Question

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently, the mean collision time between the gas molecules changes from $t_1$ to $t_2$. If $\frac{C_p}{C_v}=\gamma$ for this gas, then a good estimate for $\frac{t_2}{t_1}$ is given by

• A. $\frac{1}{2}$
• B. ${\left(\frac{1}{2}\right)}^{\frac{\gamma +1}{2}}$
• C. ${\left(\frac{1}{2}\right)}^{\gamma }$
• D. $2$

Answer 1

The correct option is B

${\left(\frac{1}{2}\right)}^{\frac{\gamma +1}{2}}$

The explanation for the correct option

Step 1: Solve for the relation between temperatures

Given: In an adiabatic process, the volume of an ideal gas gets doubled.

Mean free path $\lambda =\left(\frac{RT}{\sqrt{2}{\mathrm{\pi d}}^2{\mathrm{N}}_{\mathrm{A}}\mathrm{P}}\right)$, where

$R=$ideal gas constant

$T=$temperature

$N_A=$Avogadro's constant

$P=$pressure

$V_{RMS}=\sqrt{\frac{3RT}{M}}$, where $M=mass$

As we know that $t=\frac{\lambda }{V_{RMS}}$

$$\begin{gathered} \Rightarrow t=\frac{\frac{RT}{\sqrt{2}{\mathrm{\pi d}}^2{\mathrm{N}}_{\mathrm{A}}\mathrm{P}}}{\sqrt{{\displaystyle \frac{3RT}{M}}}} \\ \Rightarrow t\propto \frac{\sqrt{T}}{P} \end{gathered}$$

$$\begin{gathered} \frac{t_1}{t_2}=\frac{\sqrt{T_1}}{P_1}\times \frac{P_2}{\sqrt{T_2}} \\ \Rightarrow \frac{t_1}{t_2}=\frac{P_2}{P_1}\times \frac{\sqrt{T_1}}{\sqrt{T_2}}....................\left(1\right) \end{gathered}$$

Where, $t_1$ is the initial mean collision time.

$t_2$ is the final mean collision time.

$P_1$ is the initial pressure.

$P_2$ is the final pressure.

$T_1$ is the initial temperature.

$T_2$ is the final temperature.

Now we know

$$\begin{gathered} PV\propto T \\ \Rightarrow \frac{P_1}{P_2}\times \frac{V_1}{V_2}=\frac{T_1}{T_2}.............\left(2\right) \end{gathered}$$

$V_1$ is the initial volume and $V_2$ is the final volume.

As, $V_1=V$and $V_2=2V$ (Given)

For an adiabatic process, $P{V}^{\gamma }=$constant,

Therefore, $$\begin{gathered} P_1{V^{\gamma }}_1=P_2{V^{\gamma }}_2 \\ \frac{P_1}{P_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma } \\ \frac{P_1}{P_2}={\left(\frac{2V}{V}\right)}^{\gamma } \\ \frac{P_1}{P_2}=2^{\gamma } \end{gathered}$$.

Hence, $\frac{P_1}{P_2}=2^{\gamma }.........\left(3\right)$

And $V_1=V$and $V_2=2V$

Putting these values in $\left(2\right)$, we get

$$\begin{gathered} \Rightarrow 2^{\gamma }\times \frac{V}{2V}=\frac{T_1}{T_2} \\ \Rightarrow 2^{\gamma -1}=\frac{T_1}{T_2}..........\left(4\right) \end{gathered}$$

Step 2: Solve for the relation between the mean collision time of gases

Putting the values from $\left(3\right)$ and $\left(4\right)$ in $\left(1\right)$, we get

$$\begin{gathered} \Rightarrow \frac{t_1}{t_2}=\frac{1}{2^{\gamma }}\times \sqrt{2^{\left(\gamma -1\right)}} \\ \Rightarrow \frac{t_1}{t_2}=\frac{1}{2^{\gamma }}\times 2^{\frac{\left(\gamma -1\right)}{2}} \\ \Rightarrow \frac{t_1}{t_2}=2^{\frac{\left(\gamma -1\right)}{2}-\gamma } \\ \Rightarrow \frac{t_1}{t_2}=2^{\frac{\gamma -1-2\gamma }{2}} \\ \Rightarrow \frac{t_1}{t_2}=2^{\frac{-\gamma -1}{2}} \\ \Rightarrow \frac{t_1}{t_2}=2^{-\left(\frac{\gamma +1}{2}\right)} \\ \Rightarrow \frac{t_1}{t_2}={\left(\frac{1}{2}\right)}^{\left(\frac{\gamma +1}{2}\right)} \end{gathered}$$

Hence, option B is correct.