First slide

Kinetic theory of ideal gases

Question

In the case of hydrogen and oxygen, at NTP which of the following quantities is / are same?
(1) Average momentum per molecule
(2) Average kinetic energy per molecule
(3) Kinetic energy per unit volume
(4) Kinetic energy per unit mass

Moderate

A

(1) and (2)
B

(2) and (3)
C

(3) and (4)
D

All are true

Solution

$v_{R M S} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 k T}{m}}; v_{A V G} = \sqrt{\frac{8 R T}{π M}} = \sqrt{\frac{8 k T}{π m}}$
Both these depend on nature of gas.
Average momentum
= $m (v_{A V G}) = \sqrt{\frac{8 k T m}{π}}$ (depends on nature of gas).
Average kinetic energy per molecule
= $\frac{1}{2} m v_{R M S}^{2} = \frac{1}{2} m \frac{3 k T}{m} = \frac{3}{2} k T$ (independent of nature of gas, depends on temperature).
Kinetic energy per unit volume
= $\frac{N (\frac{1}{2} m v_{R M S}^{2})}{V} = \frac{N (\frac{1}{2} m (\frac{3}{m} k T))}{V} = \frac{3}{2} k T (\frac{N}{V})$
Using PV = nRT and $n = \frac{N}{N_{A}}$ , we get, $n = \frac{N}{V} = \frac{P N_{A}}{R T}$
Since, pressure and temperature are same correspondingly, So, $\frac{N}{V}$ is same.
Kinetic energy per unit mass
= $\frac{N (\frac{1}{2} m v_{R M S}^{2})}{m a s s o f g a s} = \frac{N (\frac{1}{2} m (\frac{3}{m} k T))}{m N} = \frac{3}{2 m} k T$ (depends on nature of gas and temperature).

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