Q: The ideal gas equation for n moles of a gas is:

PV = nRT. PV = nRT, where R = 8.314\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} is the universal gas constant. In terms of the number of molecules N it reads PV = Nk_B T.

Q: Boyle's law, valid at constant temperature for a fixed mass of gas, states that:

PV = \text{constant}. At constant T and n, the ideal gas equation gives PV = constant, so pressure and volume are inversely proportional.

Q: According to kinetic theory, the molecular motion of an ideal gas would cease entirely at:

0\text{ K}. Average kinetic energy is \dfrac{3}{2}k_B T, which vanishes only when T = 0\text{ K} — absolute zero, or -273.15^\circ\text{C}.

Question 1

The root-mean-square velocity of gas molecules at absolute temperature T is directly proportional to:

Accepted Answer

\sqrt{T}. v_{rms} = \sqrt{\dfrac{3RT}{M}}, so the rms speed varies as the square root of the absolute temperature.

Question 2

The root-mean-square speed of the molecules of an ideal gas of molar mass M at absolute temperature T is:

Accepted Answer

\sqrt{\dfrac{3RT}{M}}. v_{rms} = \sqrt{\dfrac{3RT}{M}}. The other expressions are the most probable speed \sqrt{2RT/M} and the mean speed \sqrt{8RT/\pi M}.

Question 3

The average translational kinetic energy of a single molecule of an ideal gas at absolute temperature T is:

Accepted Answer

\dfrac{3}{2}k_B T. A molecule has three translational degrees of freedom, each carrying \dfrac{1}{2}k_B T, so the total translational kinetic energy is \dfrac{3}{2}k_B T.

Question 4

The average kinetic energy associated with each translational degree of freedom of a gas molecule at absolute temperature T is:

Accepted Answer

\dfrac{1}{2}k_B T. By the law of equipartition of energy, every independent degree of freedom carries an average energy of \dfrac{1}{2}k_B T.

Question 5

If the absolute temperature of a gas is doubled, the rms speed of its molecules becomes:

Accepted Answer

\sqrt{2} times. v_{rms} \propto \sqrt{T}, so doubling T multiplies the rms speed by \sqrt{2} \approx 1.41.

Question 6

The ideal gas equation for n moles of a gas is:

Accepted Answer

PV = nRT. PV = nRT, where R = 8.314	ext{ J}\cdot	ext{mol}^{-1}\cdot	ext{K}^{-1} is the universal gas constant. In terms of the number of molecules N it reads PV = Nk_B T.

Question 7

Boyle's law, valid at constant temperature for a fixed mass of gas, states that:

Accepted Answer

PV = 	ext{constant}. At constant T and n, the ideal gas equation gives PV =  constant, so pressure and volume are inversely proportional.

Question 8

Charles' law states that, at constant pressure, the volume of a fixed mass of gas is:

Accepted Answer

Directly proportional to its absolute temperature. At constant P, PV = nRT gives \dfrac{V}{T} =  constant, so V \propto T with T measured on the Kelvin scale.

Question 9

According to kinetic theory, the molecular motion of an ideal gas would cease entirely at:

Accepted Answer

0	ext{ K}. Average kinetic energy is \dfrac{3}{2}k_B T, which vanishes only when T = 0	ext{ K} — absolute zero, or -273.15^\circ	ext{C}.

Question 10

Boltzmann's constant k_B is related to the universal gas constant R and Avogadro's number N_A by:

Accepted Answer

k_B = \dfrac{R}{N_A}. R is the gas constant per mole and k_B the gas constant per molecule, so k_B = \dfrac{R}{N_A} \approx 1.38 	imes 10^{-23}	ext{ J/K}.

Question 11

According to kinetic theory, the pressure exerted by an ideal gas of density \rho whose molecules have rms speed v_{rms} is:

Accepted Answer

\dfrac{1}{3}\rho v_{rms}^2. Averaging the momentum delivered to the walls over the three coordinate directions gives P = \dfrac{1}{3}\rho v_{rms}^2.

Question 12

The rms speed of gas molecules in terms of the pressure P and density \rho of the gas is:

Accepted Answer

\sqrt{\dfrac{3P}{\rho}}. Rearranging P = \dfrac{1}{3}\rho v_{rms}^2 gives v_{rms} = \sqrt{\dfrac{3P}{\rho}}.

Question 13

If the rms speed of the molecules of a gas is doubled while its density is kept constant, the pressure becomes:

Accepted Answer

4 times. From P = \dfrac{1}{3}\rho v_{rms}^2, pressure varies as the square of the rms speed, so doubling it multiplies the pressure by 4.

Question 14

If the density of a gas is doubled while the rms speed of its molecules is unchanged, its pressure:

Accepted Answer

Doubles. P = \dfrac{1}{3}\rho v_{rms}^2 is directly proportional to density at fixed molecular speed.

Question 15

The pressure exerted by a gas on the walls of its container arises from:

Accepted Answer

The rate of change of momentum of molecules colliding with the walls. Each molecule rebounding elastically from a wall reverses its normal momentum. The total rate of momentum transfer per unit area is exactly the pressure.

Question 16

A molecule of mass m moving with speed v strikes a wall normally and rebounds elastically. The magnitude of the change in its momentum is:

Accepted Answer

2mv. The momentum changes from +mv to -mv, so the change has magnitude 2mv. This factor of two is what enters the kinetic-theory derivation of pressure.

Question 17

For an ideal gas kept at constant volume, doubling the absolute temperature causes the pressure to:

Accepted Answer

Double. At constant V and n, PV = nRT reduces to P \propto T, so the pressure doubles.

Question 18

In deriving the expression for gas pressure, the collisions of molecules with the container walls are assumed to be:

Accepted Answer

Perfectly elastic, so that no kinetic energy is lost. If collisions were inelastic the gas would continuously lose energy to the walls and could never remain in thermal equilibrium at constant temperature.

Question 19

The number of degrees of freedom of a monatomic gas molecule is:

Accepted Answer

3. A monatomic molecule is a point mass, so it can only translate along three independent directions. It has no rotational or vibrational degrees of freedom.

Question 20

The number of degrees of freedom of a rigid diatomic molecule at ordinary temperatures is:

Accepted Answer

5. Three translational plus two rotational (about the two axes perpendicular to the bond) gives f = 5. Rotation about the bond axis contributes negligibly.

Kinetic Theory of Gases & Radiation — MH-CET Physics MCQs with Solutions

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