# A change in internal energy in the Joule-Thomson effect

I have trouble understanding Joule-Thomson effect, may be i am missing some basic concepts of thermodynamics. We have two chambers. When we compress ideal gas to low volume by external work (adiabatic process), We increase its internal energy and hence temperature. And then we release this compressed gas through plug to vacuum. Here is the problem arise. Final Internal energy is same as before. But how? Compressed gas is released to vacuum and hence no work is done by the gas.

I think you answered your question. If no work is done and no heat transfer occurs then by first law there is no change in internal energy

So my question is does the internal energy of the gas before work is done on it (compressing it) as same as after it is expended to vacuum?

No. The internal energy before the compression work is done on it is less than the internal energy of the gas after it expanded into the vacuum.

The internal energy of the gas increased when it was compressed. When it expanded into the vacuum, its internal energy does not change because the gas does no work when it expands into the vacuum. It keeps the increase given to it by the compression. In order for the gas to do work when it expands it must expand against some resistance (force). The vacuum offers no resistance to the expansion.

Hope this helps

• We do work in first place so ideal gas internal energy is increased but in second place, gas in released to vacuum hence so no work in done in second place. Dec 23, 2019 at 16:05
• Whole process is adiabatic Dec 23, 2019 at 16:06
• And...? In the first place an external (to the system) force does work adding energy to the gas. In the free expansion of the gas it is isolated from the surroundings. Expansion in a vacuum involves no work by the gas. First law $\Delta U=Q-W$. $W=0$, $Q=0$ (adiabatic). Therefore $\Delta U =0$. What about this don’t you understand? Dec 23, 2019 at 16:15
• So my question is does the internal energy of the gas before work is done on it (compressing it) as same as after it is expended to vacuum? Dec 23, 2019 at 16:37
• I have trouble understanding joule Thomson effect. Can you explain it to me? Dec 23, 2019 at 16:40