First slide

Thermodynamic processes

Question

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $τ_{1}$ to $τ_{2}$ . If $\frac{C_{P}}{C_{V}} = γ$ for this gas then a good estimate for $\frac{τ_{1}}{τ_{2}}$ is given by:

Difficult

A

$\frac{1}{2}$
B

2
C

${(\frac{1}{2})}^{γ}$
D

${(\frac{1}{2})}^{\frac{γ + 1}{2}}$

Solution

$M e a n f r e e t i m e = \frac{λ}{V_{r m s}} = \frac{\frac{R T}{\sqrt{2} π d^{2} N_{A} P}}{\sqrt{\frac{3 R T}{M}}}$
$t \propto \frac{\sqrt{T}}{P} \propto \frac{\sqrt{T}}{\frac{R T}{V}} \propto \frac{V \sqrt{T}}{T} \propto \frac{V}{\sqrt{T}}$
$∴ t \propto \frac{V}{\sqrt{\frac{1}{V^{γ - 1}}}} \propto \frac{V}{\frac{1}{V^{\frac{γ - 1}{2}}}} \propto V V^{\frac{γ - 1}{2}} \propto V^{1 + \frac{γ - 1}{2}} \propto V^{\frac{2 + γ - 1}{2}}$
$M e a n C o l l i s i o n t i m e, t \propto \frac{V}{\sqrt{T}} . . . . . . . . . . (1)$
$F o r a d i a b a t i c p r o c e s s, T V^{γ - 1} = constant .................(2)$
$\Rightarrow t \propto V^{\frac{γ + 1}{2}} \Rightarrow \frac{t_{2}}{t_{1}} = {(\frac{V_{2}}{V_{1}})}^{\frac{γ + 1}{2}}$
$\Rightarrow \frac{t_{2}}{t_{1}} = {(\frac{2}{1})}^{\frac{γ + 1}{2}} \Rightarrow \frac{t_{1}}{t_{2}} = {(\frac{1}{2})}^{\frac{γ + 1}{2}}$

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