Isobaric process

Remarks

If the volume changes from $V_1$ to $V_2$, the work done on the fluid is $W = -p (V_2 - V_1)$. The heat absorbed is then $Q = \Delta U + p \Delta V = \Delta H$, where $U$ is the internal energy and $H = U + pV$ is the enthalpy. For an ideal gas, an isobaric process satisfies $V / T$ constant, where $T$ is the thermodynamic temperature. Heating a gas in a rigid closed container is not isobaric: its volume is fixed, so its pressure rises.