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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 $3$ times
B.Decreases by$9$ times,
C.Increases by $\sqrt 3 $ times
D.Does not change

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Last updated date: 20th Jun 2024
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Answer
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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.

Step by step answer: Given:
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.