Join WhatsApp Icon JEE WhatsApp Group
Question 7

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently, the mean collision time between the gas molecules changes from $$\tau_1$$ to $$\tau_2$$. If $$\frac{C_p}{C_v} = \gamma$$ for this gas then a good estimate for $$\frac{\tau_2}{\tau_1}$$ is given by

The expression for mean collision time $$\tau$$ is given by:

$$\tau = \frac{1}{\sqrt{2} \pi d^2 n v_{\text{avg}}}$$

Since number density $$n = \frac{N}{V} \propto \frac{1}{V}$$ and average speed $$v_{\text{avg}} \propto \sqrt{T}$$, we have:

$$\tau \propto \frac{V}{\sqrt{T}}$$

For an adiabatic process:

$$T V^{\gamma-1} = \text{constant} \implies T \propto V^{-(\gamma-1)}$$

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

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

The volume is doubled ($$V_2 = 2V_1$$):

$$\frac{\tau_2}{\tau_1} = \left(\frac{V_2}{V_1}\right)^{\frac{\gamma+1}{2}} = (2)^{\frac{\gamma+1}{2}} = \left(\frac{1}{2}\right)^{-\frac{\gamma+1}{2}}$$

Get AI Help

Ask AI