Question

a) consider gas container of volume v, (3.0 l, 1.0 ml = 1 cm³) at atmospheric pressure p, (1.01×10⁵ pa) and temperature t₁ (22°c). if temperature of the container changes to t₂(100.0°c) and double the volume of the container find the ratio of mean free paths of gas molecules?
b) what is the ratio of average time between successive collisions (mean free time) for any given molecule at room - temperature t₁ to t₂(100.0°c)? assume volume of the container remains constant.

Explanation:

Step1: Recall ideal - gas law

The ideal - gas law is $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal - gas constant, and $T$ is temperature in Kelvin.

Step2: For part (a)

We know that $P_1V_1=nRT_1$ and $P_2V_2=nRT_2$. Given $V_2 = 2V_1$, $T_1=(22 + 273)K=295K$, $T_2=(100+273)K = 373K$. From $P_1V_1=nRT_1$ and $P_2V_2=nRT_2$, we can express $P_2$ as $P_2=\frac{P_1V_1T_2}{V_2T_1}$. Substituting the values: $P_2=\frac{P_1\times V_1\times373K}{2V_1\times295K}=\frac{373}{590}P_1\approx0.632P_1$.

Step3: For part (b)

The mean - free time $\tau=\frac{m}{4\sqrt{2}\pi r^{2}p\sqrt{\frac{k_BT}{m}}}=\frac{m^{3/2}}{4\sqrt{2}\pi r^{2}p\sqrt{k_BT}}$, where $m$ is the mass of a gas molecule, $r$ is the radius of the gas molecule, $p$ is the pressure, $k_B$ is the Boltzmann constant, and $T$ is the temperature. Since the volume is constant, from $PV = nRT$, $P\propto T$. The mean - free time $\tau\propto\frac{1}{P\sqrt{T}}$. Let $\tau_1$ be the mean - free time at $T_1$ and $P_1$, and $\tau_2$ be the mean - free time at $T_2$ and $P_2$. Since $P_1V_1=nRT_1$ and $P_2V_1=nRT_2$ (constant volume, $V_1 = V_2$), $P_2=\frac{T_2}{T_1}P_1$. Then $\frac{\tau_1}{\tau_2}=\frac{P_2\sqrt{T_2}}{P_1\sqrt{T_1}}$. Substituting $P_2=\frac{T_2}{T_1}P_1$ into the above formula, we get $\frac{\tau_1}{\tau_2}=\sqrt{\frac{T_2^{3}}{T_1^{3}}}$. Plugging in $T_1 = 295K$ and $T_2 = 373K$, $\frac{\tau_1}{\tau_2}=\sqrt{(\frac{373}{295})^{3}}\approx1.37$.

Answer:

a) The new pressure $P_2\approx0.632P_1$.
b) The ratio of the mean - free time $\frac{\tau_1}{\tau_2}\approx1.37$.