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

Given: Oxygen molecules are at 47^ C and their r.m.s. speed is equal to that of hydrogen molecules. Mass ratio is 322. Find: The temperature of hydrogen molecules in ^ C. For r.m.s. speed, v rms = 3RTM If the speeds are equal, then T1M1 = T2M2 Convert oxygen temperature to Kelvin: TO = 273 + 47 = 320 K For oxygen, MO = 32 and for hydrogen, MH = 2. So, 32032 = TH2 10 = TH2 TH = 20 K Now convert back to Celsius: tH = TH - 273 = 20 - 273 = -253^ C Therefore, the hydrogen molecules are kept at -253^ C. The correct option is B.

The r.m.s. speed of oxygen molecules at 47^ C is equal to that of… — Solution | JEE Physics

MCQEasyJEE 2026Molecular Speeds (rms, Average, Most Probable)

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

A
$-20$
B
$-253$
C
$-235$
D
$-100$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Oxygen molecules are at $47^\circ \text{C}$ and their r.m.s. speed is equal to that of hydrogen molecules. Mass ratio is $\frac{32}{2}$ .

Find: The temperature of hydrogen molecules in $^\circ \text{C}$ .

For r.m.s. speed,

v_{\text{rms}} = \sqrt{\frac{3RT}{M}}

If the speeds are equal, then

\frac{T_1}{M_1} = \frac{T_2}{M_2}

Convert oxygen temperature to Kelvin:

T_O = 273 + 47 = 320 \, \text{K}

For oxygen, $M_O = 32$ and for hydrogen, $M_H = 2$ . So,

\frac{320}{32} = \frac{T_H}{2}

10 = \frac{T_H}{2}

T_H = 20 \, \text{K}

Now convert back to Celsius:

t_H = T_H - 273 = 20 - 273 = -253^\circ \text{C}

Therefore, the hydrogen molecules are kept at $-253^\circ \text{C}$ . The correct option is B.

Common mistakes

Using Celsius directly in the ratio of temperatures is incorrect because the r.m.s. speed formula requires absolute temperature. Convert $47^\circ \text{C}$ to $320 \, \text{K}$ before applying the relation.
Using the mass ratio in the wrong order gives the wrong temperature. Since $\frac{T}{M}$ is constant for equal r.m.s. speeds, use $\frac{320}{32} = \frac{T_H}{2}$ , not the inverse.
Forgetting to convert the final Kelvin temperature back to Celsius leads to reporting $20 \, \text{K}$ as the answer. The question asks for $^\circ \text{C}$ , so compute $20 - 273 = -253^\circ \text{C}$ .

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Answer

Step-by-step solution

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