Question

A cubic vessel (with face 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(a) remains the same because 500 ms⁻¹ 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 $\frac{(v^{2}_{rms}+(500)^{2})}{v^{2}_{rms}}\ where\ v_{rms}$ was the original mean square velocity of the gas(d) will be different on the top wall and bottom wall of the vessel

Answer 1

Hola User

$<b>$

Option C is correct

Answer 2

$\green\star\mathfrak\orange{\large{\underline{\underline{Question!}}}}$

A cubic vessel (with face 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(a) remains the same because 500 ms⁻¹ 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 $\frac{(v^{2}_{rms}+(500)^{2})}{v^{2}_{rms}}\ where\ v_{rms}$ was the original mean square velocity of the gas(d) will be different on the top wall and bottom wall of the vessel

$\red\star\mathfrak\purple{\large{\underline{\underline{Answer!}}}}$

Option C is the answer