Molecular Motion & Kinetic Energy — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Molecular Motion & Kinetic Energy MCQs with step-by-step solutions (32 questions). Part of Kinetic Theory of Gases & Radiation. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Molecular Motion & Kinetic Energy · easy · theory
The root-mean-square velocity of gas molecules at absolute temperature $T$ is directly proportional to:
A. $T$
B. $\dfrac{1}{\sqrt{T}}$
C. $T^2$
D. $\sqrt{T}$ ✓ Correct
Solution: $v_{rms} = \sqrt{\dfrac{3RT}{M}}$, so the rms speed varies as the square root of the absolute temperature.
Q2 — Molecular Motion & Kinetic Energy · easy · theory
The root-mean-square speed of the molecules of an ideal gas of molar mass $M$ at absolute temperature $T$ is:
A. $\sqrt{\dfrac{RT}{3M}}$
B. $\sqrt{\dfrac{8RT}{\pi M}}$
C. $\sqrt{\dfrac{2RT}{M}}$
D. $\sqrt{\dfrac{3RT}{M}}$ ✓ Correct
Solution: $v_{rms} = \sqrt{\dfrac{3RT}{M}}$. The other expressions are the most probable speed $\sqrt{2RT/M}$ and the mean speed $\sqrt{8RT/\pi M}$.
Q3 — Molecular Motion & Kinetic Energy · easy · theory
The average translational kinetic energy of a single molecule of an ideal gas at absolute temperature $T$ is:
A. $k_B T$
B. $\dfrac{3}{2}k_B T$ ✓ Correct
C. $\dfrac{5}{2}k_B T$
D. $\dfrac{1}{2}k_B T$
Solution: A molecule has three translational degrees of freedom, each carrying $\dfrac{1}{2}k_B T$, so the total translational kinetic energy is $\dfrac{3}{2}k_B T$.
Q4 — Molecular Motion & Kinetic Energy · easy · theory
The average kinetic energy associated with each translational degree of freedom of a gas molecule at absolute temperature $T$ is:
A. $\dfrac{3}{2}k_B T$
B. $k_B T$
C. $\dfrac{1}{2}k_B T$ ✓ Correct
D. $\dfrac{5}{2}k_B T$
Solution: By the law of equipartition of energy, every independent degree of freedom carries an average energy of $\dfrac{1}{2}k_B T$.
Q5 — Molecular Motion & Kinetic Energy · medium · theory
The average kinetic energy of the molecules of an ideal gas depends only on:
A. Its pressure
B. The nature of the gas
C. Its volume
D. Its absolute temperature ✓ Correct
Solution: Since $\bar{E} = \dfrac{3}{2}k_B T$, helium and oxygen at the same temperature have exactly the same average molecular kinetic energy, even though their molecular masses differ.
Q6 — Molecular Motion & Kinetic Energy · medium · numerical
At the same temperature, the ratio of the rms speeds of hydrogen ($M = 2$) and oxygen ($M = 32$) molecules is:
A. $1 : 16$
B. $4 : 1$ ✓ Correct
C. $16 : 1$
D. $1 : 4$
Solution: At a common temperature $v_{rms} \propto \dfrac{1}{\sqrt{M}}$, so $\dfrac{v_{H_2}}{v_{O_2}} = \sqrt{\dfrac{32}{2}} = \sqrt{16} = 4$.
Q7 — Molecular Motion & Kinetic Energy · easy · numerical
If the absolute temperature of a gas is doubled, the rms speed of its molecules becomes:
A. $4$ times
B. $\sqrt{2}$ times ✓ Correct
C. $2$ times
D. Half
Solution: $v_{rms} \propto \sqrt{T}$, so doubling $T$ multiplies the rms speed by $\sqrt{2} \approx 1.41$.
Q8 — Molecular Motion & Kinetic Energy · medium · numerical
A gas is at $27^\circ\text{C}$. The temperature at which the rms speed of its molecules is doubled is:
A. $600\text{ K}$
B. $54\text{ K}$
C. $900\text{ K}$
D. $1200\text{ K}$ ✓ Correct
Solution: $v_{rms} \propto \sqrt{T}$, so doubling the speed needs four times the absolute temperature: $T = 4 \times 300 = 1200\text{ K}$.
