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Question

The mean free path of the molecules of a certain gas at 300 K is 2.6×105 m. The collision diameter of the molecule is 0.26 ×109m. Calculate (a) the pressure of the gas and (b) the number per unit volume of the gas.

A
5.306×102 Pa and 2.798×1023 m3
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B
5.306×102 Pa and 4.183×1023 m3
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C
10.6×102 Pa and 1.281×1023 m3
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D
5.306×102 Pa and 1.281×1023 m3
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Solution

The correct option is D 5.306×102 Pa and 1.281×1023 m3
Given,
l(mean free path)=2.6×105 m, σ=0.26×109 m, T=300 K
l=12πσ2N
N=12πσ2l
=1(1.414)(3.14)(0.26×109m)2(2.6×105m)
=1.281×1023 m3
Now, pressure P=N.kT
=(1.281×1023 m3)(1.38×1023J K1)×(300 K)
=5.306×102 Pa

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