NCERT Exemplar Solutions Class 11 Physics Kinetic Theory — Multiple Choice Questions Q1–13

NCERT Exemplar Problems, Class 11 Physics — Kinetic Theory, MCQ I and MCQ II. All 13 questions solved.

The book numbers these questions 13.1 to 13.13; here they are Questions 1 to 13. Questions 1 to 8 are MCQ I (exactly one option is correct) and Questions 9 to 13 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Pressure is momentum delivered to a wall. A molecule that bounces elastically hands over $2mv_x$; one that is absorbed hands over only $mv_x$.
• Only relative motion matters. Moving the whole container does not change how fast molecules hit its walls.
• The ideal gas law counts particles. $pV = \mu RT$, where $\mu$ is the number of moles of whatever particles are actually present, molecules or atoms.
• Equipartition. Each translational or rotational degree of freedom carries $\tfrac12 kT$ on average, so a diatomic molecule has $\tfrac32 kT$ of translational and $kT$ of rotational energy.

Key insight. In almost every question here, ask what the molecules actually do at the wall: how much momentum they hand over, and whether the wall is moving. Questions 1, 2, 10 and 13 are settled by that alone.

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

Question 1

A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of $500\ \text{m s}^{-1}$ in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground

Solution. Pressure comes from the change in momentum of molecules when they strike a wall, and that change depends on the velocity of the molecule relative to the wall. Adding the same $500\ \text{m s}^{-1}$ to every molecule and to the vessel leaves every relative velocity unchanged, so each collision delivers exactly the same impulse as before. The pressure is therefore unchanged on every face, which rules out (c) and (d). Option (a) gives the right result for the wrong reason: even if the vessel moved faster than $v_{rms}$ the pressure would still be unchanged, because it is the relative motion that counts, not the size of the speed. (The rocket is taken to move at steady speed; an accelerating rocket would be a different matter.)

Question 2

1 mole of an ideal gas is contained in a cubical volume $V$, ABCDEFGH at 300 K (Fig. 13.1). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time,

The figure shows a cube with top face ABCD and the opposite (bottom) face EFGH.

Solution. A molecule with normal velocity component $v_x$ that bounces elastically off ABCD reverses that component, so it hands the wall a momentum $2mv_x$. A molecule that strikes EFGH is absorbed: its normal velocity goes from $v_x$ to zero, so it hands over only $mv_x$. At any instant the rate at which molecules arrive at each face is the same (same number density, same speed distribution), so the force per unit area on EFGH is half that on ABCD. The pressure on EFGH is not zero, because absorption still stops the molecules, and it is certainly not double.

Question 3

Boyle’s law is applicable for an

Solution. Boyle’s law states that for a fixed mass of gas at constant temperature, $pV$ is constant. A constant-temperature process is isothermal. In an adiabatic process the temperature changes and $pV^\gamma$, not $pV$, is constant; isobaric and isochoric processes hold $p$ or $V$ fixed, so $pV = \text{constant}$ cannot describe them.

Question 4

A cylinder containing an ideal gas is in vertical position and has a piston of mass $M$ that is able to move up or down without friction (Fig. 13.2). If the temperature is increased,

The figure shows a vertical cylinder of gas closed at the top by a freely sliding piston, with a load of mass $M$ resting on it.

Solution. The piston is free and in equilibrium, so the gas pressure must balance the weight on it (plus the atmosphere above): $p = p_0 + Mg/A$. None of these change when the gas is heated, so $p$ stays fixed. The gas then obeys $V/T = \text{constant}$, and the piston rises as $T$ increases. Only $V$ changes.

Question 5

Volume versus temperature graphs for a given mass of an ideal gas are shown in Fig. 13.3 at two different values of constant pressure. What can be inferred about relation between $P_1$ and $P_2$?

Solution. For a fixed mass at constant pressure, $V = \left(\dfrac{\mu R}{p}\right)T$, so each line passes through the origin with slope $\mu R/p$. The slope is inversely proportional to the pressure. The $P_2$ line is steeper, so $P_2$ is the smaller pressure: $P_1 > P_2$. Physically, at the same temperature the gas at higher pressure is squeezed into a smaller volume, and the $P_1$ line is the lower one at every $T$.

Question 6

1 mole of $\text{H}_2$ gas is contained in a box of volume $V = 1.00\ \text{m}^3$ at $T = 300$ K. The gas is heated to a temperature of $T = 3000$ K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

Solution. The ideal gas law $pV = \mu RT$ counts the particles that are actually moving about and striking the walls. Dissociation $\text{H}_2 \to 2\text{H}$ turns 1 mole of molecules into 2 moles of atoms, so $\mu$ doubles. The temperature rises by a factor of $3000/300 = 10$, and $V$ is fixed. Hence

$$\frac{p_f}{p_i} = \frac{\mu_f T_f}{\mu_i T_i} = \frac{2 \times 3000}{1 \times 300} = 20.$$

Question 7

A vessel of volume $V$ contains a mixture of 1 mole of Hydrogen and 1 mole of Oxygen (both considered as ideal). Let $f_1(v)\,dv$ denote the fraction of molecules with speed between $v$ and $(v + dv)$ with $f_2(v)\,dv$, similarly for oxygen. Then

Solution. In equilibrium both gases share the same temperature, so each species has its own Maxwell distribution at that temperature, with its own molecular mass. Hydrogen molecules are 16 times lighter, so their distribution is spread to much higher speeds ($v_{rms}$ four times larger); the two curves are different, which rules out (d). Adding the two curves gives a two-humped function that is not a Maxwell distribution for any single mass, which rules out (a). Each one separately is Maxwellian.

