Why $dW=pdV$ is an inexact differential? I remember an exact differential as:
$$A=M(x,y)dx+N(x,y)dy $$
and the condition for be exact is:
$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}. $$
Can I use that definition to proof that $dW=pdv$ is not an exact differential?
I was thinking in use $W=W(p,V)$ and calculate
$$dW=\frac{\partial W}{\partial p}dp+\frac{\partial W}{\partial V}dV$$
and try to find a way to refute the idea of an exact differential for $pdV$. Am I right?
 A: Work depends on the path between final and initial states, so by stating $W=W(P,V)$ you are ignoring that path dependence. Work isn't an exact differential because it's not only a function of variables; it's also a function of path.
A: You could do that - you're basically right there.  What would $M$ and $N$ be, and would they satisfy the condition you quote?

Here's an alternative way to look at it. If there were some function $F$ such that $dF = pdV$, then it would follow that the integral of $pdV$ along any contour in the $(p,V)$ plane would be 
$$\int_A^B p dV = \int_A^B dF = F(B)-F(A)$$
and in particular, the integral of $pdV$ along any closed contour would be zero.  Is that the case?
A: In order for dW=PdV to be an exact differential, if you had two points in P,V space, $(P_1,V_1)$ and $(P_2,V_2)$, the integral of dW=PdV would have to be independent of the path between these two points.  If you could think of more than just one single path where the integral differs, then dW could not be an exact differential. It is, of course, very easy to do this. 
A: In a common form we have that inner parameters of system $b_k$ defined by outer parameters and themperature: $$b_k=f_k(a_1, ... ,a_n; T), \qquad(1)$$ where $a_i$ - outer parameters; $T$-temperature. You can also write in this way so called thermal equations of state $$A_i=A_i(a_1, ... ,a_n, T),\qquad(2)$$ where $A_i$ - generalized forces conjugates with outer parameters $a_i$. 
In common expression for elementary work $$\delta W = \sum_{i}{A_i d{a_i}} \qquad(3)$$  we have no differential of temperature (as if coefficient before $d{T}$ equals to zero). 
So if expression (3) is full differential of someone function of state we will have 
$$\frac{\partial A_i}{\partial T}=\frac{\partial 0}{\partial a_i}=0. \qquad (compare\  with\ your \quad \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} ) $$
Last means that generalized forces (e.g. pressure) do not depend on temperature. And this contradicts base thermodynamic consumption that (2) exist.
