NCERT Solutions Class 11 Physics Chapter 11 — Thermodynamics

NCERT Class 11 Physics — Thermodynamics, Chapter 11 Exercises. All 8 questions solved.

The exercises at the end of Chapter 11 are about energy moving into and out of a system as heat and as work. Questions 11.1 to 11.3 are about heat capacity — how much heat a given temperature change takes. Questions 11.5 to 11.8 apply the first law of thermodynamics, and Question 11.4 is an adiabatic compression. The working runs on three results:

• The first law — $\Delta Q = \Delta U + \Delta W$, where $\Delta Q$ is the heat supplied to the system, $\Delta U$ the rise in its internal energy and $\Delta W$ the work done by the system.
• Heat and temperature — $Q = ms\,\Delta T$ for a body of mass $m$ and specific heat $s$; for $n$ moles of a gas at constant pressure $Q = nC_p\,\Delta T$, with $C_p = \tfrac72R$ for a diatomic gas at room temperature.
• Work and adiabatic change — the work done by a gas is the area under its $P$–$V$ curve, $W = \int P\,dV$; in a quasi-static adiabatic process $PV^\gamma$ stays constant, with $\gamma = 1.4$ for a diatomic gas.

Key insight. Internal energy belongs to the state of a system; heat and work belong to the process that takes it there. So $\Delta U$ between two states is fixed, however the system travels between them, while $\Delta Q$ and $\Delta W$ change from route to route. Question 11.5 uses exactly this: one route fixes $\Delta U$, and the first law then gives the work on the other. In the free expansion of 11.6 there is neither heat nor work, so $\Delta U$ must be zero. And in 11.8 the work depends on the path drawn on the $P$–$V$ diagram, because it is the area under that path.

Question 11.1

A geyser heats water flowing at the rate of $3.0$ litres per minute from $27\ ^\circ\text{C}$ to $77\ ^\circ\text{C}$. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0\times10^4\ \text{J/g}$?

Solution. Work out the heat needed each minute, then how much fuel supplies it. The heat of combustion is given per gram, so keep the water in grams too. A litre of water has a mass of $1000\ \text{g}$, so $3000\ \text{g}$ of water passes through each minute, and each gram is warmed by $77 – 27 = 50\ ^\circ\text{C}$. With the specific heat of water $s = 4.2\ \text{J g}^{-1}\,{}^\circ\text{C}^{-1}$:

$$Q = ms\,\Delta T = 3000\times4.2\times50 = 6.3\times10^5\ \text{J per minute}$$

Assuming all the heat released by burning the fuel goes into the water, the fuel burnt per minute is

$$\frac{6.3\times10^5\ \text{J/min}}{4.0\times10^4\ \text{J/g}} = 15.75\ \text{g/min} \approx 16\ \text{g/min}$$

Question 11.2

What amount of heat must be supplied to $2.0\times10^{-2}\ \text{kg}$ of nitrogen (at room temperature) to raise its temperature by $45\ ^\circ\text{C}$ at constant pressure? (Molecular mass of $\text{N}_2 = 28$; $R = 8.3\ \text{J mol}^{-1}\,\text{K}^{-1}$.)

Solution. The heat depends on the process, so the molar heat capacity to use is the one at constant pressure. First the number of moles, $20\ \text{g}$ of nitrogen at $28\ \text{g mol}^{-1}$:

$$n = \frac{20}{28} = 0.714\ \text{mol}$$

Nitrogen is diatomic. At room temperature its molecules translate and rotate but do not vibrate appreciably, which gives $C_v = \tfrac52R$, and at constant pressure the gas also does work as it expands, adding $R$: $C_p = C_v + R = \tfrac72R = 29.05\ \text{J mol}^{-1}\,\text{K}^{-1}$. A rise of $45\ ^\circ\text{C}$ is a rise of $45\ \text{K}$, so

$$Q = nC_p\,\Delta T = 0.714\times29.05\times45 = 934\ \text{J}$$

Question 11.3

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 $(T_1 + T_2)/2$. (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.

Question 11.3 (a)

Solution. When the bodies reach a common temperature $T$, the heat lost by the hotter one equals the heat gained by the colder one:

$$m_1s_1(T_1 – T) = m_2s_2(T – T_2) \quad\Rightarrow\quad T = \frac{m_1s_1T_1 + m_2s_2T_2}{m_1s_1 + m_2s_2}$$

This is an average weighted by the heat capacities $m_1s_1$ and $m_2s_2$. It equals the plain mean $(T_1 + T_2)/2$ only if the two heat capacities are equal; otherwise the final temperature lies closer to that of the body with the larger heat capacity.

