Let there be a box filled with an ideal gas. Now apply force on it.

Consider three cases:

  1. Acceleration from rest.

  2. Deceleration from some velocity.

  3. No acceleration at all.

How will the internal energy change? According to me, for acceleration, the normal force will do work on the gas and due to collisions with the walls of the box. But how much the change will be? Will it be equal to $\frac{1}{2} mv^2$? ($m$ is the mass of gas and $v$ is the velocity attained by the same).


People asking for research effort:-

According to me, in the first two cases, the block receives excess energy from the accelerating and decelerating forces, not in the third case. The force doing work on the gas particle is the normal force after the gas strikes the wall of the container. Although I can predict whether it will decrease or not (according to my explanation). I am confused about the magnitude of the change on the internal energy.

  • 1
    $\begingroup$ From whose frame of reference? $\endgroup$ – lemon Feb 18 '17 at 15:06
  • $\begingroup$ @lemon Take it from ground. $\endgroup$ – Reeshabh Ranjan Feb 18 '17 at 15:16
  • $\begingroup$ @sammygerbil I think that will be enough (see the update) $\endgroup$ – Reeshabh Ranjan Feb 18 '17 at 16:34
  • $\begingroup$ First of all I want to ask is the box is rigid( by the word I mean perfectly rigid)? My second question is do you consider adiabatic situation? If the following assumptions I mentioned in the questions are true then there will be no external influences into the ideal gas so the internal energy won't change. Agian if you consider that the initial and final temperatures are not same then internal energy of the gas will certainly change. $\endgroup$ – Sahil96 Feb 18 '17 at 20:16

If you are not compressing the system, then the velocity of the center of mass (with total mass $M$) is the same as the velocity $v$ of the point at which you are applying force. The rate at which you are adding energy $Fv$ is exactly equal to $$\frac{d}{dt}\left(\frac{1}{2}Mv^2\right)=Mva=Fv$$ so in this case the internal energy does not change.

But as a counterexample, take two non-interacting particles of equal mass initially at rest. Exert a force $F$ on one, bringing it up to velocity $2v$. Now the center of mass of the system is moving at velocity $v$, but the total energy is $\frac{1}{2}m(2v)^2$ not $\frac{1}{2}(2m)v^2$. So yes you can change the internal energy in general.

In general if the point at which you are applying force is moving faster than the center of mass then you are compressing the gas (maybe internally they are bunched up against the rear wall) and increasing the internal energy. If the point is moving slower, the internal energy is decreasing. In that case the gas is using some of its internal energy to expand against the point you are applying force, just like you can use the internal energy in your muscles to push off against a wall.


When the box accelerates from rest, the gas will quickly compress and, assuming an adiabatic process, its temperature and, therefore its internal energy will increase.

In addition, its kinetic energy will continue to grow as the box keep speeding up.

The compression of the gas won't be uniform: it will be the highest near the back wall and drop exponentially toward the front wall. We could use the math for calculation of the atmospheric pressure for an ideal gas and just replace g by α.

When the box decelerates from some (presumably, constant) velocity, the gas will be compressed against the front wall, with a similar pressure distribution, and its temperature and internal energy will increase accordingly.

If, at some point, the box stops, the compressed air will start oscillating and, in the ideal case, will continue to oscillate forever.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.