- #1
v_pino
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Can I please check if I've converted these units right? I'm doing a question and got the answer a few order of magnitudes out.
Question: Argon has a collision cross section of 0.36nm^2. At what pressure is the mean free path of argon atoms equal to the length of the side of a 1cm^3 cubic container in which they are kept at a temperature of 10K?
Given info
==========
cross section = 0.36 nm^2
volume of cube = 1 cm^3
Conversions
==========
cross section = 0.36 * 10^-18 m^2
volume of cube = 10^-6 m^3
length of one side of cube = 0.01 m
Attempt
=======
mean free path = 1 / ((2^0.5) * cross section area * number of molecules )
number of molecules =1 / ((2^0.5) * (0.36*10^-18) *0.01) = 1.96*10^20 molecules
N = number of molecules
pV = NkT
p = (NkT)/V = ((1.96*10^20)*(1.38*10^-23)*10) / 10^-6
= 2.71*10^4 Pa
Given Solution
============
0.027 Pa
Question: Argon has a collision cross section of 0.36nm^2. At what pressure is the mean free path of argon atoms equal to the length of the side of a 1cm^3 cubic container in which they are kept at a temperature of 10K?
Given info
==========
cross section = 0.36 nm^2
volume of cube = 1 cm^3
Conversions
==========
cross section = 0.36 * 10^-18 m^2
volume of cube = 10^-6 m^3
length of one side of cube = 0.01 m
Attempt
=======
mean free path = 1 / ((2^0.5) * cross section area * number of molecules )
number of molecules =1 / ((2^0.5) * (0.36*10^-18) *0.01) = 1.96*10^20 molecules
N = number of molecules
pV = NkT
p = (NkT)/V = ((1.96*10^20)*(1.38*10^-23)*10) / 10^-6
= 2.71*10^4 Pa
Given Solution
============
0.027 Pa