Question 11.3 (b)

Solution. A coolant’s job is to carry heat away from the hot parts. From $Q = ms\,\Delta T$, a liquid of high specific heat absorbs a large amount of heat for a small rise in its own temperature. So a given mass of it removes more heat before it gets too hot to keep cooling, and the plant stays safer.

Question 11.3 (c)

Solution. While the car moves, the tyre flexes continuously and rubs against the road, and the work done against friction and in deforming the rubber is turned into heat. The air inside warms up. Its volume is nearly fixed by the tyre, so by the gas law $P \propto T$ at constant volume, and the pressure rises.

Question 11.3 (d)

Solution. Water has a much higher specific heat than sand or rock. By day the sea absorbs a great deal of heat with little rise in temperature, and by night it gives that heat back with little fall, so the air over a harbour town is kept from getting very hot or very cold; sea and land breezes carry this moderating effect inland. Desert sand, with its low specific heat, heats up quickly in the sun and cools quickly at night, so a desert town has a very large range between day and night temperatures.

Question 11.4

A cylinder with a movable piston contains $3$ moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Solution. The insulated walls and the sand on the piston stop any heat entering or leaving the gas, so the compression is adiabatic, and $PV^\gamma$ stays constant — not $PV$, as it would at constant temperature. Hydrogen is diatomic, so $\gamma = C_p/C_v = \tfrac{7/2}{5/2} = 1.4$. Then

$$P_1V_1^\gamma = P_2V_2^\gamma \quad\Rightarrow\quad \frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma = 2^{1.4} = 2.64$$

The $3$ moles and the starting conditions do not enter: only the volume ratio and $\gamma$ matter. The factor is more than the $2$ Boyle’s law would give because the work done on the gas also heats it.

Question 11.5

In changing the state of a gas adiabatically from an equilibrium state $A$ to another equilibrium state $B$, an amount of work equal to $22.3\ \text{J}$ is done on the system. If the gas is taken from state $A$ to $B$ via a process in which the net heat absorbed by the system is $9.35\ \text{cal}$, how much is the net work done by the system in the latter case? (Take $1\ \text{cal} = 4.19\ \text{J}$.)

Solution. The adiabatic route fixes $\Delta U$. No heat is exchanged, so $\Delta Q = 0$. The $22.3\ \text{J}$ of work is done on the gas, so the work done by it is $\Delta W = -22.3\ \text{J}$. The first law gives

$$\Delta U = \Delta Q – \Delta W = 0 – (-22.3) = 22.3\ \text{J}$$

The second route has the same $\Delta U$. Internal energy depends only on the state, and both routes go from $A$ to $B$, so $\Delta U$ is again $22.3\ \text{J}$. The heat absorbed is $\Delta Q = 9.35\times4.19 = 39.2\ \text{J}$, so

$$\Delta W = \Delta Q – \Delta U = 39.2 – 22.3 = 16.9\ \text{J}$$

The answer is positive, so on this route the gas does work on its surroundings.

Question 11.6

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 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?

This is a free expansion: the gas rushes into a vacuum. It is easiest to settle (b) and (c) first, and (a) follows from them.

Question 11.6 (a)

Solution. As (b) and (c) show, the gas ends at the same temperature it started at, so Boyle’s law applies between the initial and final equilibrium states. The gas that filled $A$ now fills $A$ and $B$, twice the volume:

$$P_1V = P_2(2V) \quad\Rightarrow\quad P_2 = \frac{P_1}{2} = \frac{1\ \text{atm}}{2} = 0.5\ \text{atm}$$

Question 11.6 (b)

Solution. The system is thermally insulated, so $\Delta Q = 0$. The gas expands into a vacuum, where there is nothing to push against, so it does no work: $\Delta W = 0$. By the first law, $\Delta U = \Delta Q – \Delta W = 0$.

Question 11.6 (c)

Solution. The internal energy of an ideal gas depends only on its temperature. Since $\Delta U = 0$, the temperature does not change. (A real gas cools very slightly in a free expansion, because its molecules do work against their mutual attractions as they move apart; for an ideal gas the change is zero.)

Question 11.6 (d)

Solution. No. The expansion is sudden and uncontrolled. While the gas is rushing through the stopcock, its pressure and temperature differ from one part of the cylinders to another, so the gas as a whole has no single value of $P$ or $T$. Such a non-equilibrium state cannot be a point on the $P$-$V$-$T$ surface, which is made up of equilibrium states satisfying the gas equation. Only the initial and final states lie on it; the gas does return to an equilibrium state once it settles.

