A fixed mass of gas at constant pressure occupies a volume $V$ . The gas undergoes a rise in temperature so that the root mean square velocity of its molecules is doubled. The new volume will be.

\frac{V}{2}

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\frac{V}{\sqrt{2}}

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2 V

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4 V

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Solution

The correct option is C $4 V$
Since $V_{r m s}$ is doubled by increasing the temp. so by
$V_{r m s}$ = $\sqrt{\frac{3 K T}{m}}$ , the temp. increase by four times.
Now for constant pressure
$\frac{V_{1}}{T_{1}}$ = $\frac{V_{2}}{T_{2}}$
$V_{1}$ = $V . T_{1}$ = $T^{0} K$
$T_{2}$ = $4 T^{0} K$ , $V_{2}$ = ?
$V_{2} = 4 V$

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A fixed mass of gas at constant pressure occupies a volume V. The gas undergoes a rise in temperature so that the root mean square velocity of its molecules is doubled. The new volume will be.

The correct option is C 4VSince Vrms is doubled by increasing the temp. so by Vrms = √ 3KTm, the temp. increase by four times.Now for constant pressure V1T1 = V2T2V1 = V.T1 = T0K T2 = 4T0K, V2 = ?V2=4V

A fixed mass of gas at constant pressure occupies a volume $V$ . The gas undergoes a rise in temperature so that the root mean square velocity of its molecules is doubled. The new volume will be.