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?
