# Why pressure of gas even after given some non zero constant speed remains same?

Consider a situation where a cubic vessel (with horizontal and vertical faces) contains an ideal gas at normal temperature and pressure. The vessel is being carried by a rocket at speed of $$v$$ in the vertical direction. Why will the pressure of the gas inside the vessel as seen by an observer on the ground remain the same as it was initially when the gas was not moving? Will not the temperature of the gas change inside as such speed is suddenly given to the gas? And also will not there be a pressure difference between the top and bottom walls of the vessel as such a cubic vessel will have some height too?

• Is the rocket in a gravitational field? Commented Dec 13, 2021 at 15:40
• Yeah Sir , but the problem just ask what happens to the pressure of gas when its given that speed of v , suppose 500m/s from intial point when it was having no speed (container) , afterwards affect of gravity not considered . Commented Dec 13, 2021 at 16:45
• So we are not talking about during acceleration from zero to $v$, or the static effect of gravity on pressure distribution, just comparing constant vertical velocity with the rocked sitting on the pad. Correct? Commented Dec 13, 2021 at 17:10
• > "the pressure of the gas inside the vessel as seen by an observer on the ground" How would observer on the ground manage to "see pressure" inside a moving vessel? Pressure cannot be seen, it has to be measured using manometer or calculated from other quantities measured. These measurements usually take place in the frame of the gas, so in this case, on the rocket. Commented Dec 13, 2021 at 17:17
• Yeah @Bob_D Sir Commented Dec 14, 2021 at 7:41

Will not the temperature of the gas change inside as such speed is suddenly given to the gas?

No. Temperature of gas depends on average molecule speed relative to container barycenter. So giving constant $$\Delta v$$ to container does not change the way molecules moves inside container. However if you would constantly shake the container to the left and right with high frequency,- due to walls inducing additional pressure $$\Delta P$$ which accelerates/decelerates molecules- this can and probably would change the way molecules moves and consequently - average speed of molecules.

• Is there a nice proof of this statement "So giving constant Δv to container does not change the way molecules moves inside container" ? Sir and in case of liquid can temperature be changed ? Is the freedom of moving very much to gas molecules is what makes its temperature not changed but in case of liquid it can change ? Commented Dec 14, 2021 at 8:50
• Proof is simple vector algebra : $\overline {\vec v}_{molecule} = \overline {\vec v}_{molecule_{relative}} - \vec v_{barycenter}$. Second proof is that all physical laws in different inertial frames are the same. If you would detect different gas temperature in a moving container with constant velocity, then this would break The Maxwell-Boltzmann Distribution law, cause now it would depend on exact inertial reference frame. Thus gas temperature (or average speed of molecules) does not depend on container constant speed, but rather on local speed of molecules inside container. Commented Dec 14, 2021 at 10:03
• Hmm i agree Sir what about the case instead of gas its liquid ? Also gas temperature in case when a gas container is at rest but we are observing it from a frame which is moving with v speed ? What would happen in that case as both the container and gas will this time have v added to them isnt? Unlike the case i discussed above where just the container is given v speed ? Commented Dec 14, 2021 at 17:14
• It does not matter who moves at const velocity - container or observer or both. This according to Galilean relativity principle will just change molecule and container barycenter relative velocities. However, if you'll subtract one from another, you'll get same local molecule velocity. Situation would change only in case container moves with acceleration or is in a gravitational field, which would induce particles concentration gradient, and thus - pressure and/or temperature gradient. Commented Dec 14, 2021 at 17:40
• Reason for a down-vote ? Commented Dec 21, 2021 at 17:13

Why will the pressure of the gas inside the vessel as seen by an observer on the ground remain the same as it was initially when the gas was not moving?

I'm not sure how an observer on the ground can measure the gas pressure in the rocket, but there will be no difference between the gas pressure in a rocket sitting on the ground versus the gas pressure in the rocket moving at a constant vertical velocity $$v$$. That's because the vertical velocity $$v$$ does not introduce any additional (to gravity) external forces on the gas.

It's the same for a car traveling at constant horizontal velocity $$v$$ versus it not moving. The gas pressure (and temperature) measurement will be the same. Both are inertial frames of reference where the laws of physics will be the same.

Will not the temperature of the gas change inside as such speed is suddenly given to the gas?

Now you're talking about what happens during acceleration, not constant velocity.

During the time the rocket accelerates from rest to velocity $$v$$, the gas is subjected to inertial (pseudo) forces (so called $$g$$ forces), which can result in temporary pressure (and temperature) gradients in the gas. The temperature and pressure may increase towards the bottom of the container and decrease towards the top. Once constant velocity is reached, the gradients will disappear and the equilibrium pressure (and temperature) return to the launch pad levels.

In the answer given in original problem it was said " its because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls " May u tell what it means

It is correct only if the "motion of the vessel" referred to involves a constant speed in a straight line.

