"Kinetic Theory of Gases" — Practice

Question 1

Find the mean free path of oxygen at $27^{\circ}$C and $2$ atm, taking the molecular diameter as $3.0	imes10^{-10}$ m. State how the answer changes if the pressure is halved at constant temperature.

Accepted Answer

$\lambda=\dfrac{1.38	imes10^{-23}	imes300}{1.414	imes3.1416	imes\left(3.0	imes10^{-10}ight)^{2}	imes2.02	imes10^{5}}$

Denominator $=1.414	imes3.1416	imes9	imes10^{-20}	imes2.02	imes10^{5}=8.08	imes10^{-14}$

$\lambda=\dfrac{4.14	imes10^{-21}}{8.08	imes10^{-14}}=5.1	imes10^{-8}$ m

At constant $T$, $\lambda\propto1/P$, so halving the pressure **doubles** $\lambda$ to $1.02	imes10^{-7}$ m.

Question 2

A gas is contained in a rigid sealed vessel and its absolute temperature is quadrupled. State what happens to the mean free path, the mean speed and the collision frequency, with reasons.

Accepted Answer

The vessel is rigid, so $n$ is unchanged and therefore $\lambda=1/\sqrt2\,n\pi d^{2}$ is **unchanged**.

$\bar v\propto\sqrt T$, so it doubles.

$z=\bar v/\lambda$ therefore also doubles.

The pressure quadruples, which is why the alternative form $\lambda=kT/\sqrt2\pi d^{2}P$ still returns the same value — the $T$ and $P$ increases cancel exactly.

Question 3

Estimate the time for a molecule with $\lambda=6	imes10^{-8}$ m and $\bar v=500$ m s$^{-1}$ to diffuse a net distance of $1$ m, and compare it with the time to travel $1$ m in a straight line.

Accepted Answer

$N=\left(\dfrac{1}{6	imes10^{-8}}ight)^{2}=2.8	imes10^{14}$ steps

$z=\dfrac{500}{6	imes10^{-8}}=8.3	imes10^{9}$ s$^{-1}$

$t=\dfrac{2.8	imes10^{14}}{8.3	imes10^{9}}=3.4	imes10^{4}$ s, about $9$ hours.

In a straight line it would take $1/500=2$ ms — a factor of roughly $17$ million.

Question 4

For nitrogen at $400$ K, find $v_p$, $\bar v$ and $v_{rms}$, taking $M=28$ g mol$^{-1}$.

Accepted Answer

$v_p=\sqrt{\dfrac{2	imes8.314	imes400}{0.028}}=\sqrt{2.376	imes10^{5}}=487$ m s$^{-1}$

$\bar v=1.128	imes487=550$ m s$^{-1}$

$v_{rms}=1.225	imes487=597$ m s$^{-1}$

The ordering $v_p<\bar v<v_{rms}$ always holds, because the distribution has a long high-speed tail that pulls the higher moments up.

Question 5

Compare the number of molecules with speeds near $	frac12v_p$ with the number near $v_p$ at the same temperature.

Accepted Answer

$\dfrac{f\left(v_p/2ight)}{f\left(v_pight)}=\dfrac{\left(v_p/2ight)^{2}e^{-1/4}}{v_p^{2}e^{-1}}=\dfrac14 e^{3/4}$

$=0.25	imes2.117=0.529$

About **53%** as many.

Below the peak the $v^{2}$ factor dominates the fall, while above it the exponential does — which is what makes the curve asymmetric.

Question 6

A mixture of helium and methane effuses through a pinhole. Find the ratio of their effusion rates, and state which gas is enriched in the escaping stream. Take $M=4$ and $16$ g mol$^{-1}$.

Accepted Answer

$\dfrac{r_{He}}{r_{CH_4}}=\sqrt{\dfrac{16}{4}}=2$

Helium effuses **twice** as fast, so the escaping stream is enriched in helium and the residue is enriched in methane.

A factor of four in mass gives only a factor of two in rate, which is the square root doing what it always does to separation problems.

