For nitrogen gas at 298K and a molar volume of 24.45 L (approximately corresponding to 1.00 atm pressure):

a. Find the mean free path.

b. Find the mean collision time and the mean molecular collision rate.

Question

For nitrogen gas at 298K and a molar volume of 24.45 L (approximately corresponding to 1.00 atm pressure):

a. Find the mean free path.

b. Find the mean collision time and the mean molecular collision rate.

Accepted Answer

a. [latex]N=\frac{N}{V} =\frac{6.022×10^{23} mol^{−1}}{0.02445m^{3} mol^{−1}} =2.463×10^{25} m^{−3}[/latex]

[latex]\left\langle v\right\rangle =\Bigl(\frac{8(8.3145 J K^{−1} mol^{−1})(298.15K)}{π(0.0280 kg mol^{−1})} \Bigr)^{1/2}=475ms^{−1}[/latex]

[latex]\left\langle v_{rel}\right\rangle =\sqrt{2} \left\langle v\right\rangle =671ms^{−1}[/latex]

[latex]λ=\frac{1}{\sqrt{2} π(3.7 × 10^{−10} m^{3})^{2}(2.463 × 10^{25} m^{−3})}=6.7 × 10^{−8} m[/latex]

b.

[latex]τ_{coll}=\frac{λ}{〈v〉} =\frac{6.7×10^{−8}m}{475ms^{−1}} =1.4 × 10^{−10} s[/latex]

[latex]z=\frac{1}{1.4 × 10^{−10} s} =7.1×10^{9} s^{−1}[/latex]

Question 9.18: For nitrogen gas at 298K and a molar volume of 24.45 L (appr...