d T=(\frac{1}{n R})[f(V)+V f^{\prime}(V)] d V$

Now from Eq. (i)

$$ \frac{d Q}{d V}=n C_{V} \frac{d T}{d V}+p \frac{d V}{d V}=n C_{V} \frac{d T}{d V}+p $$

$$ =\frac{n C_{V}}{n R} \times[f(V)+V f^{\prime}(V)]+p $$

$$ =\frac{C_{V}}{R}[f(V)+V f^{\prime}(V)]+f(V) $$

$ \rightarrow [\frac{dQ}{dV}] _ {V=V _ 0} $=

$ \frac{C_V}{R} $ $[ f(V_0) + (V_0) f^{\prime}]$ $(V_0)+f(V_0)$

$ f(V_0)[\frac{C_{V}}{R}+1]+V_0 f^{\prime}(V_0) \frac{C_{V}}{R} $

$\because \quad C_{V} =\frac{R}{\gamma-1} \Rightarrow \frac{C_{V}}{R}=\frac{1}{\gamma-1}$

$\Rightarrow \quad[\frac{d Q}{d V}]_{V=V_0} =[\frac{1}{\gamma-1}+1] f(V_0)+\frac{V_0 f^{\prime}(V_0)}{\gamma-1} $

$ =\frac{\gamma}{\gamma-1} p_0+\frac{V_0}{\gamma-1} f^{\prime}(V_0)$

Heat is absorbed where $\frac{d Q}{d V}>0$, when gas expands

Hence, $\gamma p_0+V_0 f^{\prime}(V_0)>0$ or $f^{\prime}(V_0)>(-\gamma \frac{p_0}{V_0})$

**7.** Consider one mole of perfect gas in a cylinder of unit

cross-section with a piston attached (figure). 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 value from $V_0$ to $V_1$.

(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 $Q, p_{a}, V, V_0$ and $k$.

## Show Answer

**Thinking Process**

We will assume the piston is massless, hence, at equilibrium atmospheric pressure and inside pressure will be same.

**Answer**
(a) Initially the piston is in equilibrium hence, $p_{i}=p_{a}$

(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 unit cross-sectional area hence, extension in the spring

$$ x=\frac{V_1-V_0}{\text { Area }}=V_1-V_0 $$

$\therefore \quad$ Force exerted by the spring on the piston

Hence,

$$ =F=k x=k(V_1-V_0) $$

$$ \begin{aligned} \text { final pressure } & =p_{f}=p_{a}+k x \\ & =p_{a}+k \times(V_1-V_0) \end{aligned} $$

(c) From first law of thermodynamics $d Q=d U+d W$

If $T$ is final temperature of the gas, then increase in internal energy

$$ d U=C_{V}(T,-T_0)=C_{V}(T,-T_0) $$

We can write,

$$ T=\frac{p_{f} V_1}{R}=[\frac{p_{a}+k(V_1-V_0)}{R}] \frac{V_1}{R} $$

Work done by the gas $=p d V+$ increase in PE of the spring

$$ =p_{a}(V_1-V_0)+\frac{1}{2} k x^{2} $$

Now, we can write $d Q=d U+d W$

$$ \begin{aligned} & =C_{V}(T-T_0)+p_{a}(V-V_0)+\frac{1}{2} k x^{2} \\ & =C_{V}(T-T_0)+p_{a}(V-V_0)+\frac{1}{2}(V_1-V_0)^{2} \end{aligned} $$

This is the required relation.