Question 7

How many molecules strike each square metre of a wall per second in argon at $300$ K and $0.5$ atm? Take $\bar v=399$ m s$^{-1}$.

Accepted Answer

$n=\dfrac{P}{kT}=\dfrac{0.5	imes1.01	imes10^{5}}{1.38	imes10^{-23}	imes300}=1.22	imes10^{25}$ m$^{-3}$

$\Phi=	frac14\left(1.22	imes10^{25}ight)(399)=1.22	imes10^{27}$ m$^{-2}$ s$^{-1}$

Steady pressure is a statistical average over this number of impacts, and the fractional fluctuation goes as $1/\sqrt N$ — utterly negligible at this scale.

Question 8

A gas has $\gamma=1.5$. Find the number of active degrees of freedom, and identify a molecular arrangement consistent with it.

Accepted Answer

$1.5=1+\dfrac2f\qquad\Rightarrow\qquad f=4$

Four active quadratic terms — for instance three translational modes plus one rotational mode, or three translational plus a half-excited pair of rotations.

A rigid diatomic at room temperature gives $f=5$ and $\gamma=1.4$; a monatomic gives $f=3$ and $\gamma=1.67$. A value of $1.5$ therefore lies between the two, indicating partial excitation.

Question 9

Sketch in words how the molar heat capacity at constant volume of hydrogen varies from $50$ K to $5000$ K, and give the three plateau values with the corresponding $\gamma$.

Accepted Answer

Below about $100$ K only translation is active: $C_V=	frac32R$, $\gamma=1.67$.

Around room temperature rotation is fully excited: $C_V=	frac52R$, $\gamma=1.40$.

Above a few thousand kelvin vibration contributes its two quadratic terms: $C_V=	frac72R$, $\gamma=1.29$.

The curve rises in steps rather than smoothly, and classical equipartition offers no explanation at all for the steps — the resolution is quantum.

Question 10

A vessel holds $2$ mol of helium and $4$ mol of a rigid diatomic gas. Find $C_V$, $C_P$ and $\gamma$ for the mixture.

Accepted Answer

$C_{V,mix}=\dfrac{2\left(	frac32Right)+4\left(	frac52Right)}{6}=\dfrac{3R+10R}{6}=2.167R$

$C_{P,mix}=3.167R$

$\gamma=\dfrac{3.167}{2.167}=1.462$

The value lies between $1.40$ and $1.67$ and closer to the diatomic end, as it must, since the diatomic supplies two thirds of the moles.

Question 11

Assuming an isothermal atmosphere at $250$ K with $M=29$ g mol$^{-1}$, find the height at which the density falls to one tenth of its sea-level value.

Accepted Answer

$H=\dfrac{RT}{Mg}=\dfrac{8.314	imes250}{0.029	imes9.8}=7315$ m

$\dfrac{n}{n_0}=e^{-h/H}=0.1\qquad\Rightarrow\qquad h=H\ln10$

$h=7315	imes2.303=1.68	imes10^{4}$ m, about $17$ km.

The real value is lower, since temperature falls with altitude and the isothermal assumption overestimates the scale height.

Question 12

A van der Waals gas has $a=1.36$ atm L$^{2}$ mol$^{-2}$ and $b=3.18	imes10^{-2}$ L mol$^{-1}$. Find its critical temperature, critical pressure and Boyle temperature.

Accepted Answer

$T_c=\dfrac{8a}{27Rb}=\dfrac{8	imes1.36}{27	imes0.0821	imes0.0318}=\dfrac{10.88}{0.0705}=154$ K

$P_c=\dfrac{a}{27b^{2}}=\dfrac{1.36}{27	imes1.011	imes10^{-3}}=49.8$ atm

$T_B=\dfrac{a}{Rb}=\dfrac{1.36}{0.0821	imes0.0318}=521$ K

Note that $T_B=	frac{27}{8}T_c$ for every van der Waals gas — a useful consistency check.

Practice — "Kinetic Theory of Gases"

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