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10 Heat and Matter
∴ m =
2kT
v 2
mp
=
2 × 1.38 × 10 −23 × 1200
(850) 2
= 4.584 × 10
−26 kg
=
4.584 × 10 −26
1.66 × 10 −27 amu = 27.6 amu
Therefore the gas is N 2 .
In 5 mol of gas total number (N ) of gas molecules will be
N =
5
28
× 6.02 × 10
23
= 1.075 × 10
23
Number of molecules N (v) dv in the interval v and v + dv will be
N (v)dv = 4πN
m
2πkT
3/2 v
2 exp
−
mv 2
2kT
dv (Maxwellian distribution)
(2)
The mean value of the interval is v =
800 + 900
2
= 850 m/s
which happens to be identical with v mp found in (a). In this case (2) is
reduced to a simpler form
N (v)dv =
4N dv
√ π e v mp
Number of N 2 molecules in 5 mol will be
N =
5
28
× 6.02 × 10
23
= 1.075 × 10
23
The speed interval
dv = 900 − 800 = 100 m/s
Thus the required number of molecules is
N (v)dv =
4 × 1.075 × 10 23 × 100
√
π × 2.718 × 850
= 1.05 × 10
22