I know that the internal energy of an ideal gas system doesn't change during an isothermal process, because an isothermal process requires an equal and opposite amount of work and heat to be exchanged between the system and its surroundings. For example, if the system does some amount of work on its surroundings, then it will absorb the same amount of heat from its surroundings. My question is, how does the system exchange this exact balance of heat and work to maintain the same internal energy? How come the heat doesn't change the temperature, but instead always just exactly replaces the internal energy lost by the system to work? What restrictions make the system undergo such a process?
[Physics] During an isothermal process for an ideal gas, how does the heat absorbed only replace exactly the work done by the system
thermodynamics
Related Solutions
As stated in other answers, the internal energy $U$ depends on other variables, e.g. volume, if internal interactions are not negligible.
On the other hand, there's a simple experiment which shows that for nearly ideal gases, such as air at standard conditions, $U$ is a function of only temperature $T$. (It is therefore assumed for ideal gases, that $U$ is exactly a function of $T$.)
In the experiment, we have an adiabatic container with a wall that divides it in two compartments: one side is filled with air at atmospheric pressure, and the other is a vacuum. Then, we remove the wall and let the gas expand. If we measure the temperature of the gas during the process, we observe the temperature practically stays constant.
Now let's look at the whole process. Since the gas is thermally insulated, it doesn't exchange heat with the surroundings: $$Q=0.$$ The work done on the gas during the process is zero, since there's no force that keeps the gas from expanding:$$W=0.$$ So, by the conservation of energy: $$\Delta U = Q+W=0.$$ The net result is a change in the volume occupied by the gas (with an inverse change in pressure), while $\Delta U =0$. So, we conclude internal energy is independent of the volume for ideal gases.
A P-V diagram is a good way to show what happens as the area under the graph is the work done and the gradient for an isothermal change is always larger (less negative) than that for an adiabatic change (the gradients are negative).
Compression from a volume $V_i$ to a volume $V_f$ results in a higher final pressure for an adiabatic change $P_{fa}$ than for an isothermal change$P_{fi}$ and hence more work is done during the adiabatic change.
Because the temperature has increased after the adiabatic change the internal energy of the system after the adiabatic change is greater than that after the isothermal change for which there is no change.
Best Answer
In an isothermal process, as you say, the temperature of the system doesn't change. If your question is "why?" the answer is simple: by definition.
If your question is "how?", then there are two possible answers. The first is it doesn't matter, whatever process you come up with, which makes the temperature constant, will be called isothermal.
A more useful answer, however is that typically this applies to systems which are in contact with a large heat reservoir. So, the "condition" that you mention is given by the zeroth law of thermodynamics, which imposes thermal equilibrium of systems in contact with each other. To ensure this happens during the whole process, the process has to be slow enough to guarantee the equilibrium is reached at all times.