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Thermodynamics Part 2

Section: Physics

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Question : 66 of 100

Marks: +1, -0

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

\tau_2

\frac{C_p}{C_v}=\gamma

for this gas then a good estimate for

\frac{\tau_2}{\tau_1}

[7 Jan 2020 Shift 2]

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

Solution:

\; \text{ Mean free path, }\; \lambda = \left( \frac{RT}{\sqrt{2}\pi d^{2} N_A P} \right)

V_{rms} = \sqrt{ \frac{3RT}{M} }

\therefore \tau = \frac{\lambda}{V_{rms}} = \frac{RT}{\sqrt{2}\pi d^{2} N_A P} \times \sqrt{\frac{M}{3RT}}

\tau \propto \frac{\sqrt{T}}{P}

\frac{\tau_1}{\tau_2} = \frac{\sqrt{T_1}}{P_1} \times \frac{P_2}{\sqrt{T_2}} = \left( \frac{P_2}{P_1} \right) \sqrt{\frac{T_1}{T_2}}

PV = nRT

\frac{P_1}{P_2} \times \frac{V_1}{V_2} = \frac{T_1}{T_2}

\Rightarrow 2^{\gamma} \times \frac{V}{2V} = \frac{T_1}{T_2}

\Rightarrow \frac{T_1}{T_2} = (2^{\gamma-1})

\therefore \frac{\tau_1}{\tau_2} = \left( \frac{P_2}{P_1} \right) \times \sqrt{\frac{T_1}{T_2}} = \frac{1}{2^{\gamma}} \times \sqrt{2^{\gamma-1}} = \frac{2^{\frac{\gamma-1}{2}}}{2^{\gamma}}

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

\Rightarrow \frac{\tau_1}{\tau_2} = \left( \frac{1}{2} \right)^{\frac{\gamma+1}{2}}

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