Chapter 12

THERMODYNAMICS

MCQ I

12.1 An ideal gas undergoes four different processes from the same initial state (Fig. 12.1). 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

Fig. 12.1

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

(a) $0.25 \mathrm{~kg}$

(b) $2.25 \mathrm{~kg}$

(c) $0.05 \mathrm{~kg}$

(d) $0.20 \mathrm{~kg}$

Exemplar Problems-Physics

12.3 Consider $P$ - $V$ diagram for an ideal gas shown in Fig 12.2.

Fig. 12.2

Out of the following diagrams (Fig. 12.3), which represents the $T$-P diagram?

(i)

(iii)

(ii)

(iv)

Fig. 12.3

(a) (iv)

(b) (ii)

(c) (iii)

(d) (i)

12.4 An ideal gas undergoes cyclic process $A B C D A$ as shown in given $P$-V diagram (Fig. 12.4).

The amount of work done by the gas is

(a) $6 P_{o} V_{o}$

(b) $-2 P_{o} V_{o}$

(c) $+2 P_{o} V_{o}$

(d) $+4 P_{o} V_{o}$

Fig 12.4

12.5 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 $\mathrm{B}$ is compressed to half of its original value adiabatically. The ratio of final pressure of gas in $B$ to that of gas in $\mathrm{A}$ is

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

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

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

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

12.6 Three copper blocks of masses $M_{1}, M_{2}$ and $M_{3} \mathrm{~kg}$ respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at $T_{1}, T_{2}, T_{3}\left(T_{1}>T_{2}>T_{3}\right)$. Assuming there is no heat loss to the surroundings, the equilibrium temprature $T$ 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) $T=\frac{M_{1} T_{1}+M_{2} T_{2}+M_{3} T_{3}}{3\left(M_{1}+M_{2}+M_{3}\right)}$

(d) $T=\frac{M_{1} T_{1} s+M_{2} T_{2} s+M_{3} T_{3} s}{M_{1}+M_{2}+M_{3}}$

MCQ II

12.7 Which of the processes described below are irreversible?

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

(b) A gas in a small cantainer at a temprature $T_{1}$ is brought in contact with a big reservoir at a higher temprature $T_{2}$ which increases the temprature 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 $\mathrm{W}$ is added to the piston, resulting in compression of gas.

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

(a) $\mathrm{d} U=0$

(b) $\mathrm{d} Q=0$

(c) $\mathrm{d} Q=\mathrm{d} U$

(d) $\mathrm{d} Q=\mathrm{d} W$

12.9 Figure 12.5 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.

12.10 Consider a cycle followed by an engine (Fig. 12.6)

1 to 2 is isothermal

2 to 3 is adiabatic

3 to 1 is adiabatic

Such a process does not exist because

Fig. 12.5

(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. Fig. 12.6

12.11 Consider a heat engine as shown in Fig. 12.7. $Q_{1}$ and $Q_{2}$ are heat added to heat bath $T_{1}$ 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) $Q_{1}<0, Q_{2}>0$

Fig . 12.7

12.12 Can a system be heated and its temperature remains constant?

12.13 A system goes from $P$ to $Q$ by two different paths in the $P-V$ diagram as shown in Fig. 12.8. Heat given to the system in path 1 is $1000 \mathrm{~J}$. The work done by the system along path 1 is more than path 2 by $100 \mathrm{~J}$. What is the heat exchanged by the system in path 2 ? Fig.12.8

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

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

12.16 Air pressure in a car tyre increases during driving. Explain.

SA

12.17 Consider a Carnot’s cycle operating between $T_{1}=500 \mathrm{~K}$ and $T_{2}=300 \mathrm{~K}$ producing $1 \mathrm{~kJ}$ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.

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

12.19 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 $\Delta V(\quad V)$ 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 $P_{2}$ ?

12.20 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 $1 \mathrm{~kW}$ power, and heat is transferred from $-3^{\circ} \mathrm{C}$ to $27^{\circ} \mathrm{C}$, find the heat taken out of the refrigerator per second assuming its efficiency is $50 %$ of a perfect engine.

