# Not sure how to calculate entropy change in a quasi-static isobaric expansion

I'm working on a problem that states:

Volume of $$n$$ moles of ideal monatomic gas increased from $$V_0$$ to $$2V_0$$ at constant $$P_0$$ in a quasi-static process. Calculate $$\Delta W$$, $$\Delta Q$$, $$\Delta E$$, and $$\Delta S$$.

In class, we derived $$dS$$ as such:

Consider the first law

$$dE= dQ-PdV$$ since $$dE=nC_vdT$$, where $$C_v$$ is specific heat at constant volume, $$nC_vdT=dQ-PdV$$ Since $$dS=\frac{dQ}{T}$$ $$dS= \frac{nC_v}{T}dT+\frac{P}{T}dV$$ by the ideal gas law $$\frac{P}{T}=\frac{Nk}{V}$$, $$dS=nC_v\frac{dT}{T}+Nk\frac{dV}{V}$$ integrating from initial state $$i$$ to final state $$f$$, this yields $$\Delta S =nC_v\log\frac{T_f}{T_i}+Nk\log\frac{V_f}{V_i} \tag{1}$$

However, since the process here is isobaric, I can't just use eq. 1 above, correct? Do I instead use the fact that $$dQ=nC_pdT$$ at constant pressure, thus $$dS=\frac{nC_pdT}{T}$$ integrating to yield $$\Delta S = nC_p\log\frac{T_f}{T_i}$$ since $$T=\frac{PV}{nR}$$, $$\frac{T_f}{T_i}=\frac{2P_0V_0}{nR}\frac{nR}{P_0V_0}=2$$ therefore $$\Delta S = nC_p\log2$$ Is this correct?

$$C_{p}-C_{v}=nR$$