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

λ=KT2d2P λαTP λ2λ1=T2T1×P1P2 λ2=(350300)(5×1041×104)4×10−7=2.3×10−6

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

λ=KT2d2P λαTP λ2λ1=T2T1×P1P2 λ2=(350300)(5×1041×104)4×10−7=2.3×10−6

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A certain ideal gas has a temperature 300 K and a pressure $5.0 \times 10^{4} Pa$ . 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 $1.0 \times 10^{4} Pa$ the mean free path is

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

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

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

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

answer is D.

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$λ = \frac{K T}{\sqrt{2} d^{2} P} λ α \frac{T}{P} \frac{λ_{2}}{λ_{1}} = \frac{T_{2}}{T_{1}} \times \frac{P_{1}}{P_{2}} λ_{2} = (\frac{350}{300}) (\frac{5 \times 10^{4}}{1 \times 10^{4}}) 4 \times 10^{- 7} = 2.3 \times 10^{- 6}$

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

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A certain ideal gas has a temperature 300 K and a pressure $5.0 \times 10^{4} Pa$ . 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 $1.0 \times 10^{4} Pa$ the mean free path is