The whole system is adiabatic, and no heat exchange can take place. If the volume of the gas now doubles, it should actually cool down.
That's why I don't understand $dT=0$
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1$\begingroup$ If $PV$ remains constant, why does $T$ need to change? $\endgroup$– Jon CusterCommented May 10, 2022 at 22:12
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$\begingroup$ @JonCuster so if i have a piston and i pull on it really hard and this expands the isolated gas, i have no temperature difference? $\endgroup$– iwabCommented May 10, 2022 at 22:16
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$\begingroup$ If there are no inter-molecular forces (so an ideal gas), then changing the volume changes the pressure. The molecules keep moving with the same rms velocity, so the temperature remains constant. $\endgroup$– Jon CusterCommented May 10, 2022 at 22:21
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$\begingroup$ @JonCuster So changing the volume of ideal gases in an adiabatic system does not change the average energy (kinetic and potential) of the particles? $\endgroup$– iwabCommented May 10, 2022 at 22:33
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3$\begingroup$ Keep in mind, in real life this never happens. Expanding a real gas in this way will usually result in cooling, but sometimes (e.g. helium or hydrogen) heating, depending on the sign of the gas's Joule Thompson coefficient. $\endgroup$– RC_23Commented May 10, 2022 at 23:38
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6 Answers
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if i have a piston and i pull on it really hard and this expands the isolated gas, i have no temperature difference?
In the case of the moving piston, the molecules of the gas are striking a moving wall. The pressure of the gas creates a force and the motion of the wall means this happens over a distance. So work is being done. This work comes at the expense of energy in the gas and it cools down.
In the Joule expansion, there is no moving wall for the gas to strike. There is nowhere for the energy to leave, so the temperature remains constant.
answered May 10, 2022 at 22:43
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2$\begingroup$ So theoretically, the faster I would pull the piston, the less the temperature would have to decrease? If I could do it instantaneously, the particles would have no chance of touching the piston in the second and losing momentum and thus speed because of the relative movement. $\endgroup$– iwabCommented May 10, 2022 at 23:07
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$\begingroup$ Basically, yes. This is the same as free expansion in the Joule experiment. Opening the isolation valve between the two chambers is essentially the same as having a piston separate the two and instantaneously move the piston to expose all the receiving chamber. $\endgroup$ Commented May 11, 2022 at 2:36
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$\begingroup$ Above continued. This is work by the gas. For a non-massless piston with gas on the other side (opposite side of piston from the gas under consideration), there is work required to instantaneously move the piston, but it is not work done by the gas under consideration. See physics.stackexchange.com/questions/436339/work-done-by-a-gas for details. $\endgroup$ Commented May 11, 2022 at 3:03
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$\begingroup$ Alternativeely, I always liked the explanation that the molecules in the air are (elastically) colliding with a receding surface, which reduces their speed. $\endgroup$– ArthurCommented May 11, 2022 at 20:24
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I like to think about these kinds of thermodynamic problems using kinetic theory and Newtonian mechanics, and not really worry about the ideal gas equation. If we look at problems this way, then temperature changes are really easy to understand. For example, let's consider the classic example where you adiabatically compress a gas in a cylinder. Kinetic theory tells us that we can envision a bunch of tiny point masses bouncing around with random velocities. As the shaft of the piston moves downwards, it collides with some of the molecules, and because of the equations for momentum conservation, this means that the gas molecules will ricochet off of the shaft with a higher velocity than when they entered, so they'll move faster and have a higher temperature. The same is true for when you pull a piston shaft upwards, allowing the gas to expand. The gas strikes the shaft as its moving upwards, and thus they ricochet off with a smaller velocity than when they entered. In Joule expansion, there's no momentum changes nor forces exerted on any individual molecule, and therefore there speeds can't increase or decrease, so it's the same temperature.
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The local cooling throughout the ideal gas taking place during this irreversible expansion is exactly offset by the local viscous heating of the gas (viscous stresses combined with high velocity gradients). For an ideal gas, the net effect is no temperature change. See a course in fluid mechanics.
answered May 11, 2022 at 0:50
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We now can detect is a slight change in temperature of the air used in the Joule experiment that Joule was unable to detect. This is because a gas is not an ideal gas. Assuming the gas is ideal, there is no change in temperature.
Consider the air on both the initial and final chambers as the system, a closed thermodynamic system (constant mass). Joule observed no change in temperature of the surrounding water bath and concluded no heat transfer had taken place to the air. Since the work done was also zero, he concluded from the first law of thermodynamics that the internal energy of the air was constant. Since the pressure changed, he concluded that the internal energy of a gas is a function of temperature alone; this is true for an ideal gas.
A couple of additional observations based on other responses and comments
The gas is expanding into an (essentially) vacuum with zero pressure and therefore does no work; this is free expansion and is not the same as expansion against a piston with a pressure on the other side.
Considering the initially empty chamber as a fixed volume system (open thermodynamic system), there is no work done on the fixed volume system as the gas expands to fill it.
A good textbook on thermodynamics, such as one by Sonntag and Van Wylen, evaluates the Joule experiment from three points of view: (1) air on both the initial and final chambers as the system, (2) gas expanding into a vacuum with zero pressure, and (3) considering the initially empty chamber as a fixed volume system.
answered May 11, 2022 at 0:05
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The whole system is adiabatic, no heat exchange can take place
The system also has no work done on it. The idea is that you just remove a partition and the gas freely expands. This is what defines Joule expansion. No heat and no work. Therefore, no internal energy change and no temperature change (for an ideal gas).
if the volume of the gas now doubles it should actually cool down.
No, this is wrong. An ideal gas will not "actually cool down" during Joule expansion. The volume doubles, but the pressure halves.
So, to answer your question:
Why is there no temperature difference in the Joule expansion experiment?
It is because no work is done on the gas and no heat is exchanged with the gas, therefore the energy is constant. The energy of an ideal gas is $\frac{3}{2}NkT$. The particle number $N$ is constant, so the temperature of an ideal gas is constant in Joule expansion.
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It depends on what kind of gas are you considering: Real or Ideal. For idelas gases there is not change in temperature because the particles are free with zero intermolecular forces. @Eli's answer explains it further. Whereas if you have a real gas then the temperature must change because real gas moleculars are not entirely free. They attract each other. So, when they move apart they utilise their KE to break the bonds and the thus temperature decreases.
