Why is pressure assumed constant in the work equation? Doesn't pressure change with the change with volume unless isobaric? Right now, I'm studying thermodynamics, and I am a bit confused on the differentials.
For example, the equation for work is $W=p*{\Delta}V$, and usually people change it to $dW = p*dV$.
But, as long as it is not isobaric process, pressure is changing, so can't you also do $dW=dp*V$? I know we usually do $dV$ since the expansion of volume is actually doing the work.
Or, why can't we do $dW=dp*dV$, and use maybe multivariable calculus, since both pressure and volume change?
I would greatly appreciate your help on this.
 A: We define work as dot product of Force and displacement. Now as pressure is the same in all directions, you could write $\vec{F}\cdot\vec{dx} = \frac{F}{A} Adx$ which would be equal to $Pdv$
Also note that we never write a second order differential i.e. dPdv, this is always approximated as zero.
A: I think what you're saying is that the pressure changes with volume, and you're correct. Pressure does indeed change with volume in most cases, but notice we're talking about infinitesimal work, which depends on the instantaneous external pressure producing an infinitesimal change in volume: $$dW = P_{\text{ext}} \, dV.$$ This is permissible, since we can make $dV$ arbitrarily small such that $P_{\text{ext}}$ is constant. We also assume that, throughout this process, the system is in equilibrium, so the external pressure balances out the internal pressure of the gas: $P_{\text{ext}} = -P$, meaning $$dW = -P \, dV.$$
Now, if we wanted to calculate the total work done on the gas, we would need to integrate the equation above -- and of course, we'd need $P$ to be some function of $V$ (or both $P$ and $V$ to be functions of something else). As an example, consider a system where the pressure varies with volume quadratically: $P(V) = V^2$. The external work done on this system to take it from $V = V_1$ to $V = V_2$ would be
$$ W_{12} = -\int_{V_1}^{V_2} P(V) \, dV = -\int_{V_1}^{V_2} V^2 \, dV = \frac{1}{3} \left(V_1^3 - V_2^3 \right).$$
A: It will depend upon what you choose as your independent variables.  If entropy (S) and volume are your independent variables, then
dE = TdS - PdV
And thus you get your definition of work as in your question.  But if S and P are your independent variables, then you have to first write:
d(PV) = PdV + VdP, which, rearranged, leads to PdV = d(PV) - VdP.  Substituting into the above:
dE = TdS - (d(PV) - VdP)
Rearranging:
d(E + PV) = TdS + VdP
We define E + PV as the enthalpy, H, and thus:
dH = TdS + VdP
You can play this game again with T and V as independent variables and define the Helmholtz free energy, or T and P as independent variables and define the Gibbs free energy.
So which variables you decide to choose as independent and which as dependent, will guide you in how your energy equation is written.  Sometimes your experiment decides for you.
