Question

The absolute temperature of a gas is increased 3 times. What will be the increase in the rms velocity of the gas molecule?

Answer 1

RMS Velocity of Gases

Here we have a question from the Kinetic Theory of Gases.

RMS Velocity of a gas refers to Root Mean Square Velocity.

It means that for a large number of observations of velocity, we first square them, take the mean (average) of these squares, and finally take the square root of this mean. So it is called Root-Mean-Square

[Extra Info:

Suppose for n molecules, the individual velocities are $\sf v_1,\ v_2,\ v_3, \dots , v_n$. Then the RMS Velocity would be:

$\sf \displaystyle v_{rms} = \sqrt{\frac{v_1^2+v_2^2+v_3^2+\dots+v_n^2}{n}}$

]

Coming back to the question. After using assumptions of the Kinetic Theory, we come to a simplified expression for RMS Velocity, which is:

$\boxed{\sf v_{rms}=\sqrt{\frac{3RT}{M}}}$

Here,

R = Universal Gas Constant

T = Absolute Temperature

M = Molecular Mass of Gas

Now, we are given that the absolute temperature of a gas is increased 3 times. We have to find the corresponding increase in RMS Velocity.

We see that the only variables we have here are $\sf v_{rms}$ and $\sf T$. All other factors are constant.

So, if we take $\sf T_2 = 3 T_1$ then we need $\sf v_2$ in the form of $\sf v_1$.

Let us use the expressions:

$\sf \displaystyle v_1 = \sqrt{\frac{3RT_1}{M}} \qquad v_2 = \sqrt{\frac{3RT_2}{M}} \\\\\\ \implies \frac{v_1}{v_2}=\frac{\sqrt{\frac{3RT_1}{M}}}{\sqrt{\frac{3RT_2}{M}}} \\\\\\ \implies \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \\\\\\ \textsf{Use \sf T_2 = 3T_1 }\\\\\\ \implies \frac{v_1}{v_2}=\frac{1}{\sqrt{3}}\\\\\\ \implies \huge \boxed{\sf v_2=\sqrt{3}v_1}$

Thus, The RMS Velocity increases $\bold{\sqrt{3}}$ times.

Answer 2

Answer:the increase in rms velocity will be root 3 times i.e 1.732 timesof the original velocity

Explanation: