NCERT Exemplar Solutions Class 11 Physics Thermodynamics — Multiple Choice Questions Q1–11

NCERT Exemplar Problems, Class 11 Physics — Thermodynamics, MCQ I and MCQ II. All 11 questions solved.

The book numbers these questions 12.1 to 12.11; here they are Questions 1 to 11. Questions 1 to 6 are MCQ I (exactly one option is correct) and Questions 7 to 11 are MCQ II (one or more options may be correct). Three ideas carry the whole page:

• The first law. $\Delta Q = \Delta U + W$, where $W$ is the work done by the gas. For an ideal gas $U$ depends on temperature alone.
• The four standard processes on a $P$-$V$ diagram. Isochoric is vertical, isobaric horizontal, isothermal has slope $-P/V$ and adiabatic the steeper slope $-\gamma P/V$.
• Work is the area under the $P$-$V$ curve, and the net work in a cycle is the enclosed area, positive for a clockwise cycle and negative for an anticlockwise one.

Key insight. Internal energy is a state function and work is not. Questions 8, 9 and 10 all turn on keeping those two apart.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 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?

Solution. Identify the easy ones first. Curve 1 is vertical, so volume is constant: isochoric. Curve 4 is horizontal, so pressure is constant: isobaric. That leaves curves 2 and 3, one isothermal and one adiabatic. On an isotherm $PV = \text{constant}$, so $dP/dV = -P/V$. On an adiabat $PV^\gamma = \text{constant}$, so $dP/dV = -\gamma P/V$. Since $\gamma > 1$, the adiabat falls more steeply from the same starting point. Curve 2 is the steeper one, so curve 2 is adiabatic and curve 3 is isothermal.

Question 2

If an average person jogs, he produces $14.5\times10^3$ cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires $580\times10^3$ cal for evaporation) is

Solution. All the heat produced in a minute goes into evaporating sweat, so the mass evaporated is the heat divided by the latent heat:

$$m = \frac{14.5\times10^3\ \text{cal}}{580\times10^3\ \text{cal/kg}} = 0.025\ \text{kg}.$$

That is $25$ g of sweat a minute, which is a sensible figure for hard exercise; $250$ g a minute would be absurd.

Misprint in the options. None of the printed options is $0.025$ kg. The key gives (a) $0.25$ kg, which has the right digits but is ten times too large; the options were evidently printed with a factor-of-ten slip. We mark (a) as the intended option, but the physically correct amount is $0.025$ kg.

Question 3

Consider the $P$-$V$ diagram for an ideal gas shown in Fig. 12.2: the curve $P = \text{constant}/V$ runs from state 1 at high pressure down to state 2 at low pressure. Out of the following diagrams (Fig. 12.3), which represents the $T$-$P$ diagram?

Solution. $PV = \text{constant}$ for an ideal gas means $nRT$ is constant, so the process is isothermal. On a $T$-$P$ diagram it is therefore a horizontal line; that rules out (i) and (ii). The remaining question is direction. Going from 1 to 2 the gas expands and its pressure falls, so state 2 must sit at the lower pressure, to the left of state 1. Diagram (iii) has 2 on the left and 1 on the right; diagram (iv) has them the wrong way round.

Question 4

An ideal gas undergoes the cyclic process ABCDA shown in the $P$-$V$ diagram (Fig. 12.4). The amount of work done by the gas is

Solution. Take the four legs in turn. AB is an expansion at $P_0$ from $V_0$ to $3V_0$, so the gas does work $+P_0(2V_0) = +2P_0V_0$. BC and DA are at constant volume, so no work is done. CD is a compression at $2P_0$ from $3V_0$ back to $V_0$, so the work done by the gas is $-2P_0(2V_0) = -4P_0V_0$. The total is

$$W = 2P_0V_0 + 0 – 4P_0V_0 + 0 = -2P_0V_0.$$

The same result comes from the area: the rectangle has area $P_0\times2V_0$, and the cycle runs anticlockwise (expansion at the lower pressure, compression at the higher), so the net work done by the gas is negative.

Question 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

Solution. Let both start at $P$ and $V$. For A, isothermal: $PV = P_A(V/2)$, so $P_A = 2P$. For B, adiabatic: $PV^\gamma = P_B(V/2)^\gamma$, so $P_B = 2^\gamma P$. The ratio is

$$\frac{P_B}{P_A} = \frac{2^\gamma P}{2P} = 2^{\gamma-1}.$$

Since $\gamma > 1$ this exceeds 1, as it should: in the adiabatic compression the gas also heats up, which raises its pressure further. Option (b) is the reciprocal.

Question 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)

Solution. With no loss to the surroundings, the total heat exchanged is zero:

$$M_1s(T – T_1) + M_2s(T – T_2) + M_3s(T – T_3) = 0.$$

The common factor $s$ cancels because all three blocks are copper, leaving

$$T = \frac{M_1T_1 + M_2T_2 + M_3T_3}{M_1 + M_2 + M_3},$$

a mass-weighted average. Option (a) would hold only for equal masses, (c) has a spurious factor of 3 and is not even a temperature between $T_3$ and $T_1$, and (d) has the wrong dimensions because $s$ appears only in the numerator.

Question 7

Which of the processes described below are irreversible?

Solution. A process is reversible only if it passes through equilibrium states and has no dissipative effects, so that it can be retraced exactly by an infinitesimal change.

