Ncert Exercises & Solutions

A cubic vessel (with faces horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of $500 {ms}^{- 1}$ in the vertical direction. The pressure of the gas inside the vessel as observed by us on the ground:

1.	remains the same because $500$ $\mathrm{ms^{-1}}$ is very much smaller than $v_{rms}$ of the gas.
2.	remains the same because the motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.
3.	will increase by a factor equal to $\Big(\frac{v_{rms}^2+(500)^2}{v_{rms}^2}\Big)$where $v_{rms}^2$ was the original mean square velocity of the gas.
4.	will be different on the top wall and bottom wall of the vessel.

2. Hint: The pressure of the gas depends on the motion of its molecules.

Step 1: Find the change in the relative motion of the gas molecules.

As the motion of the vessel as a whole does not effect the relative motion of the gas molecules with respect to the walls of the vessel, hence pressure of the gas inside the vessel, as observed by us, on the ground remains the same.

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1.	remains the same because \(500\) \(\mathrm{ms^{-1}}\) is very much smaller than \(v_{rms}\) of the gas.
2.	remains the same because the motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.
3.	will increase by a factor equal to \(\Big(\frac{v_{rms}^2+(500)^2}{v_{rms}^2}\Big)\)where \(v_{rms}^2\) was the original mean square velocity of the gas.
4.	will be different on the top wall and bottom wall of the vessel.