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 )

Question

Quantrex Academy · Accepted Answer

Correct answer: B. -253

The root-mean-square speed of gas molecules is:

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

, where

T

is absolute temperature and

M

is molar mass.

For oxygen at

T_1 = 47°C = 320

K with

M_{O_2} = 32

g/mol:

v_{rms,O_2} = \sqrt{\frac{3R(320)}{32}}

For hydrogen with

M_{H_2} = 2

g/mol at temperature

T_2

:

v_{rms,H_2} = \sqrt{\frac{3RT_2}{2}}

Equating the RMS speeds:

\sqrt{\frac{3R(320)}{32}} = \sqrt{\frac{3RT_2}{2}}

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

T_2 = 2 \times 10 = 20

K

Converting to Celsius:

t_2 = 20 - 273 = -253°C

, which is option (2).

The r.m.s. speed of oxygen molecules at 47^ C is equal to that of...

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