Question

The mean free path of a molecule in a gas is $300\mathrm{nm}$. What will the mean free path be if the gas temperature is doubled at (a) Constant volume and (b) Constant pressure?

Short answer

a) The mean free path be if the gas temperature is doubled at Constant volume is $\lambda =300\mathrm{nm}$

b) The mean free path be if the gas temperature is doubled at Constant pressure is

Step-by-step solution

Given Information (Part a)

Mean free path in a gas$=300nm$

Explanation (Part a)

According to the ideal gas law, the volume of the container $V$, the pressure $p$exerted by the gas, the temperature $T$of the gas, and the number of moles $n$of the gas in the container are all related.

$pV=N{k}_{\mathrm{B}}T$

localid="1648288031262" $\frac{N}{V}=\frac{p}{k_{\mathrm{B}}T}\to \left(1\right)$

localid="1648283676708" $k_B$is Boltzmann's constant and in SI unit its value is

$k_{\mathrm{B}}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

Due to collisions with other molecules, a molecule undergoing distance travel experiences a delay between diffusing to a different position due to these collisions.

The molecules need a certain amount of time to diffuse to a new position.

In equation (20.3), the mean free path $\lambda$is the distance the molecules travel between collisions on average.

localid="1648288045814" $\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}\to \left(2\right)$

Use the expression of $N/V$into equation (2) to get an equation for $\lambda$in terms of $T$and $p$by

localid="1648283857861" $\lambda =\frac{1}{4\sqrt{2}\pi \left(p/k_{\mathrm{B}}T\right)r^2}$

localid="1648288066442" $=\frac{k\mathrm{B}T}{4\sqrt{2}\pi p{r}^2}\to \left(3\right)$

Explanation (Part a)

Ideal gas law states that,

$pV=nRT,$

Increasing the temperature and volume will increase the pressure as well$T_2=2T_1$and $p_2=2p_1$.

From equation (3), the mean free path is $\lambda \propto \frac{1}{p}$and $\lambda \propto T$.

So, for two instants ${\lambda }_1$and ${\lambda }_2$we get the next

$\begin{array}{c}\frac{{\lambda }_1}{{\lambda }_2}=\left(\frac{T_1}{T_2}\right)\left(\frac{p_2}{p_1}\right)\end{array}$

$\begin{array}{c}\frac{{\lambda }_1}{{\lambda }_2}=\left(\frac{T_1}{2T_1}\right)\left(\frac{2p_1}{p_1}\right)\end{array}$

role="math" localid="1648284087661" $\frac{{\lambda }_1}{{\lambda }_2}=1$

${\lambda }_2={\lambda }_1$

Due to the fact that the mean free path does not change as the temperature decreases, so as the temperature doubles, the mean free path remains the same.

Final Answer (Part a)

Hence, the mean free path be if the gas temperature is doubled at Constant volume is$\lambda =300nm$

Given Information (Part b)

Mean free path in a gas$=300nm$

Explanation (Part b)

From equation (3), the mean free path is $\lambda \propto \frac{1}{p}$and $\lambda \propto T$.

So, for two instants ${\lambda }_1$and ${\lambda }_2$when $p_1=p_2$we get the next

$\begin{array}{c}\frac{{\lambda }_1}{{\lambda }_2}=\left(\frac{T_1}{T_2}\right)\left(\frac{p_2}{p_1}\right)\end{array}$

$\begin{array}{c}\frac{{\lambda }_1}{{\lambda }_2}=\left(\frac{T_1}{2T_1}\right)\left(\frac{p_1}{p_1}\right)\end{array}$

localid="1648284399879" $\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{2}$

${\lambda }_2=2{\lambda }_1$

It can be seen that the mean free path doubles when the temperature doubles and the pressure is constant, as can be seen from the figure.

$\lambda =2(300\mathrm{nm})=600\mathrm{nm}$

Final Answer (Part b)

Hence, the mean free path be if the gas temperature is doubled at Constant pressure is$600nm.$