NCERT Exemplar Solutions Class 11 Physics Thermodynamics — Very Short, Short and Long Answer Questions Q12–27

NCERT Exemplar Problems, Class 11 Physics — Thermodynamics, Very Short Answer Questions 12 to 16, Short Answer Questions 17 to 22 and Long Answer Questions 23 to 27. All 16 questions solved. The book numbers them 12.12 to 12.27.

Three ideas carry this page. The first law is bookkeeping: $\Delta Q = \Delta U + W$, with $\Delta U$ depending only on the end states and $W$ on the path. Each process has its own work: zero at constant volume, $P\Delta V$ at constant pressure, $nRT\ln(V_2/V_1)$ isothermally, $(P_1V_1 – P_2V_2)/(\gamma-1)$ adiabatically. The second law sets ceilings: no engine beats $1 – T_2/T_1$ and no refrigerator beats $T_2/(T_1 – T_2)$.

Question 12

Can a system be heated and its temperature remain constant?

Solution. Yes. By the first law $\Delta Q = \Delta U + W$. If the system does work on its surroundings equal to the heat supplied, $\Delta U = 0$ and, for an ideal gas, the temperature stays the same. That is exactly an isothermal expansion. A second example is a change of state: ice melting at $0\,^\circ$C absorbs latent heat with no change of temperature.

Question 13

A system goes from P to Q by two different paths in the $P$-$V$ diagram as shown in Fig. 12.8 (path 1 is the upper arc and path 2 the lower). 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?

Solution. $\Delta U = U_Q – U_P$ is the same on both paths because P and Q are fixed states. Write the first law for each:

$$\Delta U = Q_1 – W_1 = Q_2 – W_2.$$

So $Q_2 = Q_1 – (W_1 – W_2) = 1000 – 100 = 900$ J. Path 2 lies lower, does less work, and so needs less heat to reach the same final state.

Question 14

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

Solution. The room gets hotter. A refrigerator pumps heat $Q_2$ out of its cold space and dumps $Q_1 = Q_2 + W$ at the back, where $W$ is the electrical energy used by the compressor. With the door open, the cold space is part of the room, so the room as a whole loses $Q_2$ and gains $Q_2 + W$. The net effect is that the room gains $W$, the work of the motor, every cycle.

Question 15

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

Solution. Yes, by compressing it adiabatically. With $dQ = 0$ the first law gives $dU = -dW$. In a compression the gas has work done on it, so $dW$ (work by the gas) is negative and $dU$ is positive. For an ideal gas a rise in internal energy is a rise in temperature. A bicycle pump that warms up as you use it is an everyday example.

Question 16

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

Solution. As the tyre rolls, it flexes continuously and rubs against the road, and that work is converted into heat in the rubber and the air inside. The volume of the tyre hardly changes, so the air is heated at nearly constant volume. For a fixed mass of gas at constant volume, $P \propto T$ (Gay-Lussac’s law), so the pressure rises with the temperature.

Question 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.

Solution. For a Carnot engine $Q_2/Q_1 = T_2/T_1 = 3/5$, and the work is $W = Q_1 – Q_2$. So

$$W = Q_1\left(1 – \frac35\right) = \frac25 Q_1 = 1000\ \text{J} \;\Rightarrow\; Q_1 = 2500\ \text{J},\qquad Q_2 = 1500\ \text{J}.$$

The efficiency is $0.4$, the most any engine between these temperatures can manage.

Question 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 kilo calories, how many times must he go up and down to reduce his weight by 5 kg?

Solution. Going up once he does work against gravity $mgh$. Coming down he burns half as much, $\tfrac12mgh$. So one round trip costs

$$1.5\,mgh = 1.5\times60\times10\times10 = 9000\ \text{J}\quad(g = 10\ \text{m/s}^2).$$

The energy stored in 5 kg of fat is

$$5\times7000\times10^3\times4.2\ \text{J} = 1.47\times10^8\ \text{J}.$$

The number of round trips is

$$N = \frac{1.47\times10^8}{9000} \approx 1.63\times10^4.$$

With $g = 9.8$ m/s$^2$ the count is about $1.67\times10^4$. Either way he would need some sixteen thousand trips, which makes the point that exercise alone burns fat very slowly. (The calculation keeps the mass at 60 kg throughout, as the question intends.)

Question 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 $\Delta V\ (\ll V)$ 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 $P_2$?