The motion of the vessel only results in a constant velocity of the collection of the gas molecules and the vessel that contains the molecules in the direction of motion. But since both the vessel and the enclosed collection molecules collectively move in the same direction at the same speed, there is no change in the relative motion between the molecules and the vessel that contains them, and no change in the relative motions between the molecules themselves.

Imagine something (for example, a chain) dangling from the rear view mirror of a car. When the car is stopped, the chain hangs vertically from the mirror and does not move (e.g. sway) relative to other items in the car. When the car is moving at constant speed in a straight line the chain still hangs vertically from the mirror. There is no relative motion between the chain and other things in the car. Only when the car accelerates will there be relative motion (sway) between the chain and other things in the car.

Also in your answer u mean pressure will be a constant value after acc happened and constant velocity attained but will pressure be same at top and bottom ? I thought it would be slightly different if cube vessel has some height ?

Yes, but unless the height of the vessel (height of the column of gas) is extremely high, the difference in pressure between the top and bottom will be extremely small.

For example, consider atmospheric pressure. Let's say the bottom of the vessel is at sea level where the absolute pressure is 1 atmosphere (about 101325 Pa). Using the following equation from the Engineering Toolbox web site, once can calculate the pressure in atmospheres as a function of height $$h$$ in meters.

p = 1 atm x (1-2.225577 10$$^{-5}$$h)$$^{5.25588}$$

For $$h$$ = 1 meter

p=0.99988 atm

which for all practical purposes is an immeasurable difference.

Temperature will be unique ? If yes why ?

I'm not sure what you mean by the temperature being "unique". At equilibrium, the temperature throughout the gas is the same and has a unique value. If it is an ideal gas, the temperature is given by the ideal gas equation

$$Pv=nRT$$

$$T=\frac{Pv}{nR}$$

And what energy converts to the internal energy (gradient in Temperature) during acceleration process Sir?

Even though there may be temperature gradients within the gas during acceleration, there is no change in the internal energy of the gas during acceleration. To explain:

The change in internal energy $$\Delta U$$ for a closed system (one that does not exchange mass with its surroundings) is given by the first law

$$\Delta U=Q-W$$

Where $$Q$$ is heat and is positive if added to the system, and $$W$$ is work and is positive if work is done on the system (compression of the gas) and negative if work is done by the system (expansion of the gas).

Let us assume the vessel in question is rigid, i.e., it can neither expand or contract. For such a vessel there can be no expansion or compression of the gas. Thus $$W=0$$.

Let us further assume the vessel is surrounded by perfect thermal insulation. That means there can be no heat transfer between the gas and outside the vessel (surroundings). Thus $$Q=0$$.

Therefore, from the first law, for this vessel $$\Delta U=0$$.

So for this vessel, the change in internal energy is necessarily zero, regardless of whether the vessel is accelerating or moving at constant speed in a straight line.

For an ideal gas, any process, the change in internal energy is a function of temperature only, or

$$\Delta U=nC_{v}\Delta T$$

Keep in mind that $$\Delta T$$ is the change in the equilibrium temperature of the gas, which in this case is the temperature before and after the acceleration.

Bottom Line:Although during acceleration there may temperature (and pressure) gradients due to the inertia of the gas molecules causing them to pile up (compress) at the bottom and move apart (expand) at the top, the average temperature and pressure of the gas as a whole will be the same as the equilibrium temperature and pressure before and after the acceleration. And that is simply because there is no change in the internal energy of the gas before, during, or after the acceleration of the vessel '

Hope this helps.

• In the answer given in original problem it was said " its because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls " May u tell what it means @Bob_D Sir ? Also in your answer u mean pressure will be a constant value after acc happened and constant velocity attained but will pressure be same at top and bottom ? I thought it would be slightly different if cube vessel has some height ? Temperature will be unique ? If yes why ? And what energy converts to the internal energy (gradient in Temperature) during acceleration process Sir? Commented Dec 14, 2021 at 7:56
• .. and how it get lost Sir finally from the gas ? So as to reach the same temperature? And one more thing Sir can we say in short that intially there was just g force acting downwards for all the gas particles and finally too as there is not net force(constant velocity) acting on the vessel there also just the g force acting on the gas particles hence temperature will be same although it might take a little bit time for that to be attained ? (Acoustics) Commented Dec 14, 2021 at 8:00
• @Orion_Pax I will update my answer to respond, when I get the chance. Commented Dec 14, 2021 at 16:04
• Yeah Sure Sir :) Commented Dec 14, 2021 at 17:10
• @Orion_Pax I have updated my answer to respond to your first group of comments above. Regarding your second group of comments I'm afraid I'm having a hard time following them. Considering they may be related to my response to the first group of comments, perhaps my answers to those above will address the last group of comments. In any case, I'm sorry, but I've already spent a considerable amount of time on this and have no more time to devote to it. Commented Dec 14, 2021 at 17:53