0
$\begingroup$

$1 \text{ mol}$ 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?

$\endgroup$
4
$\begingroup$

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)

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

$\endgroup$
1
$\begingroup$
  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 $\ln p_2 = \frac{5}{2}\ln\frac{T_2}{T_1}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.