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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answer is 4.
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Detailed Solution
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
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