# When differential of $p$-$V$ $dW$ work equals $-pdV$?

The differential form of the first law takes the form $$dU=\delta Q + \delta W$$ or $$dU=TdS-pdV.$$ First, we know that $$TdS=\delta Q$$ for reversible process. So we can't say that $$dU=\delta Q -pdV$$. Assuming only $$p$$-$$V$$ work can we always say that $$\delta W=-pdV$$? Also in many textbooks when they calculate the work for a process they state that: $$W=-\int_i^fpdV$$ without stating if the process is reversible or not. I came up with this question when I read these notes Muddiest Points.

• "So we can't say that $dU=\delta Q -pdV$." Did you mean "why can't we say....?" Apr 3, 2020 at 21:20
• I believe the negative sign in your equation for work applies to the chemistry version of the first law, i.e., $\Delta U=Q+W$ , and not the engineering and physics version $\Delta U=Q-W$. Apr 3, 2020 at 21:38

You cannot. The quantity $$PdV$$ only makes sense when you are talking of a reversible process, as irreversible process may not even have a relevant $$P-V$$ plot to talk about the quantity $$PdV$$.
This is because one can only draw a $$P-V$$ plot for states of equilibrium of a system. So, an irreversible process can have a $$P-V$$ plot consisting of just $$2$$ points - the initial and final state, as that is when the system is in equilibrium.
• What if the pressure in the $-pdV$ is the external pressure? Can we find the total work from $$W=-\int_{V_1}^{V_2}p_{ext}dV$$? Apr 9, 2020 at 14:57
The equations of state hold only for such reversible processes, for only then can these macroscopic parameters be uniquely specified. In your specific example, the idea is the gas is compressed by $$dV$$, working against the internal pressure($$P(V,T)$$), slowly enough(or, left for sufficient time) so that the compressed gas has attained equilibrium too, with the pressure now being $$P(V-dV,T+dT)$$. You then compress against THIS pressure. Clearly, the total work will become an integral of $$P(V)dV$$, as you keep doing this.
• If the process goes sufficiently slow, even with friction, isn't it true there is still well defined pressure $P$ and one can calculate $PdV$? Feb 15 at 22:48