Q9 — Molecular Motion & Kinetic Energy · easy · theory
The ideal gas equation for $n$ moles of a gas is:
A. $PV = nkT$
B. $PV = nRT$ ✓ Correct
C. $P = nRTV$
D. $PV = \dfrac{nRT}{2}$
Solution: $PV = nRT$, 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 $PV = Nk_B T$.
Q10 — Molecular Motion & Kinetic Energy · easy · theory
Boyle's law, valid at constant temperature for a fixed mass of gas, states that:
A. $PT = \text{constant}$
B. $\dfrac{P}{T} = \text{constant}$
C. $PV = \text{constant}$ ✓ Correct
D. $\dfrac{V}{T} = \text{constant}$
Solution: At constant $T$ and $n$, the ideal gas equation gives $PV = $ constant, so pressure and volume are inversely proportional.
Q11 — Molecular Motion & Kinetic Energy · easy · theory
Charles' law states that, at constant pressure, the volume of a fixed mass of gas is:
A. Inversely proportional to its absolute temperature
B. Proportional to the square of its absolute temperature
C. Directly proportional to its absolute temperature ✓ Correct
D. Independent of its temperature
Solution: At constant $P$, $PV = nRT$ gives $\dfrac{V}{T} = $ constant, so $V \propto T$ with $T$ measured on the Kelvin scale.
Q12 — Molecular Motion & Kinetic Energy · medium · numerical
An ideal gas at $27^\circ\text{C}$ is heated at constant volume until its pressure doubles. Its final temperature is:
A. $600^\circ\text{C}$
B. $54^\circ\text{C}$
C. $150^\circ\text{C}$
D. $327^\circ\text{C}$ ✓ Correct
Solution: At constant volume $\dfrac{P}{T} = $ constant, so $T_2 = 2 \times 300 = 600\text{ K} = 327^\circ\text{C}$. Note the temperature must be in kelvin before doubling.
Q13 — Molecular Motion & Kinetic Energy · easy · theory
According to kinetic theory, the molecular motion of an ideal gas would cease entirely at:
A. $0^\circ\text{C}$
B. $0\text{ K}$ ✓ Correct
C. $100\text{ K}$
D. $-100^\circ\text{C}$
Solution: Average kinetic energy is $\dfrac{3}{2}k_B T$, which vanishes only when $T = 0\text{ K}$ — absolute zero, or $-273.15^\circ\text{C}$.
Q14 — Molecular Motion & Kinetic Energy · medium · theory
Which of the following is NOT a postulate of the kinetic theory of gases?
A. The molecules are in continuous random motion
B. The molecules exert strong attractive forces on one another at all times ✓ Correct
C. Collisions between molecules are perfectly elastic
D. The volume of the molecules is negligible compared with the volume of the gas
Solution: An ideal gas is assumed to have negligible intermolecular forces except during the brief instants of collision. Real gases depart from ideal behaviour precisely because such forces exist.
Q15 — Molecular Motion & Kinetic Energy · hard · theory
For the molecules of a gas, the correct order of the root-mean-square speed $v_{rms}$, the mean speed $\bar{v}$ and the most probable speed $v_{mp}$ is:
A. $v_{rms} = \bar{v} = v_{mp}$
B. $v_{rms} > \bar{v} > v_{mp}$ ✓ Correct
C. $\bar{v} > v_{rms} > v_{mp}$
D. $v_{mp} > \bar{v} > v_{rms}$
Solution: Their ratios are $\sqrt{3} : \sqrt{8/\pi} : \sqrt{2}$, i.e. $1.73 : 1.60 : 1.41$ in units of $\sqrt{RT/M}$, so the rms speed is the largest.
