83 free MCQs on Kinetic Theory, each with its own worked answer and explanation. Kinetic theory of gases connects molecular motion to pressure and temperature, explaining gas laws, specific heats, and molecular speeds.
83 practice questions on Kinetic Theory, sorted Easy → Hard. Try each one first, then open its answer page for the worked explanation. Want the full theory first? Read the Kinetic Theory notes.
Not all gas molecules move at the same speed - the Maxwell-Boltzmann distribution shows the spread, with the most probable speed vp slightly less than the average vavg, which is slightly less than the root-mean-square speed vrms used in the pressure formula.
Easy - 24 questions
Q1.
The Boltzmann constant k equals the gas constant R divided by:
The number of molecules per unit volume of an ideal gas at 300 K and pressure 1.0 × 10<sup>5</sup> Pa is about (k = 1.38 × 10<sup>-23</sup> J/K):
Two gases, hydrogen (M=2 g/mol) and oxygen (M=32 g/mol), are at the same temperature. The ratio of their rms speeds (v<sub>H</sub><sub>2</sub> / v<sub>O</sub><sub>2</sub>) is:
For a gas mixture of monatomic and diatomic molecules with degrees of freedom f<sub>1</sub> and f<sub>2</sub> respectively in equal number of moles, the average degrees of freedom of the mixture is:
If a gas obeys PV = (2/3)E where E is the total translational kinetic energy of the gas, this kinetic theory result combined with PV = nRT shows that E equals:
The mean free path of the molecules of a gas kept in a rigid closed container is λ. If the absolute temperature is doubled while the volume stays fixed, the mean free path:
One mole of a monatomic gas (C<sub>v</sub> = 3R/2) is mixed with two moles of a diatomic gas (C<sub>v</sub> = 5R/2). The molar specific heat at constant volume of the mixture is:
A gas of N molecules is enclosed in a container. If the container volume is suddenly doubled at constant temperature (free expansion, no heat exchange), the rms speed of the molecules:
A Doubles, as if the rms speed scaled directly with the container volume
B Halves, as if the rms speed scaled inversely with the container volume
C Remains the same since temperature is unchanged
D Becomes zero, as if all molecular motion stopped after expansion
Using kinetic theory, derive the relationship: if a gas has n moles and the total translational KE is (3/2)nRT, then for a diatomic gas with rotational energy also included, the total internal energy U at temperature T (f=5) is:
A vessel contains N<sub>2</sub> gas at temperature T. If the temperature is raised such that the gas molecules begin to dissociate into individual N atoms at very high T, the degrees of freedom of the system effectively:
A Increases from 5 to 6, since atoms mainly have translational freedom but more molecules exist in routine practice
B Decreases from 5 to 3 per resulting particle, since atoms only have translational degrees of freedom
C Stays at 5 overall in most cases under typical conditions according to standard textbooks in general practice
D Becomes 7 as frequently described in most textbook accounts during normal conditions as generally observed
Two ideal gas samples, A (monatomic) and B (diatomic), have equal moles and are at the same temperature. The ratio of their total internal energies U<sub>A</sub> : U<sub>B</sub> is:
If the most probable speed of gas molecules is v<sub>p</sub> = √(2RT/M), and the rms speed is v<sub>rms</sub> = √(3RT/M), the ratio v<sub>rms</sub>/v<sub>p</sub> is:
A closed rigid container has an ideal monatomic gas. Heat Q is supplied at constant volume, raising the temperature by ΔT. The fraction of heat that goes into increasing the rotational kinetic energy of the molecules is:
A 0, since monatomic gas molecules have no rotational degrees of freedom
B 2/5, the fraction associated with two rotational degrees of freedom in a diatomic gas
C 3/5, the fraction associated with translational energy in a diatomic gas
D 1/2, an arbitrary even split between translational and rotational energy
The pressure of an ideal gas is given by P = (1/3)nm(v<sub>rms</sub>)², where n is number density and m is the mass of each molecule. If the gas is compressed isothermally to half its volume, the number density n doubles. The new pressure compared to the original is:
Mean free path of a gas molecule is found to be λ at pressure P and temperature T. If the pressure is doubled at constant temperature, the new mean free path becomes:
An ideal gas undergoes a process in which its pressure and volume are related by PV² = constant. Starting from an initial temperature T, if the volume of the gas doubles, the final temperature of the gas is: