Question

The temperature at which the average speed ofthe gas molecules is double to that at atemperature of 27 $^\circ$C is

• A. 54$^\circ$C
• B. 108$^\circ$C
• C. 327$^\circ$C
• D. 927$^\circ$C

Answer 1

Answer: (D)

927$^\circ$C

Answer 2

$\bar{\upsilon} = \left( \frac{8RT}{\pi \, M } \right)^{1/2}$ or $\bar{\upsilon} \propto (T_1^{1/2}$ for a gas Now $\frac{\bar{\upsilon_1}}{\bar{\upsilon_1}} = (T_1 /T_2)^{1/2}$ or $\frac{\upsilon_1}{1\upsilon_1} \left( \frac{T_1}{T_2} \right)^{1/2} = \left( \frac{300}{T_2} \right)^{1/2}$ or $\frac{1}{4} = \frac{300}{T_2}$ or $T_2$ = 1200 K = 1200 - 273 = 927$^\circ$C