Question

The temperature at which the root mean square velocity of the gas molecules would become twice of its value at $0^{\circ }C$ is

Answer 1

Given,

$V_2=2V_1$

$T_1=0^{0}C=273.15K$

Root means square velocity,

$V_{rms}=V=\sqrt{\frac{3RT}{m}}$

$V_{rms}=V\propto \sqrt{T}$

$\frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}$

$T_2=T_1(\frac{V_2}{V_1}{)}^2$

$T_2=273.15\times (\frac{2V_1}{V_1}{)}^2$

$T_2=1092.6K$