Q.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 m s$^{-1}$ in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground

The key idea here is the Kinetic Theory Explanation of Pressure.

Pressure exerted by an ideal gas on the walls of its container results from the continuous collisions of gas molecules with the walls. This pressure depends on the average kinetic energy of the molecules and their collision rate with the walls.

1. Pressure is determined by the relative motion of gas molecules with respect to the container walls.
2. When the vessel moves at a constant velocity (like the rocket moving at $500\ \text{m s}^{-1}$), this motion is a uniform translation applied to the entire system—both the gas molecules and the walls.
3. A uniform translational motion does not alter the relative velocities between the gas molecules and the container walls.
4. Since the relative velocities and thus the collision dynamics (frequency and impulse per collision) remain unchanged, the pressure exerted by the gas on the walls also remains unchanged.

Kinetic Theory Explanation

Pressure in an ideal gas arises from the continuous collisions of gas molecules with the walls of the container. Each collision involves a change in momentum of the molecule, which exerts a force on the wall. The sum of these forces over time, averaged over the area of the wall, gives the pressure.

Crucially, this force, and thus the pressure, depends on the relative velocity of the gas molecules with respect to the container walls. If the entire system (vessel and gas) is moving at a constant velocity, the relative velocities between the molecules and the walls remain unchanged. This is a direct consequence of Galileo's principle of relativity: the laws of physics are the same in all inertial frames of reference.

Step-by-step Solution

1. Understanding Pressure from Kinetic Theory: The pressure $P$ exerted by an ideal gas is given by the formula:

where $N$ is the number of molecules, $V$ is the volume of the vessel, $m$ is the mass of a single molecule, and $\langle v^2 \rangle$ is the mean square speed of the gas molecules. This $\langle v^2 \rangle$ represents the average of the square of the *random thermal velocities* of the molecules. It is these random motions that cause collisions with the walls.

2. Gas in a Stationary Vessel:

When the vessel is stationary, the gas molecules move randomly with thermal velocities. The pressure is determined by these random motions and the frequency and force of their collisions with the stationary walls.

3. Gas in a Moving Vessel (Rocket): Now, consider the vessel moving with a constant velocity $\vec{V}_{rocket}$ (500 m s$^{-1}$ vertically upwards) relative to the ground. Every gas molecule inside the vessel, in addition to its random thermal velocity $\vec{v}_{thermal}$, also possesses this bulk velocity $\vec{V}_{rocket}$. So, the absolute velocity of a molecule $\vec{v}_{abs}$ as observed from the ground frame is:

4. Analyzing Collisions with Walls: The walls of the vessel are also moving with the velocity $\vec{V}_{rocket}$. When a gas molecule collides with a wall, the change in momentum, and thus the force exerted on the wall, depends on the velocity of the molecule relative to the wall. Let $\vec{v}_{molecule}$ be the velocity of a gas molecule and $\vec{v}_{wall}$ be the velocity of the wall, both measured from the ground frame. The velocity of the molecule relative to the wall is $\vec{v}_{rel} = \vec{v}_{molecule} - \vec{v}_{wall}$. Since the entire vessel is moving uniformly, $\vec{v}_{wall} = \vec{V}_{rocket}$. And, as established in Step 3, $\vec{v}_{molecule} = \vec{v}_{thermal} + \vec{V}_{rocket}$. Therefore, the relative velocity is:

This shows that the velocity of the gas molecules *relative to the walls* is exactly their random thermal velocity.

5. Conclusion on Pressure:

Since the relative velocities of the gas molecules with respect to the container walls remain unchanged by the uniform motion of the vessel, the frequency and force of collisions with the walls also remain unchanged. Consequently, the pressure exerted by the gas on the walls, as observed from the ground, remains the same. The uniform motion of the rocket simply shifts the reference frame; it does not alter the internal dynamics of the gas relative to its container.

> [!WARNING]
> Do not confuse the absolute velocity of the gas molecules in the ground frame with their velocity relative to the container walls. Pressure is determined by the latter. If we were to calculate the mean square speed $\langle v_{abs}^2 \rangle$ in the ground frame, it would indeed be different from $\langle v_{thermal}^2 \rangle$. However, this is not the relevant quantity for pressure *on the walls of the moving vessel*. The pressure formula $P = \frac{1}{3} \frac{N}{V} m \langle v^2 \rangle$ implicitly uses the mean square speed *relative to the container*.

6. Evaluating the Options:

* (A) "remains the same because $500\ \text{m s}^{-1}$ is very much smaller than $v_{rms}$ of the gas." While $500\ \text{m s}^{-1}$ might be smaller than $v_{rms}$ for many gases at NTP (e.g., for air, $v_{rms}$ is around $500\ \text{m s}^{-1}$ at $0^\circ\text{C}$), this is not the fundamental reason. Even if the rocket speed were much larger, as long as it's uniform, the pressure would remain the same.

* (B) "remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls." This is the correct and fundamental reason, as explained above.

* (C) "will increase by a factor equal to $\left(v_{rms}^2 + (500)^2\right)/v_{rms}^2$ where $v_{rms}$ was the original mean square velocity of the gas." This option incorrectly assumes that the pressure depends on the absolute mean square velocity in the ground frame, which would include the rocket's velocity. This is a common misconception.

* (D) "will be different on the top wall and bottom wall of the vessel." This would only happen if there were an acceleration (e.g., due to gravity or rocket acceleration) causing a pressure gradient within the gas column. Since the rocket is moving at a constant speed, there is no such acceleration effect on the pressure distribution.

The correct explanation is that the pressure depends on the relative motion between the gas molecules and the container walls, which is unaffected by the uniform translation of the entire system.

The one-line physical reason: pressure only cares about the relative velocity between gas molecules and the wall, and adding the same constant velocity to both the gas and its container (Galilean relativity — hop into the rocket's own rest frame) leaves that relative velocity completely unchanged. Equivalently: sit inside the rocket and the gas looks exactly as it would on a lab bench — no experiment inside a uniformly-moving, sealed vessel can detect the vessel's own speed. This is why option (c)'s instinct to add $(500)^2$ onto $v_{\text{rms}}^2$ is a classic trap: it uses the ground-frame speed where only the container-frame (relative) speed is physically relevant to pressure.