Expansion of an ideal gas at constant pressure I approach these expansion problems like so: 
The gas and the surroundings(piston+outside) are at the same pressure at first. We heat the gas. The pressure rises inside the syringe a bit. The gas expands so the pressure remains constant. Then I use P(the constant pressure of the gas) *dV. What I want to confirm is my reasoning on using this equation. It was derived assuming P(internal) = constant. But it does change momentarily. Is the reason we ignore it in the "a bit" nature?
Also for compression, the force exerted on the gas by surroundings (piston+outside)  is taken as the force the gas exerts on the piston. Is this Newton's third law? 
 A: 
As in the as soon as pressure rises because of the temp rise the volume expands to "counter it". So pressure never really changes much. So it's okay to ignore it for calculations 

If the piston moves by a little bit, then the pressure is literally the same. Considering $\Delta V \to 0$ which would mean $\Delta V \approx dV$.

Also for compression, the force exerted on the gas by surroundings (piston+outside) is taken as the force the gas exerts on the piston. Is this Newton's third law? 

Yes, this is true.
A: If the heat addition occurs very slowly such that the pressure and temperature gradients in the gas approach zero, the process can be considered quasi-static and the pressure of the gas will always be very close to the external pressure. If the process is also frictionless, then we can say it is a reversible isobaric expansion. At every point along the process the ideal gas equation applies
$$\frac{V}{T}=\frac{nR}{P}$$
since $P$ is constant
$$\frac{V}{T}=constant$$
Since the external force applied to the gas by the piston and atmosphere is always approximately equal to and opposite the force the piston and atmosphere apply to the gas, Newton;s third law applies.
Hope this helps.  