Solution. In one stroke the air in the tube, at pressure $P$, together with the small volume $\Delta V$ pushed in, is compressed adiabatically into the fixed volume $V$, and the pressure rises by $\Delta P$:

$$P(V + \Delta V)^\gamma = (P + \Delta P)V^\gamma.$$

Since $\Delta V \ll V$, expand to first order: $P\left(1 + \gamma\dfrac{\Delta V}{V}\right) = P\left(1 + \dfrac{\Delta P}{P}\right)$, which gives

$$\frac{dV}{dP} = \frac{V}{\gamma P}.$$

The work done in pushing the air in is $\int P\,dV$, so

$$W = \int_{P_1}^{P_2} P\,\frac{V}{\gamma P}\,dP = \frac{(P_2 – P_1)V}{\gamma}.$$

Question 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 is 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.

Solution. The temperatures are $T_2 = 270$ K inside and $T_1 = 300$ K outside. A perfect (Carnot) refrigerator has coefficient of performance

$$\beta_{\text{Carnot}} = \frac{T_2}{T_1 – T_2} = \frac{270}{30} = 9.$$

At 50% of this, $\beta = 4.5$. The motor supplies $W = 1$ kJ each second, so the heat removed from the inside per second is

$$Q_2 = \beta W = 4.5\times1\ \text{kJ} = 4.5\ \text{kJ},$$

and the heat delivered to the room is $Q_1 = Q_2 + W = 5.5$ kJ per second.

Printed key error. The key takes the Carnot engine efficiency $1 – 270/300 = 1/10$, halves it to $1/20$, sets $W/Q_1 = 1/20$ and reports $Q_1 = 20$ kJ and $19$ kJ removed per second. That makes the refrigerator remove 19 kJ for each 1 kJ of work, a coefficient of performance of 19, more than twice the Carnot limit of 9. A real refrigerator cannot outperform a perfect one, so the key’s figure is impossible. Halving the engine efficiency makes the device better, not worse; the 50% must be applied to the coefficient of performance.

Question 21

If the coefficient of performance of a refrigerator is 5 and it operates at the room temperature (27 °C), find the temperature inside the refrigerator.

Solution. Treat it as an ideal refrigerator, $\beta = T_2/(T_1 – T_2)$, with $T_1 = 300$ K:

$$5 = \frac{T_2}{300 – T_2} \;\Rightarrow\; 1500 – 5T_2 = T_2 \;\Rightarrow\; T_2 = 250\ \text{K}.$$

Question 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?

Solution. Both start at $(P_i, V_i)$ and end at volume $V_f$.

• (a) Isothermal: $PV = P_iV_i$, a falling hyperbola that ends at $P_f = P_iV_i/V_f < P_i$. Work $W_a = P_iV_i\ln(V_f/V_i)$.
• (b) Isobaric: a horizontal line at $P_i$ from $V_i$ to $V_f$. Work $W_b = P_i(V_f – V_i)$.

On the $P$-$V$ diagram the horizontal line lies above the hyperbola everywhere between $V_i$ and $V_f$, so the area under it is larger. Algebraically, with $x = V_f/V_i > 1$, $W_b/W_a = (x – 1)/\ln x > 1$ because $\ln x < x – 1$.

Question 23

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: the curve $PV^{1/2} = \text{constant}$ from state 1 $(P_1, V_1, T_1)$ to state 2 $(P_2, V_2, T_2)$.

(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 $V_2 = 2V_1$? (c) Given the internal energy for one mole of gas at temperature $T$ is $(3/2)RT$, find the heat supplied to the gas when it is taken from state 1 to state 2, with $V_2 = 2V_1$.

Solution. Let $PV^{1/2} = A$, so $P = AV^{-1/2}$.

(a) $$W = \int_{V_1}^{V_2}AV^{-1/2}\,dV = 2A\left(\sqrt{V_2} – \sqrt{V_1}\right) = 2P_1V_1^{1/2}\left(V_2^{1/2} – V_1^{1/2}\right).$$

Since $A\sqrt{V} = PV$, this is also $2(P_2V_2 – P_1V_1)$.

