Question

An ideal gas is enclosed in a cylinder at pressure and temperature, $300\text{K}$. The mean time between two successive collisions is $6\times {10}^{-8}\text{s}$. If the pressure is doubled and temperature is increased to $500\text{K}$, the mean time between two successive collisions will be close to:

• A. $3\times {10}^{-6}\text{s}$
• B. $0.5\times {10}^{-8}\text{s}$
• C. $4\times {10}^{-8}\text{s}$
• D. $2\times {10}^{-7}\text{s}$

Answer 1

The correct option is C $4\times {10}^{-8}\text{s}$

The mean time between two successive collision is given by,

$\tau =\frac{1}{2n\pi d^2{v}_{rms}}$

$\text{Here,}n=\frac{N}{V}$

From the ideal gas equation,

$PV=NkT\Rightarrow n=\frac{N}{V}=\frac{P}{kT}$

$\text{Also,}{v}_{rms}=\sqrt{\frac{3RT}{M_o}}$

$\therefore \tau \propto \frac{\sqrt{T}}{P}$
$\Rightarrow \frac{{\tau }_2}{{\tau }_1}=\frac{T_2}{P_2}\times \frac{P_1}{T_1}$

$\Rightarrow \frac{{\tau }_2}{6\times {10}^{-8}}=\frac{\sqrt{500}}{2P}\times \frac{P}{\sqrt{300}}$

$\Rightarrow {\tau }_2\approx 4\times {10}^{-8}\text{s}$

Hence, option $\left(B\right)$ is correct.