"Kinetic Theory of Gases" — Practice

Question 1

Two gases are at the same temperature. Compare their average translational kinetic energies per molecule and their rms speeds.

Accepted Answer

**Kinetic energies: equal**, at $	frac32k_BT$ per molecule. This depends on temperature alone, not on which gas it is.

**Speeds: different.** From $	frac12mv_{rms}^{2}=	frac32k_BT$, $v_{rms}\propto1/\sqrt{m}$, so the heavier gas moves more slowly.

Helium and xenon at 300 K carry identical energy per molecule, but xenon (33 times heavier) moves $\sqrt{33}\approx5.7$ times slower.

Question 2

Arrange the most probable, mean and root mean square speeds in order and give their ratios.

Accepted Answer

$$v_{mp}<v_{avg}<v_{rms}$$

Ratio $\sqrt2:\sqrt{8/\pi}:\sqrt3 = 1:1.128:1.224$

They differ **because the Maxwell distribution is asymmetric** — it has a long tail toward high speeds. A symmetric distribution would put all three at the same place.

The rms is largest because squaring weights the fast tail most heavily.

Question 3

How many degrees of freedom does a rigid diatomic molecule have, and why is rotation about the bond axis excluded?

Accepted Answer

**Five** — three translational plus two rotational, about the two axes perpendicular to the bond.

Rotation about the **bond axis itself** is excluded because the mass sits in two essentially point-like nuclei on that axis, so the moment of inertia about it is negligible. With $I\approx0$, that mode can store no meaningful energy.

It is the same reason a monatomic atom has no rotational modes at all.

Question 4

A gas at $10^{5}\ 	ext{Pa}$ has density $1.2\ 	ext{kg/m}^3$. Find its rms speed.

Accepted Answer

$P=	frac13ho v_{rms}^{2}\Rightarrow v_{rms}=\sqrt{\dfrac{3P}{ho}}$

$v_{rms}=\sqrt{\dfrac{3	imes10^{5}}{1.2}}=\sqrt{2.5	imes10^{5}}=\mathbf{500\ 	ext{m/s}}$

Notice what was **not** needed: the molar mass, the temperature, or the number of molecules. The density form of the pressure equation gives molecular speeds from purely bulk measurements.

Question 5

At what temperature will oxygen molecules have the same rms speed as hydrogen at $300\ 	ext{K}$? Take molar masses 32 and 2.

Accepted Answer

$v_{rms}=\sqrt{3RT/M}$, so equal speeds require equal $T/M$:

$\dfrac{T_{O_2}}{32}=\dfrac{300}{2}\Rightarrow T_{O_2}=\mathbf{4800\ 	ext{K}}$

Oxygen must be **sixteen times hotter** to match hydrogen's speed.

This is also why Earth has lost most of its hydrogen: at any shared temperature, the lightest molecules are the fastest, and the fastest are the ones that exceed escape velocity.

(Grams are safe here only because this is a **ratio** — in an absolute calculation, $M$ must be in kg/mol.)

Question 6

A vessel holds 1 mole of helium and 1 mole of nitrogen. Find $C_V$ and $\gamma$ for the mixture.

Accepted Answer

$C_V=\dfrac{1(1.5R)+1(2.5R)}{2}=\mathbf{2R}$

Mayer: $C_P=C_V+R=3R$, so $\gamma=\dfrac{3}{2}=\mathbf{1.5}$

The mixture behaves as though $f=4$ — a value **no real molecule has**. That is fine: $f$ here is an effective parameter for the mixture, not a count of modes.

Averaging the two gammas would give $\dfrac{1.67+1.40}{2}=1.535$, which is wrong.

Question 7

The mean free path of a gas is 100 nm at 300 K and 1 atm. Find it at 600 K and 0.5 atm, and state what happens to the collision frequency.

Accepted Answer

$\lambda\propto\dfrac{T}{P}$: doubling $T$ doubles it, halving $P$ doubles it again.

$\lambda=100	imes2	imes2=\mathbf{400\ 	ext{nm}}$

**Collision frequency** $=\dfrac{v_{avg}}{\lambda}$. The speed rises only as $\sqrt{T}$, so by $\sqrt2$, while $\lambda$ rose by 4.

Frequency changes by $\dfrac{\sqrt2}{4}\approx0.35$ — it **falls to about a third**, despite the gas being hotter.

Question 8

Predict $C_V$ for nitrogen at 300 K from equipartition with all modes active, compare with the measured $2.5R$, and explain the difference.

