Kinetic Theory

JEE Main level, single correct, medium

22 of 168

Question 22 of 168

The r.m.s. speed of oxygen molecules at 47∘C47^{\circ} \mathrm{C} is equal to that of the hydrogen molecules kept at ____\_\_\_\_ ∘C{ }^{\circ} \mathrm{C}. (Mass of oxygen molecule/mass of hydrogen molecule =32/2=32 / 2)

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Answer: (B) -253

The root-mean-square speed of gas molecules is: vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}, where TT is absolute temperature and MM is molar mass.

For oxygen at T1=47°C=320T_1 = 47°C = 320 K with MO2=32M_{O_2} = 32 g/mol:
vrms,O2=3R(320)32v_{rms,O_2} = \sqrt{\frac{3R(320)}{32}}

For hydrogen with MH2=2M_{H_2} = 2 g/mol at temperature T2T_2:
vrms,H2=3RT22v_{rms,H_2} = \sqrt{\frac{3RT_2}{2}}

Equating the RMS speeds:
3R(320)32=3RT22\sqrt{\frac{3R(320)}{32}} = \sqrt{\frac{3RT_2}{2}}

32032=T22\frac{320}{32} = \frac{T_2}{2}

T2=2×10=20T_2 = 2 \times 10 = 20 K

Converting to Celsius: t2=20−273=−253°Ct_2 = 20 - 273 = -253°C, which is option (2).

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