(b) For one mole $T = PV/R = A\sqrt V/R$, so $T \propto \sqrt V$ and

$$\frac{T_1}{T_2} = \sqrt{\frac{V_1}{V_2}} = \frac{1}{\sqrt2}.$$

(c) With $T_2 = \sqrt2\,T_1$:

$$\Delta U = \tfrac32R(T_2 – T_1) = \tfrac32RT_1(\sqrt2 – 1),\qquad W = 2(P_2V_2 – P_1V_1) = 2R(T_2 – T_1) = 2RT_1(\sqrt2 – 1).$$

By the first law,

$$Q = \Delta U + W = \tfrac72RT_1(\sqrt2 – 1).$$

Question 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. A to B: volume constant; B to C: adiabatic; C to D: volume constant; D to A: adiabatic; $V_C = V_D = 2V_A = 2V_B$.

(a) In which part of the cycle is heat supplied to the engine from outside? (b) In which part of the cycle is heat 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? [$\gamma = 5/3$ for the gas, $C_v = \tfrac32R$ for one mole]

Solution. This is the Otto cycle.

(a) On the adiabats no heat flows. On A to B the volume is fixed and the pressure rises, so the temperature rises and heat must be supplied: A to B.

(b) On C to D the volume is fixed and the pressure falls, so the temperature falls and heat is rejected: C to D.

(c) The constant-volume legs do no work. The adiabatic legs do $W = (P_iV_i – P_fV_f)/(\gamma – 1)$. Along B to C, $P_C = P_B(V_B/V_C)^\gamma = 2^{-\gamma}P_B$; along D to A, $P_D = 2^{-\gamma}P_A$. Then

$$W_{BC} = \frac{P_BV_A – 2^{-\gamma}P_B\cdot2V_A}{\gamma – 1} = \frac{P_BV_A(1 – 2^{1-\gamma})}{\gamma – 1},\qquad W_{DA} = -\frac{P_AV_A(1 – 2^{1-\gamma})}{\gamma – 1}.$$

Adding and putting $\gamma – 1 = 2/3$:

$$W = \frac32\left[1 – \left(\frac12\right)^{2/3}\right](P_B – P_A)V_A.$$

(d) The heat supplied is on A to B only: $Q_{AB} = C_v(T_B – T_A) = \tfrac32(P_B – P_A)V_A$. So

$$\eta = \frac{W}{Q_{AB}} = 1 – \left(\frac12\right)^{2/3} \approx 0.37.$$

Question 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 ($C_v = \tfrac32R$). AB: constant volume; BC: constant pressure; CD: adiabatic; DA: constant pressure.

Solution. Use $Q = nC\Delta T$ with $n = 1$ and $PV = RT$, so that $R\,\Delta T$ can be written in pressures and volumes. Note $V_B = V_A$, $P_C = P_B$ and $P_D = P_A$.

• AB (constant volume). No work, so $Q_{AB} = C_v(T_B – T_A) = \tfrac32V_A(P_B – P_A)$. Positive: heat is absorbed.
• BC (constant pressure). $Q_{BC} = C_p(T_C – T_B) = \tfrac52P_B(V_C – V_B) = \tfrac52P_B(V_C – V_A)$. This is $\Delta U = \tfrac32P_B(V_C – V_B)$ plus work $P_B(V_C – V_B)$. Positive: heat is absorbed.
• CD (adiabatic). $Q_{CD} = 0$.
• DA (constant pressure). $Q_{DA} = C_p(T_A – T_D) = \tfrac52P_A(V_A – V_D)$. Since $V_D > V_A$, this is negative: heat is rejected.

Question 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 adiabat passing through $(P_0, V_0)$.

Solution. The adiabat through $(V_0, P_0)$ is $PV^\gamma = K$, whose slope is $dP/dV = -\gamma K V^{-\gamma-1} = -\gamma P_0/V_0$.

Along the process, $dQ = nC_v\,dT + P\,dV$. From $nRT = Vf(V)$,

$$n\,dT = \frac1R\left[f(V) + Vf'(V)\right]dV.$$

Using $C_v/R = 1/(\gamma – 1)$, at $V = V_0$:

$$\frac{dQ}{dV} = \frac{P_0 + V_0f'(V_0)}{\gamma – 1} + P_0 = \frac{\gamma P_0 + V_0f'(V_0)}{\gamma – 1}.$$

The gas is expanding, so $dV > 0$, and it absorbs heat when $dQ/dV > 0$. Since $\gamma – 1 > 0$, that needs

$$\gamma P_0 + V_0f'(V_0) > 0 \;\Longleftrightarrow\; f'(V_0) > -\frac{\gamma P_0}{V_0},$$

that is, the slope of $P = f(V)$ exceeds the slope of the adiabat. Physically, a curve that falls less steeply than the adiabat keeps the pressure higher than an adiabatic expansion would, and the extra energy for that must come in as heat.

