$1 \text{ mol}$ 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?


2 Answers 2


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

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


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

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


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


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