Q16 — Molecular Motion & Kinetic Energy · easy · theory
Boltzmann's constant $k_B$ is related to the universal gas constant $R$ and Avogadro's number $N_A$ by:
A. $k_B = \dfrac{R}{N_A}$ ✓ Correct
B. $k_B = R + N_A$
C. $k_B = \dfrac{N_A}{R}$
D. $k_B = R N_A$
Solution: $R$ is the gas constant per mole and $k_B$ the gas constant per molecule, so $k_B = \dfrac{R}{N_A} \approx 1.38 \times 10^{-23}\text{ J/K}$.
Q17 — Molecular Motion & Kinetic Energy · medium · theory
Two different ideal gases are kept at the same temperature. The quantity that is necessarily the same for both is:
A. The mean free path of the molecules
B. The density of the gas
C. The average kinetic energy per molecule ✓ Correct
D. The rms speed of the molecules
Solution: Average kinetic energy $\dfrac{3}{2}k_B T$ depends only on temperature. The rms speed also involves molar mass, so it differs between the two gases.
Q18 — Molecular Motion & Kinetic Energy · hard · numerical
The rms speed of oxygen molecules ($M = 0.032\text{ kg/mol}$) at $300\text{ K}$ is approximately ($R = 8.314$):
A. $342\text{ m/s}$
B. $484\text{ m/s}$ ✓ Correct
C. $1930\text{ m/s}$
D. $968\text{ m/s}$
Solution: $v_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3 \times 8.314 \times 300}{0.032}} = \sqrt{2.34 \times 10^5} \approx 484\text{ m/s}$.
Q19 — Molecular Motion & Kinetic Energy · hard · numerical
The rms speed of a gas at $27^\circ\text{C}$ is $400\text{ m/s}$. Its rms speed at $127^\circ\text{C}$ is approximately:
A. $462\text{ m/s}$ ✓ Correct
B. $533\text{ m/s}$
C. $800\text{ m/s}$
D. $346\text{ m/s}$
Solution: $v \propto \sqrt{T}$, so $v_2 = 400\sqrt{\dfrac{400}{300}} = 400 \times 1.155 \approx 462\text{ m/s}$.
Q20 — Molecular Motion & Kinetic Energy · medium · numerical
The average translational kinetic energy of a gas molecule at $300\text{ K}$ is ($k_B = 1.38 \times 10^{-23}\text{ J/K}$):
A. $4.14 \times 10^{-21}\text{ J}$
B. $6.21 \times 10^{-21}\text{ J}$ ✓ Correct
C. $2.07 \times 10^{-21}\text{ J}$
D. $1.04 \times 10^{-20}\text{ J}$
Solution: $\bar{E} = \dfrac{3}{2}k_BT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}\text{ J}$.
Q21 — Molecular Motion & Kinetic Energy · medium · numerical
At the same temperature, the ratio of the rms speeds of helium ($M = 4$) and oxygen ($M = 32$) molecules is approximately:
A. $0.35$
B. $2.83$ ✓ Correct
C. $8.00$
D. $1.41$
Solution: $\dfrac{v_{He}}{v_{O_2}} = \sqrt{\dfrac{32}{4}} = \sqrt{8} \approx 2.83$.
Q22 — Molecular Motion & Kinetic Energy · medium · numerical
The total translational kinetic energy of $3\text{ moles}$ of an ideal gas at $400\text{ K}$ is ($R = 8.314$):
A. $3.00 \times 10^4\text{ J}$
B. $4.99 \times 10^3\text{ J}$
C. $9.98 \times 10^3\text{ J}$
D. $1.50 \times 10^4\text{ J}$ ✓ Correct
Solution: $E = \dfrac{3}{2}nRT = 1.5 \times 3 \times 8.314 \times 400 \approx 1.497 \times 10^4\text{ J}$.
Q23 — Molecular Motion & Kinetic Energy · hard · numerical
The rms speed of nitrogen molecules ($M = 0.028\text{ kg/mol}$) at $300\text{ K}$ is approximately ($R = 8.314$):
A. $267\text{ m/s}$
B. $517\text{ m/s}$ ✓ Correct
C. $484\text{ m/s}$
D. $1934\text{ m/s}$
Solution: $v_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3 \times 8.314 \times 300}{0.028}} = \sqrt{2.67 \times 10^5} \approx 517\text{ m/s}$.