Question 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 $V_0$ to $V_1$. Atmospheric pressure is $P_a$.

(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 $k$.

Solution. The spring lies inside the cylinder, joining the piston to the bottom. Take the piston as light.

(a) The spring is unstretched, so it exerts no force; the gas pressure balances the atmosphere: $P_0 = P_a$.

(b) The piston rises by $x = (V_1 – V_0)/A = V_1 – V_0$ (unit area), stretching the spring by $x$. The spring now pulls the piston down with force $kx$, so on unit area the gas must balance $P_a$ plus $kx$:

$$P_1 = P_a + k(V_1 – V_0).$$

(c) The gas does work against the atmosphere and in stretching the spring:

$$W = P_a(V_1 – V_0) + \tfrac12k(V_1 – V_0)^2.$$

Its temperature also changes, from $T_0 = P_aV_0/R$ to $T_1 = [P_a + k(V_1 – V_0)]V_1/R$, so $\Delta U = C_v(T_1 – T_0)$. The first law gives

$$Q = P_a(V_1 – V_0) + \tfrac12k(V_1 – V_0)^2 + \frac{C_v}{R}\Big[\big(P_a + k(V_1 – V_0)\big)V_1 – P_aV_0\Big].$$

Printed key slip. The key’s prose for (c) says all the heat becomes mechanical energy with no change in internal energy. That is wrong: the gas ends at a higher pressure and a larger volume, so its temperature, and therefore its internal energy, rises. The key’s own formula includes the $C_v(T – T_0)$ term, and the formula is the correct answer.

Common mistakes

• Taking $\Delta U$ as path-dependent in Question 13. It is fixed by P and Q; only work and heat change with the path.
• Thinking an open refrigerator cools the room in Question 14. It rejects $Q_2 + W$ to the room while removing only $Q_2$.
• Treating heat and temperature rise as inseparable in Question 15. Adiabatic compression raises the temperature with no heat at all.
• Halving the engine efficiency in Question 20. For a refrigerator it is the coefficient of performance that is halved; halving $W/Q_1$ gives an impossible COP of 19.
• Using $T_1/(T_1 – T_2)$ in Question 21. That is the heat pump’s COP; a refrigerator uses $T_2/(T_1 – T_2)$.
• Forgetting heat is supplied only on A to B in Question 24. The efficiency divides by $Q_{AB}$, not by all the heat exchanged.
• Using $C_v$ on the constant-pressure legs in Question 25. Those need $C_p = \tfrac52R$, because the gas also does work.
• Saying $\Delta U = 0$ in Question 27. The gas ends hotter, so $\Delta U$ must be kept in the first law.

Practise next

• A Carnot engine works between 600 K and 300 K and absorbs 1200 J from the hot reservoir each cycle. Find the work done per cycle and the heat rejected.

Show answer

$W = 600$ J and $600$ J rejected. $\eta = 1 – 300/600 = 0.5$, so $W = 0.5\times1200 = 600$ J and $Q_2 = 1200 – 600 = 600$ J.

• A refrigerator keeps its inside at $-13\,^\circ$C in a room at $27\,^\circ$C, and its coefficient of performance is 40% of the Carnot value. If its motor delivers 1 kW, how much heat does it remove each second?

Show answer

$2.6$ kJ per second. The Carnot COP is $260/(300-260) = 6.5$, 40% of which is $2.6$, so $Q_2 = 2.6\times1\ \text{kJ} = 2.6$ kJ each second.

• One mole of a monatomic ideal gas at temperature $T_1$ expands along $PV^{1/2} = \text{constant}$ until its volume is four times its initial volume. Find the heat supplied in terms of $R$ and $T_1$.

Show answer

$Q = \tfrac72RT_1$. $T \propto \sqrt V$, so $T_2 = 2T_1$. Then $\Delta U = \tfrac32R(T_2 – T_1) = \tfrac32RT_1$ and $W = 2(P_2V_2 – P_1V_1) = 2R(T_2 – T_1) = 2RT_1$, giving $Q = \tfrac72RT_1$.

• Exemplar Thermodynamics: Multiple Choice Questions — the same ideas in objective form.
• NCERT textbook Chapter 11 — Thermodynamics for the theory of the first and second laws and the Carnot engine.
• Exemplar Kinetic Theory — the next unit, which explains $C_v$ and $\gamma$ from molecular motion.