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Question

Physics Question on Kinetic molecular theory of gases

The molecules of a given mass of a gas have r.m.s velocity of 200ms1200 \, ms^{-1} at 27C27^{\circ} C and 1.0×105Nm21.0 \times 10^5 \, Nm^{-2} pressure. When the temperature and pressure of the gas are respectively, 127C127^{\circ} C and 0.05×105Nm20.05 \times 10^5 \, Nm^{-2}, the r.m.s. velocity of its molecules in ms1ms^{-1} is

A

4003\frac{400}{\sqrt{3}}

B

10023\frac{100 \sqrt{2}}{3}

C

1003\frac{100}{3}

D

1002100 \sqrt{2}

Answer

4003\frac{400}{\sqrt{3}}

Explanation

Solution

vTv200=400300v=200×23m/sv \propto \sqrt{T} \Rightarrow \frac{v}{200} = \sqrt{\frac{400}{300}} \Rightarrow v = \frac{200 \times2}{\sqrt{3}} m/s
v=4003m/s\Rightarrow v = \frac{400}{\sqrt{3}} m/s