Question 11.7

An electric heater supplies heat to a system at a rate of $100\ \text{W}$. If the system performs work at a rate of $75$ joules per second, at what rate is the internal energy increasing?

Solution. Apply the first law to what happens in each second. Of the $100\ \text{J}$ of heat supplied every second, $75\ \text{J}$ leaves again as work done by the system; the rest stays as internal energy:

$$\frac{\Delta U}{\Delta t} = \frac{\Delta Q}{\Delta t} – \frac{\Delta W}{\Delta t} = 100 – 75 = 25\ \text{J s}^{-1}$$

Question 11.8

A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. 11.11. Its volume is then reduced to the original value from $E$ to $F$ by an isobaric process. Calculate the total work done by the gas from $D$ to $E$ to $F$.

Note on the figure number. The question’s text refers to “Fig. (11.13)”, but the figure printed with it is captioned Fig. 11.11, and the chapter has no Figure 11.13; the figure meant is the one above.

Solution. The work done by a gas is the area under its path on the $P$–$V$ diagram — counted positive when the volume increases and negative when it decreases. From the figure, $D$ is at $(2.0\ \text{m}^3,\ 600\ \text{N/m}^2)$, $E$ at $(5.0\ \text{m}^3,\ 300\ \text{N/m}^2)$ and $F$ at $(2.0\ \text{m}^3,\ 300\ \text{N/m}^2)$.

From $D$ to $E$ the gas expands along a straight line, so the area under it is a trapezium with parallel sides $600$ and $300$ and width $5.0 – 2.0 = 3.0\ \text{m}^3$:

$$W_{DE} = \tfrac12(600 + 300)\times3.0 = 1350\ \text{J}$$

From $E$ to $F$ the gas is compressed at a constant $300\ \text{N/m}^2$, so the work done by it is negative:

$$W_{EF} = P\,\Delta V = 300\times(2.0 – 5.0) = -900\ \text{J}$$

Adding the two,

$$W = 1350 – 900 = 450\ \text{J}$$

This is just the area of the shaded triangle $DEF$, $\tfrac12\times3.0\times300 = 450\ \text{J}$: the area under $EF$ cancels the lower part of the area under $DE$.

Common mistakes

• Question 11.1: mixing kilograms and grams. The heat of combustion is per gram, so the water must be in grams and the specific heat in $\text{J g}^{-1}\,{}^\circ\text{C}^{-1}$ ($4.2$, not $4200$). Mixing the two puts the answer out by a factor of $1000$.
• Question 11.2: using $C_v$ instead of $C_p$. The question says constant pressure, where the gas also does work pushing back its surroundings as it expands. With $C_v = \tfrac52R$ the answer comes out as $667\ \text{J}$, which leaves out that work.
• Question 11.3(a): assuming the mean temperature is the rule. Heat lost equals heat gained, not “temperature lost equals temperature gained”. The body with the larger heat capacity $ms$ changes temperature less.
• Question 11.4: using Boyle’s law, or $\gamma = 5/3$. Nothing about the compression is isothermal — the gas is insulated — so the factor is not $2$. And hydrogen is diatomic, so $\gamma = 1.4$; the monatomic $5/3$ gives $3.17$.
• Question 11.5: getting the sign of the adiabatic work wrong. Work done on the gas is negative work by it, so $\Delta U = +22.3\ \text{J}$. Taking $\Delta U = -22.3\ \text{J}$ gives $61.5\ \text{J}$ instead of $16.9\ \text{J}$.
• Question 11.6(c): expecting the gas to cool because it expands. A gas cools in an adiabatic expansion only because it does work pushing a piston. Expanding into a vacuum it does no work, so an ideal gas keeps its temperature.
• Question 11.8: counting the work from $E$ to $F$ as positive. The volume falls from $E$ to $F$, so that work is $-900\ \text{J}$. Adding it as $+900\ \text{J}$ gives $2250\ \text{J}$, and stopping after $D$ to $E$ gives $1350\ \text{J}$.

Practise next

• Chapter 12, Kinetic Theory — why $C_v = \tfrac52R$ for a diatomic gas, and why the internal energy of an ideal gas depends on temperature alone, both of which this chapter’s answers rely on.
• Chapter 10, Thermal Properties of Matter — specific heat and the heat-balance calorimetry behind Questions 11.1 and 11.3.