A certain ideal gas has a temperature 300K and a pressure 5.0×104Pa. The molecules have a mean free path 4.0×10−7m. If the temperature is increased to 350K and pressure is reduced to 1.0×104Pa, the mean free path is

Mean free path, λ=KT2d2p⇒λ∝TP∴λ2=(T2T1)(P1P2)λ1=(350300)(5×104104)(4×10−7)=2.3×10−8m

A certain ideal gas has a temperature 300K and a pressure 5.0×104Pa. The molecules have a mean free path 4.0×10−7m. If the temperature is increased to 350K and pressure is reduced to 1.0×104Pa, the mean free path is

Mean free path, λ=KT2d2p⇒λ∝TP∴λ2=(T2T1)(P1P2)λ1=(350300)(5×104104)(4×10−7)=2.3×10−8m

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A certain ideal gas has a temperature 300K and a pressure $5.0 \times 10^{4} P a .$ The molecules have a mean free path $4.0 \times 10^{- 7} m .$ If the temperature is increased to 350K and pressure is reduced to $1.0 \times 10^{4} P a,$ the mean free path is

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$9.3 \times 10^{- 8} m$

$6.9 \times 10^{- 8} m$

$2.3 \times 10^{- 6} m$

$3.4 \times 10^{- 7} m$

answer is D.

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Mean free path, $λ = \frac{K T}{\sqrt{2} d^{2} p} \Rightarrow λ \propto \frac{T}{P}$

$∴ λ_{2} = (\frac{T_{2}}{T_{1}}) (\frac{P_{1}}{P_{2}}) λ_{1} = (\frac{350}{300}) (\frac{5 \times 10^{4}}{10^{4}}) (4 \times 10^{- 7}) = 2.3 \times 10^{- 8} m$

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A certain ideal gas has a temperature 300K and a pressure 5.0×104Pa. The molecules have a mean free path 4.0×10−7m. If the temperature is increased to 350K and pressure is reduced to 1.0×104Pa, the mean free path is

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A certain ideal gas has a temperature 300K and a pressure $5.0 \times 10^{4} P a .$ The molecules have a mean free path $4.0 \times 10^{- 7} m .$ If the temperature is increased to 350K and pressure is reduced to $1.0 \times 10^{4} P a,$ the mean free path is