KTGMedium
Question
The r.m.s speed of oxygen molecules at $47^{\circ}C$ is equal to that of the hydrogen molecules kept at $\_\_\_\_$ $\ ^{\circ}C$. (Mass of oxygen molecule/mass of hydrogen molecule=32/2)
Options
A.-235
B.-100
C.-253
D.-20
Solution
$V_{rms} = \sqrt{\frac{3R\text{ }T}{M}}$
$${V_{{rmsO}_{2}} = V_{{rmsH}_{2}} }{T_{O_{2}} = 273 + 47 = 320\text{ }K }{\sqrt{\frac{3{RT}_{O_{2}}}{M_{O_{2}}}} = \sqrt{\frac{3{RT}_{H_{2}}}{M_{H_{2}}}} }{\frac{T_{2}}{M_{O_{2}}} = \frac{T_{H_{2}}}{M_{H_{2}}} }{\frac{320}{32} = \frac{T_{H_{2}}}{2} }{T_{H_{2}} = 20\text{ }K }{T_{H_{2}} = - 253^{\circ}C}$$
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