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The molecule diameter of nitrogen is 3.5*10^-8 cm. calculate the mean free path at temprature 27 degree centigrate and p=1 atm please sir ,answer this question..


2 years ago

Susmita
425 Points
							p=nkTOr,n=p/kTWhere p=1atm=1.013×105 Pa ,T=27+273=300K,k=1.38×10-23Js.so n=2.4×10^25Mean free path$\lambda =\frac{1}{ \sqrt{2} \pi (\sigma)^2 n }$where diameter$\sigma =3.5*10^{-8}$cm=3.5×10-10mso lamda=7.5×10-8m

2 years ago
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