# 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 &=nC\ln\left(\dfrac{T_2}{T_1}\right)+nR\ln\left(\dfrac{P_1}{P_2}\right)=0 \\ & \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 & = \frac{dU}{T} + \frac{p}{T}dV \end{aligned}

For an ideal gas

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

such that

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

Integrating

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

1. The fundamental equation of thermodynamics ignoring all work terms except PV-work is

$$dU = T dS - p dV$$

1. For an ideal gas, we know that $$dU = n C_v dT$$, i.e. an ideal gas has an internal energy that is dependent only on temperature.

2. For an isentropic process, we also know that $$dS \approx 0$$.

Combining these three items gives us

$$n C_v dT = - p dV$$

Using the ideal gas law $$V = \frac{n R T}{p}$$ again, we can substitute for $$dV$$ by taking the differential using the quotient rule.

$$dV = d(\frac{n R T}{p}) = n R d(\frac{T}{p})$$

Thus, $$dV = n R \frac{p dT - T dp}{p^2}$$

We can now substitute for $$dV$$ in the $$- p dV$$ term:

$$- p dV = -p n R \frac{p dT - T dp}{p^2} = \left (-dT + T\frac{dp}{p}\right ) nR$$

Now, substituting this into the $$n C_v dT = - p dV = - dT + T\frac{dp}{p}$$ equation and rearranging gives

$$(n C_v + nR) \frac{dT}{T} = n R \frac{dp}{p}$$

This is finally a differential equation that can be solved, with the solution being:

$$\frac{C_v + R}{R}\ln{\frac{T_2}{T_1}} = \ln{\frac{p_2}{p_1}}$$

Up until now, we haven't used the fact that the gas is monatomic, only that it is ideal. If we use the fact that monatomic ideal gases have $$C_v = \frac{3}{2}R$$, giving

$$\frac{5}{2}\ln\frac{T_2}{T_1}=\ln\frac{p_2}{p_1}$$

Since $$p_1$$ is 1 atm, the equation shows that $$\ln p_2 = \frac{5}{2}\ln\frac{T_2}{T_1}$$