The mean free path of the molecules of a certain gas at 300K is 2.6×10−5m. The collision diameter of the molecule is 0.26×10−9m. Calculate (a) the pressure of the gas and (b) the number per unit volume of the gas.
A
5.306×102Pa and 2.798×1023m−3
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B
5.306×102Pa and 4.183×1023m−3
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C
10.6×102Pa and 1.281×1023m−3
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D
5.306×102Pa and 1.281×1023m−3
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Solution
The correct option is D5.306×102Pa and 1.281×1023m−3 Given, l(mean free path)=2.6×10−5m, σ=0.26×10−9m, T=300K ∵l=1√2πσ2N N=1√2πσ2l =1(1.414)(3.14)(0.26×10−9m)2(2.6×10−5m) =1.281×1023m−3 Now, pressure P=N.kT =(1.281×1023m−3)(1.38×10−23J K−1)×(300 K) =5.306×102Pa