You are reaching an incorrect conclusion for two basic reasons.
First, the ideal gas equation
$$pV=nRT$$
Does not describe a process. It only describes the relationship between pressure, volume and temperature of an ideal gas of a closed system ($n$ = constant) at any equilibrium state.
Second, your equation
$$W=pdV$$
is not correct. It should be written
$$dW=pdV$$
and then, to calculate reversible work between two equilibrium states, you have
$$W=\int_1^2 pdV$$
Which is called "boundary work" for a closed system, i.e., the work required to expand or compress the boundary of the system (ideal gas).
In order to calculate the work using the above formula, for any process you need to know how pressure varies as a function of volume. For a reversible adiabatic process the formula for an ideal gas is
$$pV^{ϒ}=C$$
where C is a constant and ϒ is the ratio $\frac{C_p}{C_v}$. This formula can be derived by combining the equations for the ideal gas law and the first law of thermodynamics.
Rewriting this equation expressing pressure as a function of volume gives you
$$p=CV^{1-ϒ}$$
Putting this equation into the equation for the work done between two states
$$W=\int_1^{2}CV^{1-ϒ}dV$$
Which, after performing the integration, gives you
$$W=\frac{(p_{2}V_{2}-p_{1}V_{1})}{1-ϒ}$$
Now, for a constant pressure process, $p$ = constant, so the work is
$$W=\int_1^2 pdV=p\int_1^2dV=p(V_{2}-V_{1})$$
Hope this helps.