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By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled?
A
1.4
B
2.0
C
2.8
D
4.0
Detailed Solution
Average velocity of gas molecules: $v_{av} = \sqrt{\frac{8RT}{\pi M}}$
So $v_{av} \propto \sqrt{T}$
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{2T}{T}} = \sqrt{2}$
$\sqrt{2} = 1.414 \approx 1.4$
So the average velocity increases by a factor of 1.4.
So $v_{av} \propto \sqrt{T}$
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{2T}{T}} = \sqrt{2}$
$\sqrt{2} = 1.414 \approx 1.4$
So the average velocity increases by a factor of 1.4.