JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 12)
The specific heats, CP and CV of a gas of
diatomic molecules, A, are given (in units of
J mol–1 K–1) by 29 and 22, respectively.
Another gas of diatomic molecules, B, has the
corresponding values 30 and 21. If they are
treated as ideal gases, then :-
A is rigid but B has a vibrational mode
A has a vibrational mode but B has none
A has one vibrational mode and B has two
Both A and B have a vibrational mode each
Explanation
For A:
$${{{C_p}} \over {{C_v}}} = \gamma = 1 + {2 \over f} = {{29} \over {22}}$$
It gives f = 6.3 $$ \approx $$ 6 (3 translational, 2 rotational and 1 vibrational)
For B:
$${{{C_p}} \over {{C_v}}} = \gamma = 1 + {2 \over f} = {{30} \over {21}}$$
$$ \therefore $$ f = 4.67 $$ \approx $$ 5 (3 translational, 2 rotational, no vibrational)
$${{{C_p}} \over {{C_v}}} = \gamma = 1 + {2 \over f} = {{29} \over {22}}$$
It gives f = 6.3 $$ \approx $$ 6 (3 translational, 2 rotational and 1 vibrational)
For B:
$${{{C_p}} \over {{C_v}}} = \gamma = 1 + {2 \over f} = {{30} \over {21}}$$
$$ \therefore $$ f = 4.67 $$ \approx $$ 5 (3 translational, 2 rotational, no vibrational)
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