Imagine a bottle beeing filled with air through a hand pump.
If you press the handle of the hand pump, you apply the force $\vec{f}$ down to the stop along the way $\vec{s}$.

Now assume that you slowly let your muscle lose, the work which will applied against your arm can than be described as $-p \Delta V$. $^1$

$$(1) \quad \vec{f}*\vec{s} = -p \Delta V$$

$^1$ The pressure must remain constant if you want to apply this exact formula. In this example this would be possible if you supply heat to the bottle during the process. But then we wouldn´t have a closed system and the equal sign in (1) will be wrong.
To avoid this problem and allow a variable pressure over the process you can use an integral $W = \int{p(V) dV} = \int{\vec{f}_{(\vec{s})}} \cdot d \vec{s}$

Imagine a bottle beeing filled with air through a hand pump.
If you press the handle of the hand pump, you apply the force $\vec{f}$ down to the stop along the way $\vec{s}$.

Now assume that you slowly let your muscle lose, the work which will applied against your arm can than be described as $-p \Delta V$. $^1$

$$(1) \quad \vec{f}*\vec{s} = -p \Delta V$$

$^1$ The pressure must remain constant if you want to apply this exact formula. In this example this would be possible if you supply heat to the bottle during the process. But then we wouldn´t have a closed system and the equal sign in (1) will be wrong.
To avoid this problem and allow a variable pressure over the process you can use an integral $W = -\int{p(V) dV} = \int{\vec{f}_{(\vec{s})}} \cdot d \vec{s}$