A) 80 K
B) \[-73\text{ }K\]
C) 3 K
D) 20 K
Correct Answer: D
Solution :
The rms velocity of the molecule of a gas of molecular weight M at Kelvin temperature T is given by. \[{{C}_{rms}}=\sqrt{\left( \frac{3RT}{M} \right)}\] Let\[{{M}_{O}}\]and\[{{M}_{H}}\]are molecular weights of oxygen and hydrogen and\[{{T}_{O}}\]and\[{{T}_{H}}\]the corresponding Kelvin temperature at which \[{{C}_{rms}}\]is same for both gases. \[{{C}_{rms(O)}}={{C}_{rms(H)}}\] \[\sqrt{\left( \frac{3R{{T}_{O}}}{{{M}_{0}}} \right)}=\sqrt{\left( \frac{3R{{T}_{H}}}{{{M}_{H}}} \right)}\] Hence, \[\frac{{{T}_{O}}}{{{M}_{0}}}=\frac{{{T}_{H}}}{{{M}_{H}}}\] \[{{T}_{O}}=273+48=320K\] \[{{M}_{O}}=32,{{M}_{H}}=2\] \[\therefore \] \[{{T}_{H}}=\frac{2}{32}\times 320=20\,K\]
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