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I tried to derive the entropy change of an isobaric process and arrived at:
$\Delta S = C_pln(T_2/T_1)+nRln(V_2/V_1)$
However, it appears the answer is just:
$\Delta S = C_pln(T_2/T_1)$
It seems like the formula I derived would be more general and would reduce to the second formula in the case of a solid or liquid. What gives?
I guess, there is some error in your first equation, where there should be $C_v$ instead of $C_p$.
For a reversible process: $$dQ_{rev}=dU+dW$$ $$dQ_{rev}=nC_vdt+PdV$$ Dividing by $T$, and substituting $P$ from ideal gas equation $PV=nRT$ $$\frac{dQ_{rev}}{T}=nC_v\frac{dT}{T}+nR\frac{dV}{V}$$ Integrating from $T_1$ to $T_2$ and $V_1$ to $V_2$: $$\Delta S_{sys}=nC_v\text{ln}\left(\frac{T_2}{T_1}\right)+nR\text{ln}\left(\frac{V_2}{V_1}\right)...(1)$$
Now $$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$ Now substituting for $\frac{V_2}{V_1}$, we get:
$$\Delta S_{sys}=nC_v\text{ln}\left(\frac{T_2}{T_1}\right)+nR\text{ln}\left(\frac{P_1}{P_2}\right)+nR\text{ln}\left(\frac{T_2}{T_1}\right)$$ From Meyer's Relation $C_p=C_v+R$, and simplifying above equation: $$\Delta S_{sys}=nC_p\text{ln}\left(\frac{T_2}{T_1}\right)+nR\text{ln}\left(\frac{P_1}{P_2}\right)...(2)$$
From second and first equation you can find $\Delta S_{sys}$, for any kind of situation which is reversible.
It has already been pointed out to you that the first equation is incorrect. You should provide your derivation if you want to find out why.
In any case, to calculate the entropy change for any process between two points you can use any convenient reversible path that connects the points (since entropy is a state function) and apply the following definition:
$$\Delta s=\int_1^2 \frac{\delta q_{rev}}{T}$$
In this case the obvious path is a reversible isobaric process. For a reversible isobaric process, $\delta q_{rev}=C_{p}dt$. Plugging it into the definition gives you
$$\Delta s=\int_1^2 \frac{\delta q_{rev}}{T}=\int_1^2\frac{C_{p}dt}{T}=C_{p}\ln\frac{T_2}{T_1}$$
Hope this helps.