12.21 If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature $\left(27^{\circ} \mathrm{C}\right)$, find the temperature inside the refrigerator.

12.22 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 becoms $V_{\mathrm{f}}$. 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?

LA

12.3

Fig. 12.9

Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in Fig. 12.9.

(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 $V_{2}=2 V_{1}$.

12.24 A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in Fig. 12.10.

Fig. 12.10 A to B : volume constant

$\mathrm{B}$ to $\mathrm{C}$ : adiabatic

C to D : volume constant

$\mathrm{D}$ to $\mathrm{A}$ : 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?

[ $\gamma=5 / 3$ for the gas], $\left(C_{v}=\frac{3}{2} R\right).$ for one mole

12.25 A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. 12.11. 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

DA : constant pressure

Fig. 12.11

12.26 Consider that an ideal gas ( $n$ moles) is expanding in a process given by $P=f(V)$, which passes through a point $\left(V_{0}, P_{0}\right)$. Show that the gas is absorbing heat at $\left(P_{o}, V_{0}\right)$ if the slope of the curve $P=f(V)$ is larger than the slope of the adiabat passing through $\left(P_{0}, V_{o}\right)$.

12.27 Consider one mole of perfect gas in a cylinder of unit cross section with a piston attached (Fig. 12.12). 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 $\mathrm{V}{0}$ 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 $k$. Atomospheric pressure $=\mathrm{P}_{\mathrm{a}}$

Fig. 12.12

Solutions 12

12.1 (c) adiabatic

A is isobaric process, D is isochoric. Of B and C, B has the smaller slope (magnitude) hence is isothermal. Remaing process is adiabatic.

12.2 (a)

12.3 (c)

12.4 (b)

12.5 (a)

12.6 (b)

12.7 (a), (b) and (d).

12.8 (a), (d)

12.9 (b), (c)

12.10 (a), (c)

12.11 (a), (c)

12.12 If the system does work against the surroundings so that it compensates for the heat supplied, the temperature can remain constant.

12.13 $U_{p}-U_{Q}=$ W.D. in path 1 on the system $+1000 \mathrm{~J}$

$=$ W.D. in path 2 on the system $+\mathrm{Q}$

$$ Q=(-100+1000) \mathrm{J}=900 \mathrm{~J} $$

12.14 Here heat removed is less than the heat supplied and hence the room, including the refrigerator (which is not insulated from the room) becomes hotter.

12.15 Yes. When the gas undergoes adiabatic compression, its temperature increases.

$\mathrm{d} Q=\mathrm{d} U+\mathrm{d} W$

As $\mathrm{d} Q=0$ (adiabatic process)

so $\mathrm{d} U=-\mathrm{d} W$

In compression, work is done on the system So, $\mathrm{dW}=-\mathrm{ve}$

$\Rightarrow \mathrm{d} U=+\mathrm{ve}$

So internal energy of the gas increases, i.e. its temperature increases.

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

So according to Charle’s law, at constant V, P $\alpha T$.

Therefore, pressure of gas increases.

12.17 $\frac{Q}{Q_{1}}=\frac{T_{2}}{T_{1}}=\frac{3}{5}, \quad Q_{1}-Q_{2}=10^{3} \mathrm{~J}$

$Q_{1}\left(1-\frac{3}{5}\right)=10^{3} \mathrm{~J} \Rightarrow Q_{1}=\frac{5}{2} \times 10^{2} \mathrm{~J}=2500 \mathrm{~J}, Q_{2}=1500 \mathrm{~J}$

12.18 $5 \times 7000 \times 10^{3} \times 4.2 \mathrm{~J}=60 \times 15 \times 10 \times \mathrm{N}$

$$ N=\frac{21 \times 7 \times 10^{6}}{900}=\frac{147}{9} \times 10^{3}=16.3 \times 10^{3} \text { times. } $$

