# 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

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 .