JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 15)
Consider two ideal diatomic gases A and B at
some temperature T. Molecules of the gas A are
rigid, and have a mass m. Molecules of the gas
B have an additional vibrational mode, and
have a mass $${m \over 4}$$
. The ratio of the specific heats ($$C_V^A$$ and $$C_V^B$$ ) of gas A and B, respectively is :
7 : 9
5 : 7
3 : 5
5 : 9
Explanation
Degree of freedom of a diatomic molecule if
vibration is absent = 5
Degree of freedom of a diatomic molecule if vibration is present = 7
$$ \therefore $$ $$C_V^A$$ = $${5 \over 2}R$$
and $$C_V^B$$ = $${7 \over 2}R$$
$$ \therefore $$ $${{C_V^A} \over {C_V^B}} = {5 \over 7}$$
Degree of freedom of a diatomic molecule if vibration is present = 7
$$ \therefore $$ $$C_V^A$$ = $${5 \over 2}R$$
and $$C_V^B$$ = $${7 \over 2}R$$
$$ \therefore $$ $${{C_V^A} \over {C_V^B}} = {5 \over 7}$$
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