2.19 $P\left[1+\gamma \frac{\Delta v}{V}\right]=P\left(1+\frac{\Delta p}{P}\right)$

$\gamma \frac{\Delta v}{V}=\frac{\Delta p}{P} ; \frac{d v}{d p}=\frac{V}{\gamma p}$

W.D. $=\int_{P_{1}}^{P_{2}} p d v=\int_{P_{1}}^{P_{2}} p \frac{V}{\gamma p} d p=\frac{\left(P_{2}-P_{1}\right)}{\gamma} V$

12.20

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

Efficiency of refrigerator $=0.5 \eta=\frac{1}{20}$

If $Q$ is the heat/s transferred at higher temperture then $\frac{W}{Q}=\frac{1}{20}$

or $\mathrm{Q}=20 \mathrm{~W}=20 \mathrm{~kJ}$,

and heat removed from lower temperture $=19 \mathrm{~kJ}$.

12.21 $\frac{Q_{2}}{W}=5, Q_{2}=5 \mathrm{~W}, Q_{1}=6 \mathrm{~W}$

$$ \frac{T_{2}}{T_{1}}=\frac{5}{6}=\frac{T}{300}, \quad T_{2}=250 \mathrm{~K}=-23 \mathrm{C} $$

12.22 The $P$ - $V$ digram for each case is shown in the figure. In case (i) $P_{\mathrm{i}} V_{\mathrm{i}}=P_{\mathrm{f}} V_{\mathrm{f}}$; therfore process is isothermal. Work done $=$ area under the $\mathrm{PV}$ curve so work done is more when the gas expands at constant pressure.

12.23 (a) Work done by the gas (Let $\mathrm{PV}^{1 / 2}=A$ )

$ \Delta W =\int_{V1}^{V2} p d v=A \int_{V1}^{V2} \frac{d V}{\sqrt{V}}=A\left[\frac{\sqrt{V}}{1 / 2}\right]_{V1}^{V2}=2 A\left(\sqrt{V2}-\sqrt{V1}\right) $

$ =2 P_{1} V_{1}^{1 / 2}\left[V_{2}^{1 / 2}-V_{1}^{1 / 2}\right] $

(b) Since $T=p V / n R=\frac{A}{n R} \cdot \sqrt{V}$

Thus, $\frac{T_{2}}{T_{1}}=\sqrt{\frac{V_{2}}{V_{1}}}=\sqrt{2}$

(c) Then, the change in internal energy

$$ \Delta U=U_{2}-W_{1}=\frac{3}{2} R\left(T_{2}-T_{1}\right)=\frac{3}{2} R T_{1}(\sqrt{2}-1) $$

$$ \begin{aligned} & \Delta W=2 A \sqrt{V_{1}}(\sqrt{2}-1)=2 R T_{1}(\sqrt{2}-1) \ & \Delta Q=(7 / 2) R T_{1}(\sqrt{2}-1) \end{aligned} $$

12.24 (a) A to B

(b) $\mathrm{C}$ to $\mathrm{D}$

(c) $\quad W_{AB}=\int_A^B p d V=0 ; W_{C D}=0$.

Similarly. $W_{BC}=\left[\int_B^C p d V=k \int_B^C \frac{d V}{V^r}=k \frac{V^{-r+1}}{-R+1}\right]_{VB}^{VC}$

$$ =\frac{1}{1-\gamma}\left(P_c V_c-P_B V_B\right) $$

Similarly, $W_{D A}=\frac{1}{1-\gamma}\left(P_{A} V_{A}-P_{D} V_{D}\right)$

Now $P_{C}=P_{B}\left(\frac{V_{B}}{V_{C}}\right)^{\gamma}=2^{-\gamma} P_{B}$

Similarly, $P_D=P_A 2^{-\gamma}$

Totat work done $=W_{BC}+W_{DA}$

$$ \begin{aligned} & =\frac{1}{1-\gamma}\left[P_{B} V_{B}\left(2^{-\gamma+1}-1\right)-P_{A} V_{A}\left(2^{-\gamma+1}-1\right)\right] \\ & =\frac{1}{1-\gamma}\left(2^{1-\gamma}-1\right)\left(P_{B}-P_{A}\right) V_{A} \\ & =\frac{3}{2}\left(1-\left(\frac{1}{2}\right)^{2 / 3}\right)\left(P_{B}-P_{A}\right) V_{A} \end{aligned} $$

