The r.m.s speed of oxygen molecules at $47^{\circ} \mathrm{C}$ is equal to that of the hydrogen molecules kept at $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$. (Mass of oxygen molecule/mass of hydrogen molecule=32/2)

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Heat & Thermodynamics

The r.m.s speed of oxygen molecules at

4 7^{\circ} C

is equal to that of the hydrogen molecules kept at

____

^{\circ} C

. (Mass of oxygen molecule/mass of hydrogen molecule=32/2)

(A)-235
(B)-100
(C)-253✓ Correct
(D)-20

Answer: (C)

Solution

The rms speed is

v_{r m s} = \frac{3 R T}{M}

. Equating for O₂ and H₂:

\frac{3 R T _{O_{2}}}{M _{O_{2}}} = \frac{3 R T _{H_{2}}}{M _{H_{2}}}

. This gives

\frac{T _{O_{2}}}{M _{O_{2}}} = \frac{T _{H_{2}}}{M _{H_{2}}}

. With

T_{O_{2}} = 273 + 47 = 320

M_{O_{2}} = 32

M_{H_{2}} = 2

\frac{320}{32} = \frac{T _{H_{2}}}{2}

, so

T_{H_{2}} = 20

= - 25 3^{\circ}

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option

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