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)
Q. 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)
A. -100
B. -253
C. -20
D. -235
Correct Answer: -253

Explanation

Root mean square speed is given by:

\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]

Since the r.m.s. speeds are equal,

\[ \sqrt{\frac{3kT_1}{m_1}} = \sqrt{\frac{3kT_2}{m_2}} \]

Squaring both sides:

\[ \frac{T_1}{m_1} = \frac{T_2}{m_2} \]

Given:

Temperature of oxygen = 47 °C = 320 K

Mass ratio:

\[ \frac{m_{O_2}}{m_{H_2}} = \frac{32}{2} = 16 \]

Thus,

\[ \frac{320}{m_{O_2}} = \frac{T_2}{m_{H_2}} \]

\[ T_2 = 320 \times \frac{m_{H_2}}{m_{O_2}} \]

\[ T_2 = 320 \times \frac{1}{16} \]

\[ T_2 = 20 K \]

Convert to °C:

\[ T = 20 - 273 \]

\[ T = -253 °C \]

Hence hydrogen must be kept at -253 °C.

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