p vs V graph - dependence question Context:
I'm reading Fermi's Thermodynamics. On page 6, he states that the work done transforming a system from state $A$ to state $B$ (with the accompanying graph below) is given by $$W=\int_{V_A}^{V_B}p\text{ } dV.$$ 
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Question: In the graph, is $p$ dependent on $V$? If so, how and why? I can't make sense of it intuitively. Is the book already assuming the Ideal Gas Law?
 A: This is only an example. In this case, the pressure does depend on $V$, as you can directly see on the graph. If it didn't depend on V, all volumes would have the same value of $P$, that is, $P$ would be a horizontal line.
In general, $P$ depends on $V$, and the shape depends on the problem you have.
For the concrete case of an ideal gas, at constant temperature, $P\cdot V=const$, so it means that $P=\frac{const}{V}$ and it looks like the $1/x$ function.
A: Answer by @FGSUZ is correct. I just want to add a few minor points.
Pressure $p$ depends on $V$, but not only on $V$. You may choose the other thermodynamic variable to be for example temperature $T$ (any choice different from $V$ will do), and then the dependence is written $p(V,T)$. In a thermodynamic process going from state 1 to state 2, $V$ and $T$ are continuously varied. This process forms a three-dimensional curve in the $p$-$V$-$T$ diagram. What is presented in Fermi's graph is the projection of this three-dimensional curve on the $p$-$V$ plane. To calculate work this is all we need. See that no assumption was made regarding the thermodynamic process except that it is quasistatic which enables us to represent the process by a continuous curve; in particular, ideal gas behavior was not assumed.