Q24 — Molecular Motion & Kinetic Energy · hard · numerical
The rms speed of hydrogen molecules ($M = 0.002\text{ kg/mol}$) at $300\text{ K}$ is approximately ($R = 8.314$):
A. $3741\text{ m/s}$
B. $1934\text{ m/s}$ ✓ Correct
C. $484\text{ m/s}$
D. $517\text{ m/s}$
Solution: $v_{rms} = \sqrt{\dfrac{3 \times 8.314 \times 300}{0.002}} = \sqrt{3.74 \times 10^6} \approx 1934\text{ m/s}$.
Q25 — Molecular Motion & Kinetic Energy · medium · numerical
The rms speed of the molecules of a gas at $27^\circ\text{C}$ is $500\text{ m/s}$. The temperature at which it becomes $1000\text{ m/s}$ is:
A. $2400\text{ K}$
B. $900\text{ K}$
C. $1200\text{ K}$ ✓ Correct
D. $600\text{ K}$
Solution: $v_{rms} \propto \sqrt{T}$, so doubling the speed requires four times the absolute temperature: $T = 4 \times 300 = 1200\text{ K}$.
Q26 — Molecular Motion & Kinetic Energy · medium · numerical
The average translational kinetic energy of a gas molecule at $600\text{ K}$ is ($k_B = 1.38 \times 10^{-23}\text{ J/K}$):
A. $8.28 \times 10^{-21}\text{ J}$
B. $1.24 \times 10^{-20}\text{ J}$ ✓ Correct
C. $2.48 \times 10^{-20}\text{ J}$
D. $6.21 \times 10^{-21}\text{ J}$
Solution: $\bar{E} = \dfrac{3}{2}k_BT = 1.5 \times 1.38 \times 10^{-23} \times 600 = 1.242 \times 10^{-20}\text{ J}$.
Q27 — Molecular Motion & Kinetic Energy · easy · numerical
The ratio of the rms speeds of the molecules of a gas at $400\text{ K}$ and at $100\text{ K}$ is:
A. $16 : 1$
B. $1 : 2$
C. $4 : 1$
D. $2 : 1$ ✓ Correct
Solution: $v_{rms} \propto \sqrt{T}$, so the ratio is $\sqrt{\dfrac{400}{100}} = \sqrt{4} = 2$.
Q28 — Molecular Motion & Kinetic Energy · medium · numerical
The total translational kinetic energy of one mole of an ideal gas at $300\text{ K}$ is ($R = 8.314$):
A. $3741\text{ J}$ ✓ Correct
B. $2494\text{ J}$
C. $1247\text{ J}$
D. $7483\text{ J}$
Solution: $E = \dfrac{3}{2}nRT = 1.5 \times 1 \times 8.314 \times 300 \approx 3741\text{ J}$.
Q29 — Molecular Motion & Kinetic Energy · medium · numerical
Two moles of an ideal gas occupy $0.05\text{ m}^3$ at $300\text{ K}$. The pressure is ($R = 8.314$):
A. $9.98 \times 10^4\text{ Pa}$ ✓ Correct
B. $1.99 \times 10^5\text{ Pa}$
C. $2.49 \times 10^5\text{ Pa}$
D. $4.99 \times 10^4\text{ Pa}$
Solution: $P = \dfrac{nRT}{V} = \dfrac{2 \times 8.314 \times 300}{0.05} = \dfrac{4988.4}{0.05} \approx 9.98 \times 10^4\text{ Pa}$.
Q30 — Molecular Motion & Kinetic Energy · easy · numerical
An ideal gas at $300\text{ K}$ is cooled at constant volume until its pressure halves. Its final temperature is:
A. $150\text{ K}$ ✓ Correct
B. $75\text{ K}$
C. $273\text{ K}$
D. $600\text{ K}$
Solution: At constant volume $\dfrac{P}{T} = $ constant, so halving the pressure halves the absolute temperature: $T_2 = 150\text{ K}$.