# Is dH work dependent?

In the equation $\ce{d}H=\ce{d}U+\ce{d}(PV)$

Under constant pressure is $\ce{d}(PV)$ equal to the work done ($W$) on the system? (where $W= P_{\text{ext}}\mathrm{d}V$ (for irreversible process) and W = $P_{\text{int}}\mathrm{d}V$ (for reversible process)) And is this $W = \Delta{n_g}RT$?

• What is your question exactly ?? – Soumik Das Mar 17 '18 at 5:08
• I've edited it. please do check it out – J_B892 Mar 17 '18 at 5:10

## 2 Answers

Enthalpy is a physical property of a substance (including even a mixture of substances undergoing a chemical reaction). It is thus a function of state, and its change depends only on the initial and final states, and not on the process path between the initial and final states (of the substance or reacting mixture). Since work is a process-path-dependent entity, enthalpy can generally be regarded as independent of work.

When a system undergoes any thermodynamic process( without undergoing any chemical conversion), it is assumed to be a closed system, i.e mass of the system doesn't change. So, in a closed system $\mathrm{\Delta n_g =0}$.
So, in that closed system under constant pressure, $\mathrm{d(PV) =PdV}$. and also differentiating both sides of the ideal gas equation $$\mathrm{d(PV) =d(n_gRT) -> PdV = n_gRdT (as~ dn_g =0~ by~ definition)}$$So, the work in reversible process is $\ce{P_{int}dV = n_gRdT}$.
But work in a irreversible process is different which is just $\mathrm{P_{ext}(V_2 -V_1) = P_{ext}n_gR(T_2/P_2 - T_1/P_1)}$.
But, if you consider a process where a chemical conversion occurs at a constant Temperature,there no of gaseous molecules in the reactants and the products are different. There you can write,$\mathrm{dH = dU + \Delta n_gRT}$, as , $\mathrm{PV_{final} -PV_{initial} = \Delta n_gRT.}$ But I would not prefer to call the last term on the R.H.S of the equation as work .