Gas at a pressure $P_{o}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to:

Open in App

Solution

$P = \frac{1}{3} \frac{m N}{V} V_{r m s}^{2}$ $∴ P \propto m V_{r m s}^{2}$
So $\frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}} \times {(\frac{V_{2}}{V_{1}})}^{2}$
$= \frac{m_{1} / 2}{m_{1}} {(\frac{2 V_{1}}{V_{1}})}^{2} = 2$
$\Rightarrow P_{2} = 2 P_{1} = 2 P_{o}$

Suggest Corrections

Similar questions

Q. Gas at a pressure

P_{0}

is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure

P

will be equal to

Gas at a pressure $P_{0}$ , in contained is a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to

Q. A gas at a pressure

P_{0}

is contained in a vessel. If the masses of all the molecules are halved and their velocities doubled, the resulting pressure

P

would be equal to :

Q. An ideal gas at pressure

P_{0}

in a vessel. If the masses of all the molecules are halved and their RMS speeds doubled, the resulting pressure P will be:

Join BYJU'S Learning Program

Gas at a pressure Po is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to:

P=13mNVV2rms ∴P∝mV2rmsSo P2P1=m2m1×(V2V1)2 =m1/2m1(2V1V1)2=2⇒P2=2P1=2Po

Gas at a pressure $P_{o}$ is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to:

$P = \frac{1}{3} \frac{m N}{V} V_{r m s}^{2}$ $∴ P \propto m V_{r m s}^{2}$
So $\frac{P_{2}}{P_{1}} = \frac{m_{2}}{m_{1}} \times {(\frac{V_{2}}{V_{1}})}^{2}$
$= \frac{m_{1} / 2}{m_{1}} {(\frac{2 V_{1}}{V_{1}})}^{2} = 2$
$\Rightarrow P_{2} = 2 P_{1} = 2 P_{o}$