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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.

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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

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  • $\begingroup$ 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. $\endgroup$ – Muhammad Numan Dec 23 '19 at 16:05
  • $\begingroup$ Whole process is adiabatic $\endgroup$ – Muhammad Numan Dec 23 '19 at 16:06
  • $\begingroup$ 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? $\endgroup$ – Bob D Dec 23 '19 at 16:15
  • $\begingroup$ 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? $\endgroup$ – Muhammad Numan Dec 23 '19 at 16:37
  • $\begingroup$ I have trouble understanding joule Thomson effect. Can you explain it to me? $\endgroup$ – Muhammad Numan Dec 23 '19 at 16:40

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