• (a) Hammering converts work into internal energy through friction and deformation. The rod will never cool down and spontaneously lift the hammer. Irreversible.
• (b) Heat flows across a finite temperature difference $T_2 – T_1$. It cannot flow back from the gas to the hotter reservoir by itself. Irreversible.
• (c) Quasi-static means the gas is in equilibrium at every stage, and the frictionless piston means no dissipation. This is the textbook example of a reversible process.
• (d) Adding a finite weight suddenly compresses the gas rapidly, through non-equilibrium states with turbulence and pressure gradients. Removing the weight would not bring it back to the starting state. Irreversible.

Question 8

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

Solution. The internal energy of an ideal gas depends only on its temperature, and the temperature does not change, so $dU = 0$: (a) is true. The first law $dQ = dU + dW$ then gives $dQ = dW$: (d) is true. Heat does flow, since the gas does work while keeping its temperature fixed, so (b) is false; and (c) would require $dW = 0$, which happens only at constant volume.

Question 9

Figure 12.5 shows the $P$-$V$ diagram of an ideal gas undergoing a change of state from A to B. Four different paths I, II, III and IV as shown in the figure may lead to the same change of state.

Solution. Internal energy is a function of state. All four paths start at A and end at B, so $\Delta U = U_B – U_A$ is the same for every one of them: (b) is true and (a) is false.

Work is not a function of state; it is the area under the path. All four paths sweep the same range of volume from $V_A$ to $V_B$, so the path that stays highest encloses the most area. Path I is the highest curve, so it does the most work: (c) is true. The lowest path is III, which dips below the straight line II, so III, not II, has the least work: (d) is false.

Question 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. In the figure, 1 to 2 is a short expansion at the top, 2 to 3 falls steeply to a lower point 3, and 3 to 1 returns directly to 1. Such a process does not exist because

Solution. Look at the heat. The only leg on which heat flows is the isothermal one, 1 to 2, where the gas expands and absorbs heat $Q$. On the two adiabatic legs no heat flows at all. Over a complete cycle $\Delta U = 0$, so the net work equals $Q$: all the absorbed heat would be turned into work with nothing rejected. That is exactly what the Kelvin–Planck statement of the second law forbids, so (a) is a correct reason.

The geometric version of the same fact is (c). Points 3 and 1 would both lie on the adiabat 2–3 and on the adiabat 3–1, so two different adiabats would meet at point 3. Two adiabats can never intersect; if they did, the cycle above could be built and would violate the second law. Option (b) is wrong because converting work completely into heat is perfectly possible (friction does it). Option (d) is wrong because an adiabat and an isotherm certainly can intersect; they do at point 2.

Question 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. In the figure the engine sits between an upper bath $T_1$ and a lower bath $T_2$, with $Q_2$ drawn from $T_2$ into the engine, $Q_1$ drawn from the engine into $T_1$, and $W$ fed into the engine.

If $W > 0$, then possibilities are:

Solution. Energy conservation over one cycle (the working substance returns to its starting state) says that what goes in equals what comes out:

$$Q_1 = Q_2 + W.$$

With $W > 0$ this gives $Q_1 > Q_2$ in every case, which rules out (b) at once.

• (a) $Q_1 > Q_2 > 0$: heat is taken from $T_2$, work is done on the engine, and more heat is delivered to $T_1$. This is a refrigerator or heat pump. Possible.
• (c) $Q_2 < Q_1 < 0$: both signs reverse, so heat actually flows out of the engine into $T_2$ and from $T_1$ into the engine, with $|Q_2| = |Q_1| + W$. The engine takes in heat and work and dumps all of it into $T_2$, which is like friction or a resistor warming a bath. Nothing forbids that. Possible.
• (d) $Q_1 < 0,\ Q_2 > 0$ would need $Q_1 – Q_2 = W$ to be negative, contradicting $W > 0$. Impossible.

Common mistakes

• Picking the gentler curve as adiabatic in Question 1. The adiabat is steeper than the isotherm by the factor $\gamma$.
• Choosing 0.25 kg without checking in Question 2. $14.5/580 = 0.025$; the printed options are ten times too large.
• Drawing the isotherm in the wrong direction in Question 3. Expansion lowers the pressure, so state 2 is on the low-pressure side.
• Ignoring the direction of the cycle in Question 4. An anticlockwise cycle on a $P$-$V$ diagram gives negative net work by the gas.
• Inverting the ratio in Question 5. The adiabatic final pressure is the larger, so $P_B/P_A = 2^{\gamma-1} > 1$.
• Calling a sudden compression reversible in Question 7. Only quasi-static, frictionless changes are reversible.
• Saying $dQ = 0$ for an isothermal process in Question 8. Constant temperature means $dU = 0$, not that no heat flows.
• Treating work as a state function in Question 9. $\Delta U$ is the same on every path; the work done is not.

Practise next

• An ideal monatomic gas ($\gamma = 5/3$) is compressed adiabatically to one-eighth of its original volume. By what factor does its pressure increase?

• A gas is taken round a rectangular clockwise cycle on a $P$-$V$ diagram between pressures $1\times10^5$ Pa and $3\times10^5$ Pa and volumes $1$ L and $4$ L. Find the net work done by the gas per cycle.

• Exemplar Thermodynamics: Very Short, Short and Long Answer Questions — the same ideas applied to Carnot cycles, refrigerators and spring-loaded pistons.
• NCERT textbook Chapter 11 — Thermodynamics for the theory of the first law, the standard processes and the second law.
• Exemplar Thermal Properties of Matter — the previous unit, whose calorimetry appears in Questions 2 and 6.
• Exemplar Kinetic Theory — the next unit, which explains $U$ and $\gamma$ from molecular motion.