JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Heat & Thermodynamics

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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 mass

\frac{m}{4}

. The ratio of the specific heats

(C_{v}^{A}

and

C_{v}^{B})

of gas

A

and

B

, respectively is

(A) $3 : 5$
(B) $5 : 7$ ✓ Correct
(C) $7 : 9$
(D) $5 : 9$

Answer: (B)

Solution

A rigid diatomic molecule has no vibrational mode, so it carries only

5

degrees of freedom, three translational and two rotational, which gives

C_{v}^{A} = \frac{5 R}{2}

. Gas B has one extra vibrational mode, and that mode adds two more degrees of freedom (one kinetic, one potential), so B has

7

degrees of freedom in all and

C_{v}^{B} = \frac{7 R}{2}

. The molecular mass never enters the molar specific heat, so the required ratio is simply

\frac{C _{v}^{A}}{C _{v}^{B}} = \frac{5 R /2}{7 R /2} = \frac{5}{7}

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Heat & Thermodynamics

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