Consider a mixture of gas molecule of types A, B and C having masses $\mathrm{m_{A}}<\mathrm{m_{B}}<\mathrm{m_{C}}$. The ratio of their root mean square speeds at normal temperature and pressure is:

(C) $\frac{1}{v_{A}}<\frac{1}{v_{B}}<\frac{1}{v_{C}}$

(A) $\frac{1}{v_{A}}>\frac{1}{v_{B}}>\frac{1}{v_{C}}$

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Heat & Thermodynamics

Consider a mixture of gas molecule of types A, B and C having masses

m_{A} < m_{B} < m_{C}

. The ratio of their root mean square speeds at normal temperature and pressure is:

(A) $\frac{1}{v _{A}} > \frac{1}{v _{B}} > \frac{1}{v _{C}}$
(B) $v_{A} = v_{B} = v_{C} = 0$
(C) $\frac{1}{v _{A}} < \frac{1}{v _{B}} < \frac{1}{v _{C}}$ ✓ Correct
(D) $v_{A} = v_{B} \neq = v_{C}$

Answer: (C)

Solution

The root mean square speed of a gas molecule is

v_{r m s} = \frac{3 k T}{m}

, so at a common temperature

v_{r m s} \propto \frac{1}{m}

. With

m_{A} < m_{B} < m_{C}

this gives

v_{A} > v_{B} > v_{C}

. Taking reciprocals of positive quantities reverses the order, so

\frac{1}{v _{A}} < \frac{1}{v _{B}} < \frac{1}{v _{C}}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option

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