Clarification required on whether to use $C_v$ or $C_p$ in this question on finding entropy change $1 \text{ mol}$ of an ideal monoatomic gas at $300 \text{ K}$ is subjected to a reversible isentropic compression process until final temperature reaches $600 \text{ K}$. If initial pressure is $1 \text{ atm}$. Then find $\ln (\text{P}_2)$.




My Attempt

$$\begin{aligned}\Delta S &amp;=nC\ln\left(\dfrac{T_2}{T_1}\right)+nR\ln\left(\dfrac{P_1}{P_2}\right)=0 \\ &amp; \boxed{\ln(P_2)= \ln(P_1)+C\ln\left(\dfrac{T_2}{T_1}\right)}\end{aligned}$$



Now if I take $C=C_v$, I get the answer $\ln(P_2)=1.5\ln(2)$, but the answer is given to be $2.5\ln(2)$ which will be obtained when we take $C=C_p$. So how to decide which one to take?
The correct form can be identified as follows. Start from the combined 1st and 2nd laws:

$$\begin{aligned} d S &amp; = \frac{dU}{T} +  \frac{p}{T}dV \end{aligned}$$

For an ideal gas 

$$\begin{aligned} d U &amp;= nC_V dT \\ \frac{p}{T} &amp;= \frac{nR}{V} \end{aligned}$$

such that

$$\begin{aligned} d S &amp; = nC_V \frac{dT}{T} +  nR \frac{dV}{V} \end{aligned}$$

Integrating

$$\begin{aligned} \Delta S &amp; = nC_V \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{V_2}{V_1} \right) \end{aligned}$$

Substitute in the ideal gas law in the right-hand side (note $n$ is constant)

$$\begin{aligned} \Delta S &amp; = nC_V \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 T_2}{P_2 T_1} \right) \\ &amp; = n(C_V +R) \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 }{P_2 } \right) \\ &amp; = nC_p \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 }{P_2 } \right) \end{aligned}$$