Isothermal Process

An isothermal process is any process carried out in such a way that the temperature of the system remains exactly constant throughout. In practice this is achieved by keeping the gas in very good thermal contact with a large heat reservoir held at a fixed temperature -- large enough that it can absorb or supply any amount of heat exchanged during the process without its own temperature changing appreciably -- and carrying the process out slowly enough that the gas has time, at every stage, to exchange whatever heat is needed to stay at the reservoir's temperature.

Since $T$ is fixed throughout, the ideal gas equation of state for $n$ moles, $PV = nRT$, shows that the product $PV$ itself stays constant during an isothermal process:

A graph of pressure $P$ against volume $V$ for an isothermal process is therefore a curve of the form $P = \text{constant}/V$ -- a rectangular hyperbola -- and every isothermal curve for a given amount of gas at a different fixed temperature is a different hyperbola, lying further from the origin (higher up on the graph) the higher the temperature it represents.

Change in internal energy. Because the internal energy of an ideal gas depends only on its temperature, and the temperature does not change at all during an isothermal process, $\Delta U = 0$ identically, for any isothermal process, whether the gas expands or is compressed, and regardless of how much heat is exchanged. Applying the first law, $Q = \Delta U + W$, with $\Delta U = 0$, gives simply

so during an isothermal expansion, every joule of heat absorbed by the gas from the reservoir is converted, in its entirety, into work done by the gas on its surroundings -- none of it goes into raising the gas's own internal energy, since that stays fixed. Conversely, during an isothermal compression, the work done on the gas by the surroundings is given out, in its entirety, as heat rejected to the reservoir.

Work done. For $n$ moles of an ideal gas at fixed temperature $T$, expanding reversibly (i.e., slowly enough to remain effectively in equilibrium at every stage) from an initial volume $V_1$ to a final volume $V_2$, the work done by the gas is found by integrating $dW = P\,dV$ using the equation of state $P = nRT/V$:

$W = \int_{V_1}^{V_2} P\,dV = \int_{V_1}^{V_2} \frac{nRT}{V}\,dV = nRT\ln\!\left(\frac{V_2}{V_1}\right)$ …