For Ar(g), sigma=0.36nm^2 (a) Compute the mean free path of Ar(g) at 300K and 1.
ID: 891142 • Letter: F
Question
For Ar(g), sigma=0.36nm^2
(a) Compute the mean free path of Ar(g) at 300K and 1.0 atm. Ans: 810 Angstrum
(b) At what pressure does the mean free path at 300k become comprable to 10 times the diameter of an Ar atom? The diameter of Ar is 3.78 angstrum. Ans: 21.2 atm
(c) N2(g) is contained n a vessel 10.0cm in diameter at 300K. At what pressure does the mean free path become equal to the size of the container? Ans: 8.0 x 10 ^-7 atm
(d) How many molecules are in this container (from part c)? Ans: 1.0 x 10 ^16 molecules
(e) Based on calculations for parts a-d can Argon be treated as an ideal gas under these conditions of P,V,T? Discuss.
Explanation / Answer
To calculate the mean free path of Ar(g) at 300K and 1.0 atm use the following expression :
= RT/[(2)d²NP]
First calculate the d from as follows:
= pi d^2
0.36nm^2 = 3.14 *d^2
d^2= 0.114 nm^2 or 11.4 angstrum^2
= RT/[(2)d²NP]
Where R is the ideal gas constant
T is the temperature in Kelvin
d is the collision diameter
N is Avagadro's number
P is the pressure, 1atm
T = 300 K
= 0.0821 0.001 m^3-atm/mole-K* 10^30 angstrum^3 300/[(2) 3.14 *11.4 angstrum^2 *6.022*10^23*1.0]
= 810 Angstrum
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