for oxygen gas at 25 degree C ,collision collision diameter for oxygen molecule is 361 picometer mean free path at 1 atm pressure
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Explanation:
Collision diameter, d=0.361nm=3.61×10
−10
m
We know that PV=
N
A
N
RT
(a) ∴
V
N
=
RT
PN
A
=
0.08206LatmK
−1
mol
−1
×298K
1atm×6.022×10
23
mol
−1
=2.46×10
22
L
−1
=2.46×10
25
m
−3
The mean free path= λ=
2
Πd
2
(
V
N
)
1
=
2
Π(3.61×10
−10
m)
2
×2.46×10
25
m
−3
1
=6.983×10
−8
m
(b) 0.1Pa=0.987×10
−6
atm
V
N
=
RT
PNa
=
0.08206LatmK
−1
mol
−1
×298K
6.022×10
23
mol
−1
×0.987×10
−6
atm
=2.428×10
19
m
−3
∴ The mean free path= λ=
2
Πd
2
(
V
N
)
1
=
2
×Π(3.6×10
−10
m)
2
×2.428×10
19
m
−3
1
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