Why doesn't the temperature reduce after free expansion of a gas?

Question

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?

Answer 1

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.

Answer 2

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. $$T_\text{after} = T_{\text{before}}$$

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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 $\mathrm{d}T=0$