Isobaric Process

An isobaric process is one that occurs at constant pressure, so $\Delta p = 0$. Water boiling at a fixed (typically atmospheric) pressure is the standard everyday example.

Work done. Since pressure is constant, it can be pulled outside the work integral $W = \int p\,dV$, giving simply:

using the ideal gas equation to rewrite $p\,\Delta V$ in terms of temperature change, where $T_i, V_i$ and $T_f, V_f$ are the system's initial and final temperature-volume pairs.

Change in internal energy. The change in internal energy is given generally, at constant volume-heat-capacity, by

where $C_V$ is the molar specific heat at constant volume, and $\Delta T = T_f - T_i$.

Heat exchanged. Applying the First Law, $Q = \Delta U + W$, and substituting Eqs. (4.10) and (4.11):

Defining $C_p = C_V + R$ as the molar specific heat at constant pressure, this simplifies to:

So in an isobaric process, unlike an isothermal one, the temperature genuinely changes ($\Delta T \ne 0$), and consequently so does the internal energy (Eq. 4.11); the heat added (Eq. 4.12) is 'used' for two things at once — part of it raises the system's temperature/internal energy, and part of it goes into doing work as the gas expands or contracts. How this heat splits between the two depends on the value of $C_p$: gases with a larger $C_p$ need more heat input for a given temperature rise, since more of that heat is 'competing' to also do work. …