Question

Recall the rms speed of the molecules in an ideal gas for $v_0$ at temperature $T_0$ and pressure $ P_0 $. Find the RMS speed if the temperature is raised from $T_0$ = 293 K to 573 K.

Answer 1

Hint: The velocities of various molecules in a gas form a Maxwellian distribution. The root mean square speed of the gas molecules is directly proportional to the square root of the temperature of the gas.
Formula used:
The rms speed of the gas molecules is:
$v_{rms} = \sqrt{ \dfrac{3 k_B T}{m}} $.

Complete answer:
In an ideal gas, all the molecules of the gas move around with different velocities. When we plot all these velocities as a function of the number of molecules having those velocities, we observe a Maxwellian distribution. There is an average velocity, which most number of molecules in the gas have (where the peak of the plot exists). Very few particles touch extremely high or low velocities.
The rms or root mean square velocity is as the name suggests the square root of the average of squared velocities which can also be written as:
$v_{rms} = \sqrt{ \dfrac{v_1^2 + v_2^2 + ... + v_n^2 }{n}}$.
For Maxwellian distribution rms velocity comes out to be:
$v_{rms} = \sqrt{ \dfrac{3 k_B T}{m}} $
where m is the mass of one molecule in the gas and $k_B$ is Boltzmann constant.

Let us assume that rms velocity is $v_0$ when the temperature is $T_0 = 293 K$ and rms velocity is $v'_0$ when the temperature is $T'_0 = 573 K$. The ratio of the two velocities will be:
$\dfrac{v'_0}{v_0} = \sqrt{ \dfrac{T'_0}{T_0}} = \sqrt{\dfrac{ 593}{293}} = 1.39846$.
Therefore, the velocity of the rms gas at 573 K is
$v'_0 = 1.4 v_0$.

Note:
If one does not remember the expression for rms velocity, one can simply equate kinetic energy expression with the expression that we get from equipartition of energy i.e. (3/2)kT. This will produce the required expression. The average velocity should not be confused with the rms velocity as both have different expressions.