The equation for PV work is given by $$w = -P_\mathrm{ext}\Delta V$$ What work does $w$ imply - the work done by the system or on the system? Considering that the convention in chemistry is to consider things from the point of view of the system, $w$ should denote the work done by the system. However, by that definition, work done by the system would be positive during compression, rather than the work done on the system.
So does $w$ actually denote work done on the system?
There exist two conventions in thermodynamics, one used by theoreticians, one by engineers.
The theoreticians think that all energies entering the system must be positive, heat or work. So when a work is done on a gas system, the gas is compressed, and the work is positive. But as the volume of the gas decreases in this compression, $\Delta V = V_{final} - V_{initial}$ is negative, and if we want the work to be positive one must write : $\Delta w = - p\Delta V$.
The engineers have a different approach. They think that the gaseous system is part of a machinery with pistons and connecting rods. We heat this gas so that the heat added to the machinery is positive. But the piston must produce some useful work, which is positive. And when it does it repels the piston, so that the volume increase : $\Delta V$ is positive and the work is also positive $\Delta w = + p\Delta V$.
Fortunately both approaches leads to the same final result. For theoreticians, the internal energy $U$is the sum of all that enters the system. $\Delta U = \Delta Q + \Delta w = \Delta Q -p\Delta V$
For the engineers, the internal energy is the fraction of the energy (heat) that is "lost" in the system, and that has not been used for producing work. So $\Delta U$ is : $\Delta U = \Delta Q - \Delta w = \Delta Q - p\Delta V$
So whatever the convention for defining the sign of the work, the change of internal energy is the same.
