Question

The root mean square speed of gas molecules at a temperature $427k$ and pressure $1.5bar$ is $1\times {10}^{4}cm/sec$. If the temperature is raised three times, Calculate the new rms speed of gas molecules.

Answer 1

As we know that the most probable speed $\left(V\right)$ of a molecule is given by-

$V=\sqrt{\frac{2RT}{M}}$

For the same gas-

$V\propto \sqrt{T}$

$\Rightarrow \frac{V_1}{V_2}=\sqrt{\frac{T_1}{T_2}}.....\left(1\right)$

Given that the root mean square speed of the gas at $427K$ is ${10}^{4}m/s$ and we have to find the root mean square speed of same gas when the temperature is increased to $3$ times.

Therefore,

$V_1={10}^{4}m/s$

$V_2=?=V\left(\text{say}\right)$

$T_1=427K$

$T_2=3\times T_1=3\times 427=1281K$

Now from ${eq}^n\left(1\right)$, we have

$\frac{{10}^4}{V}=\sqrt{\frac{427}{1281}}$

$\Rightarrow \frac{{10}^4}{V}=\sqrt{\frac{1}{3}}$

$\Rightarrow V=\sqrt{3}\times {10}^4$

Hence the new rms speed of the gas will be $\sqrt{3}$ times of its initial rms speed, i.e., $\sqrt{3}\times {10}^{4}m/s$.