Online Test — Kinetic Theory
20 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 20
Boyle's law (constant temperature) states that:
$PV=\text{constant}$
$\frac{V}{T}=\text{constant}$
$\frac{P}{T}=\text{constant}$
$PT=\text{constant}$
Explanation: At constant $T$, $PV=\text{constant}$, so $P\propto\frac{1}{V}$.
Question 2 of 20
The ideal gas equation in terms of moles is:
$PV=nRT$
$PV=Nk_BT^2$
$PV=\frac{nRT}{2}$
$PVT=nR$
Explanation: $PV=nRT$, equivalently $PV=Nk_BT$.
Question 3 of 20
The value of the universal gas constant $R$ is about:
$1.38\times10^{-23}\ \text{J/K}$
$8.314\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$
$6.022\times10^{23}$
$9.8\ \text{N/kg}$
Explanation: $R=8.314\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$.
Question 4 of 20
Avogadro's law says that equal volumes of gases at the same $T$ and $P$ contain:
equal masses
equal numbers of molecules
equal volumes of liquid
equal densities
Explanation: Equal volumes at the same $T$ and $P$ have equal numbers of molecules.
Question 5 of 20
Which is NOT an assumption of the kinetic theory?
Random molecular motion
Elastic collisions
Negligible molecular volume
Constant strong attraction between molecules
Explanation: Kinetic theory assumes no inter-molecular force except during collisions.
Question 6 of 20
The kinetic theory expression for pressure is:
$P=\frac{1}{3}\rho\,\overline{v^2}$
$P=\frac{1}{2}\rho\,\overline{v^2}$
$P=\rho\,\overline{v^2}$
$P=\frac{2}{3}\rho\,\overline{v^2}$
Explanation: $P=\frac{1}{3}\frac{Nm\,\overline{v^2}}{V}=\frac{1}{3}\rho\,\overline{v^2}$.
Question 7 of 20
The average translational kinetic energy of a gas molecule is:
$\frac{1}{2}k_BT$
$k_BT$
$\frac{3}{2}k_BT$
$\frac{5}{2}k_BT$
Explanation: $\frac{1}{2}m\,\overline{v^2}=\frac{3}{2}k_BT$.
Question 8 of 20
Temperature of a gas is a measure of the:
total volume of molecules
average kinetic energy of molecules
number of molecules
size of molecules
Explanation: Kinetic interpretation: $T\propto$ average translational KE of the molecules.
Question 9 of 20
The rms speed of gas molecules is:
$\sqrt{\frac{3RT}{M}}$
$\sqrt{\frac{2RT}{M}}$
$\sqrt{\frac{RT}{M}}$
$\frac{3RT}{M}$
Explanation: $v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3k_BT}{m}}$.
Question 10 of 20
If the absolute temperature of a gas is quadrupled, its rms speed becomes:
double
four times
half
unchanged
Explanation: $v_{rms}\propto\sqrt{T}$; quadrupling $T$ doubles $v_{rms}$.
Question 11 of 20
At the same temperature, the gas with the highest rms speed is the one with the:
largest molar mass
smallest molar mass
largest molecules
highest pressure
Explanation: $v_{rms}\propto\frac{1}{\sqrt{M}}$, so the lightest gas is fastest.
Question 12 of 20
The number of degrees of freedom of a diatomic molecule (ordinary $T$) is:
3
5
6
7
Explanation: 3 translational + 2 rotational $=5$.
Question 13 of 20
The average energy associated with each degree of freedom per molecule is:
$\frac{1}{2}k_BT$
$k_BT$
$\frac{3}{2}k_BT$
$\frac{f}{2}k_BT$
Explanation: Law of equipartition: each degree of freedom carries $\frac{1}{2}k_BT$.
Question 14 of 20
The molar specific heat at constant volume of a monatomic gas is:
$\frac{3}{2}R$
$\frac{5}{2}R$
$3R$
$R$
Explanation: $C_v=\frac{f}{2}R=\frac{3}{2}R$ for $f=3$.
Question 15 of 20
Mayer's relation states that:
$C_p-C_v=R$
$C_p+C_v=R$
$C_p=C_v$
$C_pC_v=R$
Explanation: $C_p=C_v+R$ for one mole of an ideal gas.
Question 16 of 20
The value of $\gamma$ for a monatomic gas is:
$\frac{4}{3}$
$\frac{7}{5}$
$\frac{5}{3}$
$1$
Explanation: With $f=3$, $\gamma=1+\frac{2}{3}=\frac{5}{3}\approx1.67$.
Question 17 of 20
The mean free path of a gas molecule is given by:
$\lambda=\frac{1}{\sqrt{2}\,\pi d^2 n}$
$\lambda=\sqrt{2}\,\pi d^2 n$
$\lambda=\frac{n}{\pi d^2}$
$\lambda=\pi d^2 n$
Explanation: $\lambda=\frac{1}{\sqrt{2}\,\pi d^2 n}$, where $n$ is the number density.
Question 18 of 20
A gas with $\gamma=1.33$ is most likely:
monatomic
diatomic
polyatomic
a vacuum
Explanation: $\gamma=\frac{4}{3}\approx1.33$ corresponds to $f=6$ (polyatomic).
Question 19 of 20
A gas at $0^\circ\text{C}$ is heated to $273^\circ\text{C}$ at constant pressure. Its volume becomes:
unchanged
double
half
four times
Explanation: $T_1=273$ K, $T_2=546$ K; $V\propto T$, so the volume doubles.
Question 20 of 20
At a given temperature, the average translational KE of molecules of different gases is:
greater for heavier gases
greater for lighter gases
the same for all gases
zero for inert gases
Explanation: $\frac{3}{2}k_BT$ depends only on temperature, so it is the same for all gases.