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Kinetic Theory

Question

For a gas $\frac{R}{C_{v}}$ = 0.67. This gas is made up of molecules which are

monoatomic

polyatomic

mixture of diatomic and polyatomic molecules

diatomic

Solution

monoatomic

The molecule of a monoatomic gas has three translational degrees of freedom. Thus the average energy of a molecule at temperature T is ( 3/2 ) k_BT.

Given:-

$\frac{R}{C_{V}} = 0.67$

So, $C_{v} = \frac{R}{0.67}$

= 1.5 R

= $\frac{3}{2} R$

This value corresponds to monoatomic