Why does PV work consider total pressure instead of net pressure? In physics, we define work as: $$W = F_{net}×d$$
However, in physical chemistry, PV work is calculated as:
$$W = P×ΔV$$
Why is total pressure used instead of net pressure? For example, if a gas is compressed in a situation where the external pressure is 10 atm and the internal pressure is 9 atm, why is 10 atm used to calculate the work done, rather than the net 1 atm?
I think of it like this: if I'm pushing a block to the right with 10N, while someone else is pushing the block to the left with 9N, there is a net force of 1N to the right doing work. Similarly, if 10 atm is pushing down on the gas and 9 atm is pushing back up, wouldn't there be a net pressure of 1 atm pushing down and doing work? 
Please note, to me, my question seems quite simple and it has probably been asked before, but I can't seem to find another question asking this. Please let me know if it's duplicate. Thanks.
 A: I have a friend who could probably do a better job of explaining this, but I’ll give it a try.
The work here is called boundary work, or p dV work, and applies to a closed system typically represented by a gas in a cylinder fitted with a piston. The piston is normally considered massless so no work is done on the piston itself. So your analogy of two persons pushing on a block representing the piston and doing work on the block wouldn’t apply.
An analogy that would apply is you and I pushing horizontally on each others hands. I apply a force F on your hands. By Newton's third law you apply an equal and opposite force F on my hands. Although the forces are equal and opposite, we need to consider the net force acting on each of us. 
Let's say we both stand on different surfaces involving friction. But the coefficient of friction of the surface you stand on is less than mine. You are on ice. I am on dry concrete. If the force I apply to you exceeds the maximum static friction force of your ice that opposes my force, you will slide and I will have done work on you. If the equal and opposite force you apply to me does not exceed the maximum static friction force of the concrete I stand on and that opposes your force on me, the net force on me will be zero, and no work will be done on me.
Carry the analogy to our gas in a cylinder fitted with a massless piston. Let the pressure of the gas in the cylinder be greater than the atmospheric pressure outside. Just like the forces we exerted on each others hands were equal and opposite, per Newton's third law the external pressure and internal pressure at the piston interface is the same. However, like the higher coefficient of friction under my feet, resisting a displacement of the piston in the direction of the gas inside the cylinder is an increasing pressure gradient of the gas behind the piston. The pressure external is constant and less. Therefore the piston interface will move in the direction of expansion of the internal gas and compression of the atmosphere. The gas does work on the atmosphere.
Hope this helps.
A: Problem Statement
The first case (work done by forces) considers the net work done. The second case (work done by pressure) considers only the work done by external forces. The analogy seems not to be exact.
Why not?
Background
Forces and Work
Let's expand the definition of net work. The net is the sum of external AND internal.
The definition of work done on an object by external forces is
$$ w_{ext} = \int \vec{F}_{ext} \bullet d\vec{z} $$
When the vectors are in the same direction, the result is positive. Otherwise, it is negative.
The definition of work done on an object by internal forces is
$$ w_{int} = \int \vec{F}_{int} \bullet d\vec{z} $$
Consider a wall where the left side is defined as being INTERNAL and the right side is defined as being EXTERNAL. Suppose we have an external force of $F_{ext}$ pointing to the left and an internal force of $F_{int}$ pointing to the right. Define positive $z$ as the direction from the inside pointing out. This states that $\Delta z$ is positive when the object moves OUTWARD (to the right) and $\Delta z$ is negative when the object moves INWARD (to the left). This also means that the magnitude of $F_{ext}$ is NEGATIVE (because it points opposite to our coordinate system) (and inversely for $F_{int}$.
The net work done for any movement is
$$ w_{net} = w_{ext} + w_{int} = -F_{ext}\Delta z + F_{int}\Delta z = -( F_{ext} - F_{int})\Delta z$$
We instinctively appreciate that the wall will move from right to left (move inward) when the magnitude $|F_{ext} - F_{int}| > 0$. The net work will be positive when the wall moves from the right to the left because the net force is positive, the net displacement is negative, and the negative sign corrects.
So, what does this net force do? It moves the object. What does that cause? An acceleration leading to a velocity. Where does that lead? The net KINETIC energy of the object changes!
Note that we do not talk about the kinetic energy of the internal side or of the external side separately. It is the net of the object.
Pressure and Internal Energy
Now consider a case of a system (a gas, liquid, or solid) that is subjected to an external pressure. The pressure of the system defines its INTERNAL energy. We want to determine how the internal energy will change due to the action of applying an external pressure.
How does the internal energy of a system change due to the application of an external pressure? It changes in accord with the amount of work done on the system by the external pressure. 
$$ dU = \delta w_{ext,on}$$
The definition of work done on a system by the application of an external pressure is
$$ \delta w_{ext,on} = \vec{p}_{ext} \bullet d\vec{V} $$
Allow again that we define a piston with the coordinate system pointing out (to the right) and the internal side to the left. The vector $\vec{p}_{ext}$ is always pointing inward and is therefore negative in magnitude. The lowest value of $p_{ext}$ is zero (vacuum). We can safely write the integral as
$$ w_{ext,on} = - |{p}_{ext}| \Delta{V} $$
We recognize from our coordinate system that $\Delta V$ is positive when the piston moves from left to right (the container expands) and negative otherwise (the container collapses). From this, we find


*

*In the case that $p_{ext} = 0$, $w_{ext,on} = 0$ regardless of whether the piston moves to the left or to the right. This is free expansion/compression.

*In the case when the piston moves from left to right (expansion), the work done on the system (container) by any external force is negative. Internal energy decreases.

*In the case when the piston moves from right to left (compression), the work done on the system (container) by any external force is positive. Internal energy increases.
Summary
When forces act on an object, they may do two things. First, the net external forces may accelerate (move) the object and thereby may change its kinetic energy. Secondly, the net external forces acting on the object may change its internal energy.
In both cases, we consider the net external forces. The confusion in the first case is that all forces that move an object are by implicit definition only those external to it. The confusion in the second case is that all forces that are external to an object are by implicit definition net forces.
We can move a piston and consider the change in its kinetic energy. Alternatively we can move a piston and consider the change in the internal energy of the system that it bounds. The net external forces are either net external entirely to the piston or they are net external entirely to the bounded system. It is your choice of reference frame that matters!
Other Notes
The IUPAC definition of internal energy change is $dU = \delta q + \delta w_{ext,on}$. The Clausius definition used by engineers is $dU = \delta q - \delta w_{ext,BY}$, where $w_{ext,BY} = -w_{ext,on}$. Engineers prefer to state that a system that expands does POSITIVE work and subtract the positive work done BY the system to get the change in internal energy (rather than add the negative work done on the system).
A: I think the key point here is that a gas system needs to exert a permanent force on its surroundings to maintain its equilibrium at rest. Changing of system perspective in your example, the person pushing the block already has to overcome the 9N force acting as a resistive force to set the mass in motion, so the total work done by this person is not 1$x$, $x$ being the distance moved, but $10x$. This is similar to what a gas system would have to do to expand.
An easy example to work out is the case of a spring. Suppose a spring of constant $k$ initially balances a force of 9N ($kx_{i}=$9N) and responds to increase of force up to 10N ($kx_{f}=$10N). Suppose also the transformation is done so that the process is reversible (the applied force is always balanced with the spring reaction), the total work done is:
$$  W=\int_{x_{i}}^{x_{f}} F dx =\int_{x_{i}}^{x_{f}} kx dx$$and it may readily be seen, because $x_{i} \neq 0$, that:
$$W \neq \int_{x_{i}}^{x_{f}} k(x-x_{i})dx$$
