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The molecules of a given mass of a gas have r.m.s velocity of 200 $ms^{-1}$ at 27°C and $1.0 \times 10^{5}\ Nm^{-2}$ pressure. When the temperature and pressure of the gas are respectively, 127 °C and $0.05 \times 10^{5}\ Nm^{-2}$, the rms velocity of its molecules in $ms^{-1}$ is
A
$\frac{400}{\sqrt{3}}$
B
$\frac{100\sqrt{2}}{3}$
C
$\frac{100}{3}$
D
$100\sqrt{2}$
Explanation
$v_{rms} \propto \sqrt{T}$, independent of pressure.
Detailed Solution
$V_{rms} = \sqrt{\frac{3RT}{m}}$; $V_{rms}$ does not depend on pressure.
$\frac{(V_{rms})_2}{(V_{rms})_1} = \sqrt{\frac{T_f}{T_i}}$, with $T_i = 300$ K, $T_f = 400$ K
$(V_{rms})_2 = 200\sqrt{\frac{400}{300}} = 200\sqrt{\frac{4}{3}} = \frac{400}{\sqrt{3}}$ m/s
$\frac{(V_{rms})_2}{(V_{rms})_1} = \sqrt{\frac{T_f}{T_i}}$, with $T_i = 300$ K, $T_f = 400$ K
$(V_{rms})_2 = 200\sqrt{\frac{400}{300}} = 200\sqrt{\frac{4}{3}} = \frac{400}{\sqrt{3}}$ m/s