Question

Consider an ideal gas confined in an isolated chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as ${V^q}$, where $V$is the volume of gas. The value of $q$ is: $\left( {\gamma = \dfrac{{{C_p}}}{{{C_v}}}} \right)$

Answer 1

To solve this question, we will use the formula for average time of collision between molecules which is the function of number of molecules per unit volume and rms value of velocity of molecules. After this, we will use the relation of the number of atoms per unit and rms velocity with volume and temperature respectively. Finally, we will use the standard relation between temperature and volume for adiabatic processes to determine the required value.

Formulas used:
$\tau = \dfrac{1}{{n\pi \sqrt 2 {v_{rms}}{d^2}}}$
where, $\tau$ is the average time of collision between molecules, $n$ is the number of molecules per unit volume, ${v_{rms}}$ is the rms value of velocity of molecules and $d$ is the average diameter of molecules.
$n \propto \dfrac{1}{V}$
where $n$ is the number of molecules per unit volume and $V$ is the volume of gas
${v_{rms}} \propto \sqrt T$, where ${v_{rms}}$ is the rms value of velocity of molecules and is the temperature of gas
For adiabatic process $T{V^{\gamma - 1}} = {\text{constant}}$

Complete step by step answer:
We know that the average time of collision between molecules if given by the formula
$\tau = \dfrac{1}{{n\pi \sqrt 2 {v_{rms}}{d^2}}}$
In this relation, only the number of molecules per unit volume ($n$) and is the rms value of velocity of molecules (${v_{rms}}$) change with volume and temperature respectively which satisfy the following relations:
$n \propto \dfrac{1}{V} \Rightarrow n = {K_1}{V^{ - 1}}$, where, ${K_1}$is constant
${v_{rms}} \propto \sqrt T \Rightarrow {v_{rms}} = {K_2}{T^{\dfrac{1}{2}}}$, where, ${K_2}$ is constant
From these equations, we can say that
$\tau \propto V{T^{ - \dfrac{1}{2}}}$
But we know that for adiabatic processes, $T{V^{\gamma - 1}} = {\text{constant}}$.
Therefore, we can write
$\tau \propto V{\left( {{V^{1 - \gamma }}} \right)^{ - \dfrac{1}{2}}} \Rightarrow \tau \propto {V^{\dfrac{{\gamma + 1}}{2}}}$
But, we are given that the average time of collision between molecular increases as ${V^q}$
$\therefore q = \dfrac{{\gamma + 1}}{2}$

Note: Here, we have used the concept of the adiabatic process. It is defined as the thermodynamic process in which there is no exchange of heat from the system to its surroundings neither during expansion nor during compression. The adiabatic process can be either reversible or irreversible.