I am currently studying Thermodynamics from Khan Academy.
In one of their videos, Sal explains how the free expansion of a gas doesn't change the temperature of the system because the average kinetic energy of the molecules of the gas doesn't change during the expansion. I've a question regarding this.

Consider a fixed amount of gas inside a box. Say this box is inside a box that is 10000 times in volume than our original box. I then disappear the walls of our original box and the gas starts to expand throughout the large box. In this case also, the average kinetic energy of the molecules stay the same.

Does this mean that the temperature wouldn't be reduced?

It clearly doesn't seem so.
Say I use a thermometer having an initial reading of $T_0$ such that $T_0<T$ to test the temperature before and after expansion. The thermometer registers temperature $T$ before expansion because the gas molecules are able to increase the kinetic energy of the particles in the thermometer to report $T$. If I test temperature using the thermometer after expansion from an arbitrary point in the box, there wouldn't be as much molecules hitting the thermometer to raise its temperature to report $T$.

Would the thermometer still report $T$ after a considerable amount of time, knowing that the gas has expanded very much?
Would the temperature of the system still be $T$ after the expansion?

  • 2
    $\begingroup$ Does this answer your question? Temperature of gases $\endgroup$ Mar 12, 2022 at 21:17
  • $\begingroup$ I suggest reading through the discussions found from a search on this site of free expansion temperature and Joule expansion temperature. Temperature doesn't depend on having a certain rate of molecular collisions. $\endgroup$ Mar 12, 2022 at 21:18
  • $\begingroup$ @Chemomechanics Temperature doesn't depend on having a certain rate of molecular collisions. Isn't it important when we're practically measuring temperature? $\endgroup$
    – Silica19
    Mar 12, 2022 at 21:21
  • 1
    $\begingroup$ Yes; when we talk about a thermometer in idealized problems, however, we generally assume that its reading corresponds to the equilibrium value or—equivalently—that it has sat in its environment long enough to thermally equilibrate. Yes, this takes longer in a rarefied gas, but the final value for a given temperature is the same. $\endgroup$ Mar 12, 2022 at 22:13
  • $\begingroup$ There is an MIT OCW course, I'd say its more rigorous treatment of the issue than Khan acads vids $\endgroup$ Mar 12, 2022 at 22:14

2 Answers 2


The temperature does not reduce in free expansion of an ideal gas because the expansion cooling is exactly offset by irreversible viscous heating caused by rapid expansion (deformation) of the gas. For a real gas, the two effects do not exactly cancel, and there is a small amount of either residual cooling or residual heating.

  • $\begingroup$ Can you please elaborate what do you mean by irreversible viscous heating caused by rapid expansion of the gas? Do you mean that the surroundings which was at $0$ K before the expansion see an increase in temperature? Also, how does expansion cooling occur? $\endgroup$
    – Silica19
    Mar 13, 2022 at 11:08
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    $\begingroup$ Are you familiar with the concept of viscous heating? When a fluid shears between parallel plates or expands very rapidly, the viscous stresses in the fluid produce viscous dissipation of the mechanical energy to internal energy. This has noting to do with the surroundings. It occurs within the body of the fluid (system). In an expanding gas, this occurs simultaneously with the expansion cooling. The latter occurs when small parcels of the fluid within the system do local work on adjacent parcels of fluid. $\endgroup$ Mar 13, 2022 at 11:17
  • $\begingroup$ No, I didn't have any idea about this; do know it now. My question is more about when the gas has expanded and a thermometer is introduced to record temperature, would the magnitude be same as when tested before expansion (say $T$)? I don't think so because there wouldn't be as much molecules hitting the thermometer to raise its temperature to report $T$ if the gas has expanded into a very large volume. $\endgroup$
    – Silica19
    Mar 13, 2022 at 11:23
  • $\begingroup$ For an ideal gas the initial and final temperatures will be the same. The pressure acting on the thermometer will be less due to less molecules hitting the thermometer. $\endgroup$ Mar 13, 2022 at 11:29
  • $\begingroup$ If less molecules are hitting the thermometer, then less kinetic energy is being transferred to the particles of thermometer compared to before expansion. This means that the thermometer would record a lower temperature. $\endgroup$
    – Silica19
    Mar 14, 2022 at 5:32

Consider the thermosphere in the Earth's atmosphere.

The highly attenuated gas in this layer can reach 2,500 °C (4,530 °F) during the day. Despite the high temperature, an observer or object will experience cold temperatures in the thermosphere, because the extremely low density of the gas (practically a hard vacuum) is insufficient for the molecules to conduct heat.

That is to say, the temperature is not related to the number of gas molecules hitting the thermometer. Rather, the temperature is measured through the kinetic energy of the gas molecules. (for ideal gas) $$E = \frac{3}{2}k_BT$$ Also, noted that $T$ is an intensive quantity, instead of an extensive quantity. It has nothing to do with the volume of the container unless energy is added/extracted (such as compressing the container with constant pressure, adding work $\Delta W = P \Delta V$)

Now to answer your question, since neither work nor heat is added to the system, $\Delta U = \Delta E_k = \Delta n = 0$, where $U$ is the internal energy and $n$ is the number of molecule. The average kinetic energy per molecule is unchanged, thus the measurement of $E$ is unchanged, thus the thermometer should read the same. $$T_\text{after} = T_{\text{before}}$$

Illustrated mathematically, for ideal gas, we have $U = U(T)$ instead of $U = U(T,V)$. From the first law of thermodynamics, we have: $$\mathrm{d} U = \bar{\mathrm{d}}Q + \bar{\mathrm{d}}W$$ $$\mathrm{d} U = C_v\mathrm{d}T-p\mathrm{d}V$$ since $$-p = \big(\frac{\partial U}{\partial V}\big)_T=0$$ therefore $$\mathrm{d} U = C_v\mathrm{d}T$$ From $\mathrm{d}U=0$, $C_v$ is a constant, we have $\mathrm{d}T=0$


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