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My book has this line:

In an isothermal expansion, the gas absorbs heat and does work

But, if a gas absorbs heat, then how can its temperature remain constant? (as the process is isothermal)

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From the first law, $\Delta U=Q-W$. If the process is isothermal, then $\Delta U=0$ (assuming an ideal gas), so Q = W. The heat energy is consumed by your gas doing work, so there is no change in internal energy (temperature).

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Hopefully I can provide you with some physical intuition on this phenomenon through a simple example.

Imagine a gas heated by a flame in a closed cylinder with a piston on the top balanced by the same pressure as the gas inside. As you heat the gas, the particles move faster (gain kinetic energy) and collide with the walls of the container harder, increasing the pressure of your system. This pressure increase is then used to drive the piston out. As the particles of your hot gas collide with the piston wall to move it, they necessarily dump some of their kinetic energy into it, slowing down (i.e. cooling) in the process. On the other hand, kinetic energy is not lost when the gas particles bump into the other rigid walls of the container, since those collisions are basically elastic.

So overall you have a heat source dumping kinetic energy into the gas particles from one side of the cylinder, and the gas particles dumping their kinetic energy into the surroundings while pushing a piston on the other side. Now run this process very carefully and slowly, and the temperature of the gas can be held more-or-less constant!

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    $\begingroup$ This answer is basically the same as the shorter answer by Chet Miller on Dec. 30 '15. $\endgroup$ – Maurice Jan 22 '20 at 10:45
  • $\begingroup$ I was attempting to provide some physical insight into why the other answer makes sense, but I guess that isn't appreciated here. $\endgroup$ – CuriousChemStudent Jan 22 '20 at 17:43

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