Question

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from \( {\tau _1} \) to \( {\tau _2} \) . If \( {{{C_p}} \over {{C_v}}} = \gamma \) for this gas then a good estimate for \( {{{\tau _2}} \over {{\tau _1}}} \) is given by :

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

Hint:

Mean collision time \(\tau\propto\dfrac{\lambda}{v_{rms}}\propto\dfrac{1}{n\sqrt T}\). For an adiabatic process use \(pV^\gamma=\) const and \(TV^{\gamma-1}=\) const.

Answer 1

The Correct Option is A

Step 1: \(\tau=\dfrac{\lambda}{v_{rms}}\), with \(\lambda\propto\dfrac1n\propto V\) and \(v_{rms}\propto\sqrt T\).
Step 2: For an adiabatic process \(TV^{\gamma-1}=\) const, so \(T\propto V^{-(\gamma-1)}\) and \(v_{rms}\propto V^{-(\gamma-1)/2}\).
Step 3: \(\tau\propto V\cdot V^{(\gamma-1)/2}=V^{(\gamma+1)/2}\).
Step 4: With \(V_2=2V_1\): \(\dfrac{\tau_2}{\tau_1}=2^{\frac{\gamma+1}{2}}\).
Answer: (A) \((2)^{\frac{1+\gamma}{2}}\)