(d) Heat supplied during process A, B

$$ \mathrm{d} Q_{A B}=\mathrm{d} U_{A B} $$

$$ Q_{A B}=\frac{3}{2} n R\left(T_{B}-T_{A}\right)=\frac{3}{2}\left(P_{B}-P_{A}\right) V_{A} $$

Efficiency $=\frac{\text { Net Work done }}{\text { Heat Supplied }}=\left[1-\left(\frac{1}{2}\right)^{2 / 3}\right]$

12.25

$Q_{A B}=U_{A B}=\frac{3}{2} R\left(T_{B}-T_{A}\right)=\frac{3}{2} V_{A}\left(P_{B}-P_{A}\right)$

$\Theta_{B C}=U_{B C}+W_{B C}$

$$ \begin{aligned} & =(3 / 2) P_{B}\left(V_{C}-V_{B}\right)+P_{B}\left(V_{C}-V_{B}\right) \ & =(5 / 2) P_{B}\left(V_{C}-V_{A}\right) \end{aligned} $$

$\mathrm{Q}_{\mathrm{CA}}=0$

$Q_{D A}=(5 / 2) P_{A}\left(V_{A}-V_{D}\right)$

12.26 Slope of $P=f(V)$, curve at $\left(V_{0}, P_{0}\right)$

$$ =f\left(V_{\mathrm{o}}\right) $$

Slope of adiabat at $\left(V_{0}, P_{0}\right)$

$$ =k(-\gamma) V_o^{-1-\gamma}=-\gamma P_o / V_o $$

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

$$ \begin{aligned} \mathrm{d} Q & =\mathrm{d} V+\mathrm{d} W \ & =n C_{v} \mathrm{~d} T+P \mathrm{dV} \end{aligned} $$

Since $T=(1 / n R) P V=(1 / n R) V f(V)$

$$ \mathrm{d} T=(1 / n R)\left[f(V)+V f^{\prime}(\mathrm{V})\right] \mathrm{d} V $$

Thus

$ \frac{d Q}{d V^{{v=V_{o}}}}=\frac{C V}{R}\left[f\left(V_{o}\right)+V_{o} f^{\prime}\left(V_{o}\right)\right]+f\left(V_{o}\right) $

$ =\left[\frac{1}{\gamma-1}+1\right] f\left(V_{o}\right)+\frac{V_{o} f^{\prime}\left(V_{o}\right)}{\gamma-1} \\ =\frac{\gamma}{\gamma-1} P_{o}+\frac{V_{o}}{\gamma-1} f^{\prime}\left(V_{o}\right) $

Heat is absorbed when $\mathrm{dQ} / \mathrm{d} V>0$ when gas expands, that is when

$ \gamma P_{o}+V_o f^{\prime}\left(V_o\right)>0 $

$ f^{\prime}\left(V_o\right)>-\gamma P_o / V_o $

12.27 (a) $P_{i}=P_{a}$

(b) $\quad P_{f}=P_{a}+\frac{k}{A}\left(V-V_{o}\right)=P_{a}+k\left(V-V_{o}\right)$

(c) All the supplied heat is converted to mechanical energy. No change in internal energy (Perfect gas)

$$ \Delta Q=P_{a}\left(V-V_{o}\right)+\frac{1}{2} k\left(V-V_{o}\right)^{2}+C_{V}\left(T-T_{o}\right) $$

where $T_{o}=P_{a} V_{o} / R$,

$$ T=\left[P_{a}+(R / A)-\left(V-V_{o}\right)\right] V / R $$