A) 80 K
B) -73 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 \[{{M}_{H}}\] be molecular weights of oxygen and hydrogen and \[{{T}_{O}}\] and \[{{T}_{H}}\] the corresponding kelvin temperatures at which \[{{C}_{rms}}\] is same for both gases. i.e., \[{{C}_{rms(O)}}={{C}_{rms(H)}}\] \[\sqrt{\left( \frac{3R{{T}_{O}}}{{{M}_{O}}} \right)}=\sqrt{\left( \frac{3R{{T}_{H}}}{{{M}_{H}}} \right)}\] Hence, \[\frac{{{T}_{O}}}{{{M}_{O}}}=\frac{{{T}_{H}}}{{{M}_{H}}}\] Given, \[{{T}_{O}}=273+47=320\,K\] \[{{M}_{O}}\,=32,\,\,{{M}_{H}}=2\] \[\therefore \] \[{{T}_{H}}=\frac{2}{32}\times 320=20\,K\]
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