Correct answer is (D)

Time (t) = $\frac{V}{4\pi \sqrt{2}{r}^{2}vN}$ ....(1)

Here, v = most probable speed
= $\sqrt{\frac{2RT}{\pi M}}$

$\Rightarrow$ v $\propto$ $\sqrt{T}$

$\therefore$ From (1),

t $\propto$ $\frac{1}{\sqrt{T}}$

Correct answer is (C)

The mean free path of molecules of an ideal gas is given as:

$\lambda$ = $\frac{V}{\sqrt{2}\pi d^{2}N}$

where : V = Volume of container
N = No of molecules

Mean free path is independent of temperature hence with increasing temp since volume of container does not change (closed container), so mean free path is unchanged.

Average collision time = $\lambda$ V_av

and V_av $\propto$ $\sqrt{T}$

$\therefore$ Average collision time $\propto$ $\frac{1}{\sqrt{T}}$

Hence with increase in temperature the average collision time decreases.

Correct answer is (B)

$\lambda =\frac{1}{\sqrt{2}\pi d^{2}n}$

Mean free time, t = $\frac{\lambda }{v}$

Also v $\propto$ $\sqrt{\frac{T}{M}}$

$\therefore$ t $\propto$ $\frac{\sqrt{M}}{d}$

$\frac{t_{Ar}}{t_{xe}}$ = $\frac{d_{Xe}^2}{d_{Ar}^2}\times \sqrt{\frac{M_{Ar}}{M_{Xe}}}$

= ${\left(\frac{0.1}{0.07}\right)}^2\times \sqrt{\frac{40}{140}}$

= 1.09

$\therefore$ Nearest possible answer is 1.83.

Correct answer is (B)

Mean free time = Mean free path Average speed

= $\frac{\frac{1}{\sqrt{2}\pi D^{2}n}}{\sqrt{\frac{8RT}{\pi M}}}$

$\therefore$ t $\propto$ $\frac{1}{\sqrt{T}}$

Correct answer is (A)

$\tau$ $\propto$ $\frac{V}{\sqrt{T}}$ ....(1)

Also we know, PV^$\gamma$ = k

We know, PV = nRT

$\Rightarrow$ P $\propto$ $\frac{T}{V}$

$\therefore$ $\left(\frac{T}{V}\right)$V^$\gamma$ = k

$\Rightarrow$TV^{$\gamma$ - 1} = k

$\Rightarrow$ T $\propto$ V^{1 - $\gamma$}

Using this value in equation (1)

$\tau$ $\propto$ $\frac{V}{V^{\frac{1-\gamma }{2}}}$

$\Rightarrow$ $\tau$ $\propto$ $V^{1-\frac{1-\gamma }{2}}$

$\Rightarrow$ $\tau$ $\propto$ $V^{\frac{\gamma +1}{2}}$

$\therefore$ $\frac{{\tau }_2}{{\tau }_1}$ = ${\left(\frac{2V}{V}\right)}^{{}^{\frac{\gamma +1}{2}}}$ = ${\left(2\right)}^{\frac{1+\gamma }{2}}$