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(C) The formula for the average speed of gas molecules is given by ${v_{av}} = \sqrt{\frac{8RT}{\pi M}}$.Since the average speeds are equal,we have $\

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$-254$

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(C) The formula for the average speed of gas molecules is given by ${v_{av}} = \sqrt{\frac{8RT}{\pi M}}$.Since the average speeds are equal,we have $\sqrt{\frac{8R{T_{H_2}}}{\pi {M_{H_2}}}} = \sqrt{\frac{8R{T_{O_2}}}{\pi {M_{O_2}}}}$.Squaring both sides and simplifying,we get $\frac{{T_{H_2}}}{{M_{H_2}}} = \frac{{T_{O_2}}}{{M_{O_2}}}$,which implies ${T_{H_2}} = {T_{O_2}} 	imes \frac{{M_{H_2}}}{{M_{O_2}}}$.Given ${T_{O_2}} = 31 + 273 = 304\,K$,${M_{H_2}} = 2\,g/mol$,and ${M_{O_2}} = 32\,g/mol$.Substituting the values: ${T_{H_2}} = 304 	imes \frac{2}{32} = 304 	imes \frac{1}{16} = 19\,K$.Converting back to Celsius: $19 - 273 = -254\,^oC$.

The temperature of the hydrogen at which the average speed of its molecules is equal to that of oxygen molecules at a temperature of $31\,^oC$ is ........ $^oC$.

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