Question 8

An inflated rubber balloon contains one mole of an ideal gas, has a pressure $p$, volume $V$ and temperature $T$. If the temperature rises to $1.1\,T$, and the volume is increased to $1.05\,V$, the final pressure will be

Solution. For a fixed amount of gas, $pV/T$ is constant, so

$$p’ = p \times \frac{V}{V’} \times \frac{T’}{T} = p \times \frac{1.1}{1.05} \approx 1.048\,p.$$

This lies between $p$ and $1.1\,p$. Physically, if the balloon could not stretch the pressure would rise all the way to $1.1\,p$; because the rubber gives a little, part of the rise goes into volume instead, so the pressure rises by less.

Question 9

ABCDEFGH is a hollow cube made of an insulator (Fig. 13.4). Face ABCD has positive charge on it. Inside the cube, we have ionized hydrogen.

The figure shows the cube with the top face ABCD marked $+$, the rest of the cube uncharged.

The usual kinetic theory expression for pressure

Solution. The derivation of $p = \tfrac13 nm\overline{v^2}$ assumes the molecules move freely between collisions and that all directions are equivalent. Here the charged face repels the positive ions and attracts the electrons, so the particles feel an electric force between collisions: (b) is true. That force picks out one direction, the line joining ABCD to the opposite face, so the velocity distribution is no longer the same in every direction and the number density is no longer uniform: (d) is true. Nothing in the set-up makes the collisions with the insulating walls inelastic, so (c) is not the reason, and (a) is false.

Question 10

Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory $pV = \dfrac23 E$, $E$ is

Solution. The relation is derived by adding up the change in linear momentum of molecules at the wall, so only the velocity of the centre of mass enters, and $E$ is the translational kinetic energy $\tfrac12 Nm\overline{v^2}$. Rotation does not push on the wall. Option (a) is wrong twice: $E$ is not the total energy and is not per unit volume (it multiplies $V$ on the other side). Option (b) gives the wrong reason, since rotational energy is $kT$ per molecule, two-thirds of the translational $\tfrac32 kT$, hardly small. Option (d) is wrong because rotational kinetic energy is never negative.

Question 11

In a diatomic molecule, the rotational energy at a given temperature

Solution. Collisions keep reshuffling energy between molecules, so rotational energies, like translational ones, are spread over a distribution of the Maxwell (Boltzmann) type and are independent of the translational distribution: (a) is true and (b) is false. By equipartition, a diatomic molecule has 3 translational and 2 rotational degrees of freedom, each with average energy $\tfrac12 kT$. So on average

$$\frac{\langle E_{rot}\rangle}{\langle E_{trans}\rangle} = \frac{kT}{\tfrac32 kT} = \frac23 ,$$

which makes (d) true and (c) false.

A word on (d): “for each molecule” has to be read as “per molecule, on average”. At any instant a given molecule can have any split between rotation and translation; it is the averages that stand in the ratio $2:3$. The printed key gives (a), (d), and we agree with it on that reading.

Question 12

Which of the following diagrams (Fig. 13.5) depicts ideal gas behaviour?

Solution. Test each graph against $pV = \mu RT$.

• (a) At constant $P$, $V = (\mu R/P)\,T$: a straight line through the origin. Correct.
• (b) At constant $T$, $P = \mu RT/V$, a rectangular hyperbola. A straight falling line would make $P$ reach zero at a finite volume, which no ideal gas does. Wrong.
• (c) At constant $V$, $P = (\mu R/V)\,T$: a straight line through the origin. Correct.
• (d) $PV = \mu RT$ grows in proportion to $T$; a horizontal line would mean $PV$ is independent of temperature. Wrong.

Question 13

When an ideal gas is compressed adiabatically, its temperature rises: the molecules on the average have more kinetic energy than before. The kinetic energy increases,

Solution. A molecule bouncing elastically off a stationary wall leaves with the same speed, so the fixed parts of the wall cannot change the kinetic energy: (b) is wrong. A molecule bouncing off a piston moving inwards with speed $u$ comes back with its normal speed increased from $v$ to $v + 2u$, just as a ball speeds up when struck by a bat. Each such collision adds energy, and that is where the work done by the piston goes. In an ideal gas there are no forces between collisions, so (c) is wrong, and collisions between molecules only share energy out, they cannot raise the total, so (d) is wrong.

Common mistakes

• Adding the rocket’s speed to $v_{rms}$ in Question 1. Pressure depends on molecular velocity relative to the walls, which is unchanged.
• Taking zero pressure on the absorbing face in Question 2. Stopping a molecule still takes impulse $mv_x$; it is half, not zero.
• Forgetting that dissociation doubles the number of moles in Question 6. That gives 10 instead of 20.
• Reading the steeper V–T line as the higher pressure in Question 5. The slope is $\mu R/p$, so steeper means lower pressure.
• Including rotational energy in $pV = \tfrac23 E$ in Question 10. Only linear momentum is exchanged with the wall.
• Picking (b) or (d) in Question 13. A fixed wall returns molecules at the same speed and intermolecular collisions conserve the total; only the moving piston adds energy.

Practise next

• Problem. Find the root mean square speed of oxygen molecules at $27^\circ$C. (Molar mass of $\text{O}_2 = 32\ \text{g mol}^{-1}$, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.)

• Problem. A mixture contains 1 mole of helium and 1 mole of nitrogen. Ignoring vibration, find the ratio of specific heats $\gamma = C_p/C_v$ of the mixture.

• Exemplar Kinetic Theory: Very Short, Short and Long Answer Questions — where the same ideas are applied to rms speeds, internal energy, mean free path and leaks through small holes.
• NCERT textbook Chapter 12 — Kinetic Theory for the derivation of gas pressure, the kinetic meaning of temperature and the law of equipartition.
• Exemplar Thermodynamics — the previous unit, whose isothermal and adiabatic processes reappear here.
• Exemplar Oscillations — the next unit.