Accepted Answer

**Prediction:** 3 translational + 2 rotational + 1 vibrational counting **twice** $\Rightarrow f=7$, so $C_V=3.5R$.

**Measured:** $2.5R$ — short by exactly $R$, which is precisely the vibrational contribution ($	frac22R$).

**Why:** vibrational energy levels are widely spaced, so at 300 K almost no molecule has enough energy to reach the first excited vibrational state. The mode is **frozen out** and contributes nothing.

Classical equipartition cannot produce this — it predicts every mode is active at every temperature. The resolution is quantum, and hydrogen's stepped $C_V$ curve (1.5R, then 2.5R, then approaching 3.5R) was one of the earliest clear signs that classical physics was incomplete.

Question 9

Molecules move at around $500\ 	ext{m/s}$, yet a smell takes minutes to cross a room. Resolve the paradox quantitatively.

Accepted Answer

The mean free path in air is about $70\ 	ext{nm}$.

Collision frequency $=\dfrac{500}{70	imes10^{-9}}\approx7	imes10^{9}$ per second.

So the molecule is redirected billions of times a second, and its path is a **random walk**, not a straight line. In a random walk the net displacement grows as $\sqrt{N}$ steps rather than $N$.

Crossing 5 m in a straight line would take 10 ms. As a random walk of 70 nm steps it needs roughly $(5/70	imes10^{-9})^{2}$ steps — an enormous number, and minutes of real time.

The speed was never the limiting factor; the number of deflections is.

Question 10

Explain why a real gas is harder to compress than the ideal law predicts at high pressure, but exerts less pressure than predicted at low temperature.

Accepted Answer

**High pressure — the point-mass assumption fails.** The molecules' own volume is no longer negligible against the container, so the free space available is less than $V$. Compression meets extra resistance, and the gas is **harder** to compress than ideal. This is the excluded-volume term in van der Waals.

**Low temperature — the no-interaction assumption fails.** Molecules move slowly enough to be deflected by mutual attraction as they approach the wall, so they strike it less hard. The measured pressure is **lower** than ideal. This is the attractive-pressure term.

A gas is most ideal when it is **far from liquefying**, which is why helium and hydrogen are closest to ideal at ordinary conditions and water vapour is among the furthest.

Question 11

One mole of an ideal gas expands isothermally at $400\ 	ext{K}$ from $5\ 	ext{L}$ to $15\ 	ext{L}$. Find the work done by the gas, and state what happens to its internal energy.

Accepted Answer

$W=nRT\ln\dfrac{V_2}{V_1}=1(8.314)(400)\ln 3$

$W=3325.6(1.0986)=\mathbf{3654\ 	ext{J}}$, done **by** the gas.

**Internal energy is unchanged.** For an ideal gas $U$ depends on temperature alone, and the temperature is fixed — so $\Delta U=0$ and every joule of that work had to be supplied as heat from the surroundings.

*Sanity check on the size:* an isobaric expansion at the initial pressure would have done $P_1\Delta V=\dfrac{nRT}{V_1}\Delta V=3325.6(2)=6651$ J. Isothermal work is smaller, because the pressure falls away as the gas expands rather than being held up.

Question 12

The van der Waals constant $b$ for carbon dioxide is $0.0427\ 	ext{L/mol}$. Estimate the diameter of a carbon dioxide molecule, given that $b$ equals four times the total volume of the molecules in one mole.

Accepted Answer

$b=0.0427\ 	ext{L/mol}=4.27	imes10^{-5}\ 	ext{m}^{3}	ext{/mol}$

$b=4N_A\left(\dfrac43\pi r^{3}ight)$

$4(6.022	imes10^{23})(4.189)r^{3}=4.27	imes10^{-5}$

$1.009	imes10^{25}\,r^{3}=4.27	imes10^{-5} \Rightarrow r^{3}=4.23	imes10^{-30}$

$r=1.62	imes10^{-10}\ 	ext{m}, \qquad d=\mathbf{3.2	imes10^{-10}\ 	ext{m}}$

*Sanity check:* about 0.32 nm, the same order as the $3.7	imes10^{-10}\ 	ext{m}$ used for nitrogen in the mean free path calculation. Molecular diameters really do cluster around a few tenths of a nanometre.

This is the point of the van der Waals constants: they are not fitting parameters with no meaning, but measurable numbers that hand you molecular sizes from bulk pressure readings.

Practice — "Kinetic Theory of Gases"

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