Why does a gas do no work when expanding in a vacuum? I know it is true that there is no opposing force from other air molecules for the gas to overcome in a vacuum, so that gas molecules won't do work in the horizontal direction. But still the gas molecules have to overcome the gravity in the perpendicular direction and do work.

So I think a gas will still do work even while it's expanding in a vacuum.

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    $\begingroup$ It does do work unless it's an ideal gas . Gravity has nothing to do with the work an expanding gas does. $\endgroup$ Mar 11, 2021 at 14:12

2 Answers 2


Gas expanding into a vacuum does work on itself: the gas behind has higher pressure than the gas in front so the gas speeds up as the pressure drops ---Bernoulli in action. Internal energy is being converted into kinetic energy, so the the gas temperature drops. If it is expanding into an empty chamber, the gas will eventually hit the walls and after while friction will bring it to rest. The friction heats the gas up, and if it is an ideal gas, and the chamber is insulated so there heat has nowhere to go, the temperature will come back to to its starting value. This is the usual situation in which it is said that the gas expanding into a vacuum "does no work". It means that at the end of the day no internal energy has been lost to exteral work by the gas so $U_{\rm start}= U_{\rm end}$. Work was definitely done in intermediate steps.

It is during the frictional reheating --- by randomization of the collective flow motion --- that the increase in entropy $S_{end}-S_{start}=nR \ln (V_{end}/V_{start})$ occurs.

I'm ignoring gravity here, as it is usually insignificant.

If we are in outer space the gas just keeps going, and going, and going....


Your question can prompt answers in two seemingly separate directions. One direction considers the gas molecules as individual systems onto themselves. The other considers the entire gas as a system. Let's try to tackle both properly.

Consider an idealized picture of the gas as being self-contained even as it expands. Call that self-contained "bubble" of gas the system. The external pressure on the system is zero when that bubble is surrounded by a perfect vacuum. The thermodynamic definition of mechanical work for an expanding system is $w_{mech} = -\int p_{ext} dV = 0$. This is free expansion of the system. We say the gas as a bulk system does no work (on the surroundings).

Now step inside the system. Stand at any point in the gravitational frame of "up" and "down". Stand on a gas particle and imagine that it moves "up" in the gravitational frame. That particle has increased its potential energy in the gravitational frame. The gain in potential must come at the expense of a loss in energy specific only to the particle. If the particle is an ideal gas, it has no vibrational or rotational modes (neglect also electronic and nuclear here). So, an ideal gas particle will slow down as it moves upward in a gravitational field. Alternatively, a "real" molecule has rotational and vibrational internal energy modes. The real molecule could move to a lower rotational or vibrational state as it moves up in gravitational potential energy.

Translation, rotation, and vibration are measures of the internal energy of the gas itself. None of the changes that increase the potential energy of an individual particle at the expense of decreasing its internal energy are defined as mechanical work. They are all defined as a change in the internal energy of the system. The system is not doing mechanical work. But the gravitational field is doing other work on the individual particles.

You can now speculate about how the placement of the gas bubble relative to "bottom" and "top" walls in a box might change the considerations. What happens for example to a bubble of ideal gas that starts at the "bottom" of a box and expands (freely) "upward" compared to a bubble of ideal gas that starts at the "top" of a box and expands (freely) "downward"? Does the gas in the former get cooler because the internal energy of the gas particles decreases while the gravitational energy increases? Does it go the other way for the gas bubble that expands "downward"?

The answer depends on whether the process is adiabatic or isothermal free expansion. In the former (a perfectly adiabatic free expansion), we can predict that an ideal gas bubble that expands "upward" should be cooler at its end state compared to one that expands "downward" in a gravitational field. The gas does no work in either case. But the gravitational field does work on the individual particles. How do we make this prediction?

The ideal gas particles that expand "upward" gain potential energy. They correspondingly loose kinetic energy. This is a loss in internal energy of the gas particles. For an ideal gas, internal energy is only dependent on temperature. Lower internal energy is lower temperature.

A complementary view is to use the full expansion of the first law

$$\Delta U = q + w_{mech} + w_{oth}$$

where $w_{oth}$ is the other work. With adiabatic free expansion of the gas in a gravitational field, $\Delta U = w_{oth}$. The gas molecules do work to move upward in a gravitational field. Therefore $w_{to}$ is negative. The internal energy of the particles decreases as they move "up hill". Assuming that this internal energy change is dissipated over the entire system, we conclude that the internal energy of the bulk gas decreases as well.

  • $\begingroup$ so in a molecular level, gas indeed does work in a vacuum? $\endgroup$ Mar 13, 2021 at 9:21
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    $\begingroup$ Gas molecules that go “up” in potential energy must do other work against the gravitational field. This is true whether the gas is expanding in vacuum or against a constant external pressure. This is true whether the gas is ideal or real. If this were not the case, our atmosphere would simply go away. But gravity holds it in place. $\endgroup$ Mar 13, 2021 at 14:34

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