As you said, the work done on system is defined as $$\delta W= -pdV$$

And for Isobaric process, the work done by the system will be $$\delta W = p(V_f-V_i)$$

To specify state of system uniquely, We need at least two variable. As you are taking the pressure to be constant, then one variable is already fix. The other one is, of course, the volume ( better in this case).

Now as you did, Suppose you have two different gases with different equation of state,

$$P=f(V,T) \ \ \ \mathrm{and} \ \ \ P=g(V,T)$$

Now if the pressure is fixed during adiabatic process, say $P_0$, then $$P_0=f(V,T)=g(V,T)$$

So in your case during the adiabatic process, You must have $$P_0=\frac{RT}{V-b}=\frac{RT}{V}\left(1-\frac{B}{V}\right)$$

So that If the volume boundary are same, then the work done in both cases will be same. But the Temperature at boundaries will be different.

$$\delta W=P_0(V_f-V_i)=\frac{RT}{V-b}(V_f-V_i)=\frac{RT}{V}\left(1-\frac{B}{V}\right)(V_f-V_i)$$

If you draw the trajectory in $V-T$ space, then the trajectory will be different but both have same value of pressure at each point. Here

As you said, the work done on system is defined as $$\delta W= -pdV$$

And for Isobaric process, the work done by the system will be $$\delta W = p(V_f-V_i)$$

To specify state of system uniquely, We need at least two variable. As you are taking the pressure to be constant, then one variable is already fix. The other one is, of course, the volume ( better in this case).

Now as you did, Suppose you have two different gases with different equation of state,

$$P=f(V,T) \ \ \ \mathrm{and} \ \ \ P=g(V,T)$$

Now if the pressure is fixed during isobaric process, say $P_0$, then $$P_0=f(V,T)=g(V,T)$$

So in your case during the isobaric process, You must have $$P_0=\frac{RT}{V-b}=\frac{RT}{V}\left(1-\frac{B}{V}\right)$$

So that If the volume boundary are same, then the work done in both cases will be same. But the Temperature at boundaries will be different.

$$\delta W=P_0(V_f-V_i)=\frac{RT}{V-b}(V_f-V_i)=\frac{RT}{V}\left(1-\frac{B}{V}\right)(V_f-V_i)$$

If you draw the trajectory in $V-T$ space, then the trajectory will be different but both have same value of pressure at each point. Here