The pressure exerted by a gas is $P_{0}$ . if the mass of molecules becomes half and their velocites becomes double, the pressure will become

\frac{P_{0}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 P_{0}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4 P_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $2 P_{0}$
Pressure is given by $P = \frac{m N v_{r m s}^{2}}{3 V}$
so $P \propto m v_{r m s}^{2}$

Given $m \to m / 2$ and $v \to 2 v$

so $\frac{P^{'}}{P_{0}} = \frac{m v^{2}}{\frac{m}{2} \times (2 v)^{2}}$

$P^{'} = 2 P_{0}$

Hence option C is correct.

Suggest Corrections

Similar questions

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. 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:

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 is filled in a container at pressure

P_{0}

. If the mass of molecules is halved and their rms speed is doubled, then the resultant pressure would be.

Join BYJU'S Learning Program

The pressure exerted by a gas is P0. if the mass of molecules becomes half and their velocites becomes double, the pressure will become

The correct option is D 2P0Pressure is given by P=mNv2rms3Vso P∝mv2rmsGiven m→m/2 and v→2vso P′P0=mv2m2×(2v)2P′=2P0Hence option C is correct.

The pressure exerted by a gas is $P_{0}$ . if the mass of molecules becomes half and their velocites becomes double, the pressure will become

The correct option is D $2 P_{0}$
Pressure is given by $P = \frac{m N v_{r m s}^{2}}{3 V}$
so $P \propto m v_{r m s}^{2}$

Given $m \to m / 2$ and $v \to 2 v$

so $\frac{P^{'}}{P_{0}} = \frac{m v^{2}}{\frac{m}{2} \times (2 v)^{2}}$

$P^{'} = 2 P_{0}$

Hence option C is correct.