Question

What will be the ratio of the root mean square speeds of the molecules of an ideal gas at $270K$ and$30K$?

Answer 1

In order to solve this question, we are going to firstly write the formula for the root mean square speeds of the molecules of a gas. After that, the speeds of the molecules are calculated at the two different temperatures as given in the question and then, the ratio is calculated by dividing.

Formula used:
The root mean square speed of the molecules is given as:
${v_{rms}} = \sqrt {\dfrac{{3RT}}{m}}$
Where, $T$is the temperature, $m$is the mass of the molecules and $R$is the Gas constant.

Complete step-by-step solution:
It is given in the question that the values of the two temperatures are as follows:
{T_1} = 270K \\\ {T_2} = 30K \\\
We know that the root mean square speed of the molecules is given as:
${v_{rms}} = \sqrt {\dfrac{{3RT}}{m}}$
Where, $T$is the temperature, $m$is the mass of the molecules and $R$is the Gas constant.
Putting the values of the temperatures and calculating the root mean square speeds of the molecules:
{v_{rms\left( 1 \right)}} = \sqrt {\dfrac{{3R \times 270}}{m}} \\\ {v_{rms\left( 2 \right)}} = \sqrt {\dfrac{{3R \times 30}}{m}} \\\
Dividing the above two root mean square speed equations in order to get the ratio of the two, we get
\dfrac{{{v_{rms\left( 1 \right)}}}}{{{v_{rms\left( 2 \right)}}}} = \dfrac{{\sqrt {\dfrac{{3R \times 270}}{m}} }}{{\sqrt {\dfrac{{3R \times 30}}{m}} }} \\\
\Rightarrow \dfrac{{{v_{rms\left( 1 \right)}}}}{{{v_{rms\left( 2 \right)}}}} = \sqrt {\dfrac{{270}}{{30}}} = \sqrt 9 = 3 \\\

Hence, the ratio of the root mean square speeds of the molecules of an ideal gas at $270K$ and$30K$is$3:1$.

Note: It is important to note that the root-mean-square speed is the measure of the speed of particles in a gas, defined as the square root of the average velocity-squared of the molecules in a gas. The mass and the Gas constant values for a particular gas remain the same and only temperature is the variant.