IMO Practice Test — Kinetic Theory
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
The rms speeds of two gases of molar masses $M_1$ and $M_2$ at the same temperature are in the ratio:
$M_1:M_2$
$\sqrt{M_1}:\sqrt{M_2}$
$\sqrt{M_2}:\sqrt{M_1}$
$M_2:M_1$
Explanation: $v_{rms}\propto\frac{1}{\sqrt{M}}$, so $\frac{v_1}{v_2}=\sqrt{\frac{M_2}{M_1}}$, i.e. $\sqrt{M_2}:\sqrt{M_1}$.
Question 2 of 12
If both the pressure and volume of an ideal gas are doubled, the rms speed of its molecules changes by a factor of:
$1$
$2$
$\sqrt{2}$
$4$
Explanation: $PV\propto T$; doubling both quadruples $T$, and $v_{rms}\propto\sqrt{T}$, so it doubles.
Question 3 of 12
The ratio of the rms speed to the average kinetic energy temperature dependence is such that doubling $v_{rms}$ requires the temperature to be:
doubled
tripled
quadrupled
halved
Explanation: $v_{rms}\propto\sqrt{T}$; to double it, $T$ must be quadrupled.
Question 4 of 12
One mole of a monatomic gas is mixed with one mole of a diatomic gas. The $C_v$ of the mixture is:
$\frac{3}{2}R$
$\frac{5}{2}R$
$2R$
$4R$
Explanation: $C_v=\frac{n_1C_{v1}+n_2C_{v2}}{n_1+n_2}=\frac{\frac{3}{2}R+\frac{5}{2}R}{2}=2R$.
Question 5 of 12
For a gas whose molecules have $f$ degrees of freedom, the ratio $\frac{C_p}{C_v}$ equals:
$\frac{f}{2}$
$1+\frac{2}{f}$
$\frac{2}{f}$
$\frac{f+2}{2}$
Explanation: $C_v=\frac{f}{2}R$, $C_p=\frac{f+2}{2}R$, so $\gamma=1+\frac{2}{f}$.
Question 6 of 12
The total internal energy of one mole of a diatomic gas at temperature $T$ is:
$\frac{3}{2}RT$
$\frac{5}{2}RT$
$3RT$
$RT$
Explanation: $U=\frac{f}{2}RT=\frac{5}{2}RT$ for $f=5$.
Question 7 of 12
If the molecular diameter is halved at fixed number density, the mean free path becomes:
half
double
four times
one-fourth
Explanation: $\lambda\propto\frac{1}{d^2}$; halving $d$ multiplies $\lambda$ by $4$.
Question 8 of 12
Two gases A (monatomic) and B (diatomic) are at the same temperature. The ratio of the average KE per molecule of A to B is:
$3:5$
$5:3$
$1:1$
$3:2$
Explanation: Average total KE per molecule is $\frac{f}{2}k_BT$, but the average translational KE ($\frac{3}{2}k_BT$) is equal; per molecule energy share of translation is the same, ratio $1:1$.
Question 9 of 12
At a temperature of 600 K the rms speed of a gas is $v$. At 150 K the rms speed is:
$\frac{v}{2}$
$\frac{v}{4}$
$2v$
$\frac{v}{\sqrt{2}}$
Explanation: $v_{rms}\propto\sqrt{T}$; $\sqrt{\frac{150}{600}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$, so it is $\frac{v}{2}$.
Question 10 of 12
The pressure of an ideal gas is doubled while keeping the number density fixed. The rms speed becomes:
unchanged
double
$\sqrt{2}$ times
four times
Explanation: $P=\frac{1}{3}nm\overline{v^2}$ at fixed $n$ means $\overline{v^2}\propto P$, so $v_{rms}\propto\sqrt{P}$, i.e. $\sqrt{2}$ times.
Question 11 of 12
A vessel contains equal masses of helium ($M=4$) and oxygen ($M=32$) at the same temperature. The ratio of their rms speeds (He : O$_2$) is:
$1:1$
$2\sqrt{2}:1$
$1:2\sqrt{2}$
$8:1$
Explanation: $\frac{v_{He}}{v_{O_2}}=\sqrt{\frac{32}{4}}=\sqrt{8}=2\sqrt{2}$, so $2\sqrt{2}:1$.
Question 12 of 12
If a gas is compressed to half its volume at constant temperature, its mean free path becomes:
double
half
unchanged
four times
Explanation: $\lambda\propto\frac{1}{n}$; halving the volume doubles $n$, so $\lambda$ halves.