Question:

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

Updated On: Apr 20, 2025

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The Correct Option is A

The relationship between velocity ($ v $) and temperature ($ T $) is given by:

$v \propto \sqrt{T}$

Now, let's solve for the velocity in the given scenario:

$\frac{v}{200} = \sqrt{\frac{400}{300}} \Rightarrow v = \frac{200 \times 2}{\sqrt{3}} \, \text{m/s}$

Simplifying this, we get:

$v = \frac{400}{\sqrt{3}} \, \text{m/s}$

The velocity is $ \frac{400}{\sqrt{3}} $ meters per second.

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