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# RMS velocity of a gas is calculated with the formula $\sqrt {\dfrac{{3PV}}{M}}$. Volume is increased by $3$ times, the RMS velocity of the gas at constant temperature is:A.Increases by $3$ timesB.Decreases by$9$ times,C.Increases by $\sqrt 3$ timesD.Does not change

Last updated date: 20th Jun 2024
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Hint: The relationship between $PV$ and $T$ can be used here to see how one would get affected by another and in turn how it will change the RMS velocity.

The formula to calculate the RMS velocity of a gas: ${v_{rms}} = \sqrt {\dfrac{{3PV}}{M}}$
The increase in the volume: ${V_2} = 3{V_1}$
Let’s have a look at the given formula for RMS velocity of a gas:
${v_{rms}} = \sqrt {\dfrac{{3PV}}{M}}$
Let’s put subscript $1$ for initial conditions of pressure and volume to give:
${v_{rms}}_{_1} = \sqrt {\dfrac{{3{P_1}{V_1}}}{M}}$
Here, $M$ is the molar mass of the gas so that would remain constant.
Now, we will write the expression for RMS velocity after the increase in volume by using subscript $2$ as follows:

${v_{rms}}_{_2} = \sqrt {\dfrac{{3{P_2}{V_2}}}{M}}$
We are given that ${V_2} = 3{V_1}$ so we can substitute this in the above expression as follows:

${v_{rms}}_{_2} = \sqrt {\dfrac{{3{P_2}\left( {3{V_1}} \right)}}{M}}$ --- (1)
Now, we have to consider how it will affect the pressure and we can do that by using the ideal gas equation that relates pressure $\left( P \right)$ , volume $\left( V \right)$, amount $\left( n \right)$ and temperature $\left( T \right)$ of the gas as follows:
$PV = nRT$
Here, $R$ is the universal gas constant and has a fixed value.
For a given amount of gas and at constant temperature, the R.H.S. of this equation becomes constant which means L.H.S. of this equation would also be constant or we can write:
${P_1}{V_1} = {P_2}{V_2}$

We can use the given ${V_2} = 3{V_1}$ to determine the effect on pressure as follows:
$\Rightarrow {P_2}\left( {3{V_1}} \right) = {P_1}{V_1}\\ \Rightarrow {P_2} = \dfrac{{{P_1}{V_1}}}{{3{V_1}}}\\ \Rightarrow {P_2} = \dfrac{{{P_1}}}{3}$

Let’s substitute this in equation (1) as follows:
${v_{rms}}_{_2} = \sqrt {\dfrac{{3\left( {3{V_1}} \right)}}{M}\dfrac{{{P_1}}}{3}} \\ \Rightarrow {v_{rms}}_{_2} = \sqrt {\dfrac{{3{P_1}{V_1}}}{M}} \\ \Rightarrow {v_{rms}}_{_2}= {v_{rms}}_{_1}$
Hence, the RMS velocity does not change which makes option D to be the correct one

Note: We can also deduce this by establishing that R.H.S. of the given formula is constant under given conditions so RMS velocity won’t change.