# Mathematical proof of ΔQ=ΔH at constant external pressure

Consider a system with an ideal gas. A process is being carried out in which the gas is being compressed at a constant external pressure ($P_{ext}$). For this process ΔQ should be equal to ΔH, but this is where I am stuck.

Work done by the system will be: $$ΔW= -P_{ext}ΔV$$ and $$ΔH=ΔU+Δ(P_{sys}V)$$Now we can write: $$ΔQ=ΔU-ΔW = ΔH-Δ(P_{sys}V)-(-P_{ext}ΔV)$$ In this expression $Δ(P_{sys}V)$ and ($P_{ext}$ΔV) are not equal as the process is irreversible and the so the external pressure will remain constant throughout the process but the internal pressure varies. I know that at the initial and final stage the external and internal pressure will be equal as the system will be then at equilibrium with the surrounding but the magnitude of the internal pressure will be equal to the constant external pressure($P_{ext}$) only at the final stage. So how will ΔQ be equal to ΔH when the process is carried out at constant external pressure?

In an irreversible process, the total force per unit area exerted by the gas on the piston face is comprised of the local thermodynamic gas pressure at the piston face (as determined by the local gas density and temperature at the piston face) plus a viscous stress related to the rate of deformation of the gas in close proximity to the piston. So the thermodynamic pressure is not uniform throughout the gas and there are viscous stresses present. By comparison, in a reversible process, the pressure is uniform throughout the gas and the viscous stresses are negligible. In both cases, if the piston is massless, the force per unit area exerted by the gas on its side of the piston is exactly equal to $p_{ext}$, the force per unit area on the outside of the piston, and the work will be the same, as calculated from the force on either side of the piston.
If a frictionless piston has mass and the cylinder is horizontal, the piston will oscillate about its final position until the viscous stresses damp out its motion (i.e., its kinetic energy dissipates). In the end, the final gas pressure will then be equal to $p_{ext}$, and, from a transient force and energy balance on the piston, one can show that the total work done by the gas will still integrate to $p_{ext}\Delta V$.
If the compression takes place irreveribly by, at time zero, suddenly increasing the external force per unit area from the initial gas pressure $p_0$ to the constant external force per unit area $p_{ext}$, the change in enthalpy of the gas from its initial state to its final state will not be equal to Q. We will have $$W=p_{ext}(V_f-V_0)$$ and $$\Delta U = Q-W=Q-p_{ext}(V_f-V_o)$$so$$Q=\Delta U+p_{ext}(V_f-V_o)$$while$$\Delta H=\Delta U+(p_{ext}V_f-p_0V_0)$$
• Q will be equal to $\Delta H$ if the gas compression is carried out by deliberately removing heat from the system while maintaining the external pressure equal to the original gas pressure (i.e., $p_{ext}=p_0$). But, if you try to control the compression by applying any other external pressure than the original gas pressure, Q will not equal $\Delta H$. Oct 16 '15 at 19:14
• I ran out of words on my previous comment, so I want to continue. Regarding the ambiguity, sometimes they mean that $p_{ext}=p_0$, while at other times they just mean that $p_{ext}$ is constant, but not necessarily equal to $p_0$. In the former case, $Q=\Delta H$, while, in the latter case, it is not. You just need to consider the context of the discussion. Oct 17 '15 at 11:46