A cubic vessel (with face horizontal + vetical ) 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 vertical direction. The pressure of the gas inside the vessel as observed by us on the ground.

Options

A. remains the same because 500 m/s is very much smaller than `v_(rms)` of the gas
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
C. will increase by a factor equal to `[v_(rms)^(2)+(500)^(2)]//v_(rms)^(2),"where"v_(rms)` was the original mean square velocity of the gas
D. will be different on the top wall and bottom wall of the vessel.

Explanation

To solve the problem, we need to analyze the situation of the cubic vessel containing an ideal gas at Normal Temperature and Pressure (NTP) while it is being carried by a rocket moving at a speed of 500 m/s in a vertical direction. We want to determine how this motion affects the pressure of the gas inside the vessel as observed from the ground.

### Step-by-Step Solution:

1. **Understanding NTP Conditions**:
- At Normal Temperature and Pressure (NTP), the conditions are defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm (101.325 kPa). The gas is in equilibrium under these conditions.

2. **Effect of Rocket's Motion**:
- The rocket is moving vertically at a speed of 500 m/s. We need to consider how this affects the pressure of the gas inside the vessel.

3. **Pressure and Gas Molecules**:
- The pressure of a gas is determined by the collisions of gas molecules with the walls of the container. The pressure can be expressed using the ideal gas law:
\[ P = \frac{nRT}{V} \]
- Here, \(n\) is the number of moles of gas, \(R\) is the universal gas constant, \(T\) is the temperature, and \(V\) is the volume of the gas.

4. **Relative Motion**:
- The motion of the vessel as a whole (the rocket's movement) does not affect the relative motion of the gas molecules with respect to the walls of the vessel. The gas molecules continue to collide with the walls of the vessel in the same manner as before, regardless of the vessel's overall motion.

5. **Conclusion on Pressure**:
- Since the relative motion of the gas molecules with respect to the walls remains unchanged, the pressure exerted by the gas inside the vessel will also remain unchanged. The pressure as observed from the ground will be the same as it was at NTP.

6. **Final Answer**:
- Therefore, the pressure of the gas inside the vessel, as observed by us on the ground, remains the same.