Mean free path formulae

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Question

Mean free path of molecules of a sample of ideal gas is $λ$ . Now the gas undergoes an isothermal compression process from volume v to volume v/2. Then the mean free path of the gas molecules is

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Solution

For an ideal gas,
$PV = μRT \Rightarrow PV = ({MN}_{a}) (\frac{RT}{{PN}_{a}}) \Rightarrow (\frac{{μN}_{a}}{V}) = \frac{{PN}_{a}}{RT} \Rightarrow n = \frac{{PN}_{a}}{RT}$
$Now λ = \frac{1}{\sqrt{2} {πd}^{2} n} = \frac{1}{\sqrt{2} {πd}^{2}} x \frac{RT}{{PN}_{a}} \Rightarrow λ \propto \frac{1}{P} [For T = constant]$
$\frac{λ'}{λ} = \frac{P}{P'} = \frac{P}{2 P} = \frac{1}{2} \Rightarrow λ' = \frac{λ}{2}$

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Mean free path formulae

Remember concepts with our Masterclasses.

Mean free path of molecules of a sample of ideal gas is λ. Now the gas undergoes an isothermal compression process from volume v to volume v/2. Then the mean free path of the gas molecules is

For an ideal gas, PV=μRT⇒PV=MNaRTPNa⇒μNaV=PNaRT⇒n=PNaRTNow λ=12πd2n=12πd2xRTPNa⇒λ∝1P [For T=constant]λ'λ=PP'=P2P=12⇒λ'=λ2

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Mean free path of molecules of a sample of ideal gas is $λ$ . Now the gas undergoes an isothermal compression process from volume v to volume v/2. Then the mean free path of the gas molecules is