Why doesn't an ideal gas undergoing an isobaric process violate the second law of thermodynamics

Question

I have been stuck on this for the past few hours, and still have got no resolution:

Let's say that we have an isobaric process being done by $1$ mol of a gas. Accordingly, we'll be using gas laws and the first law of thermodynamics to analyze ​the situation.

So, by the ideal gas law:
$$PV=nRT$$

For sake of understanding, we take $n = 1\space \mathrm {mol}$:

$$\Rightarrow PV=RT \tag{1}$$

Now, using First Law of Thermodynamics,

$$\delta Q = \delta W + \mathrm d U$$ and $\mathrm d U = C_v\, \mathrm d T$, $\space\space\delta W=P\, \mathrm dV$

$$\Rightarrow \delta Q = P\, \mathrm d V + C_v\, \mathrm dT \tag{2}$$

Dividing $(2)$ by $T$, we get,

$$\frac{\delta Q}{T} = \frac{P\, \mathrm dV}{T} + \frac{C_v\, \mathrm d T}{T} \tag{3}$$

Using $(1)$ in $(3)$,

$$\frac{\delta Q}{T} = \frac{R\, \mathrm d V}{V} + \frac{C_v\, \mathrm dT}{T}$$

And $ \displaystyle \frac{\delta Q}{T} = \mathrm dS$,

$$\Rightarrow \mathrm d S = \frac{R\, \mathrm dV}{V} + \frac{C_v\, \mathrm d T}{T} \tag{4}$$

Integrating $(4)$, we get,

$$\Delta S = R\,\ln\left(\frac{V_2}{V_1}\right)+C_v\,\left(\frac{T_2}{T_1}\right)$$

Now, if $T$ decreases in this process, then by the ideal gas law, $V$ should also decrease. If this follows, then the both the terms $R\ln(\frac{V_2}{V_1})$ and $C_v(\frac{T_2}{T_1})$ become negative as $T_2<T_1$ and $V_2<V_1$. So, the overall $RHS$ should become negative, and consequently, $\Delta S$ should become negative.

Please help to point out my mistake in the above mentioned text. Is $\Delta S$ coming negative because I have used an ideal gas?

Answer 1

By the Sakur-Tetrode equation, the entropy of a monatomic, ideal gas is given by

$$\frac{S}{k_BN}=\ln\left[\frac VN\left(\frac{4\pi m}{3h^2}\frac{U}{N}\right)^{3/2}\right]+\frac52$$

For our purposes, it will make sense to use the ideal gas law to express $V$ in terms of $P$ and $T$, and to express $U=\frac32Nk_BT$, so we get

$$\frac{S}{k_BN}=\ln\left[\frac {k_BT}{P}\left(\frac{2\pi m}{h^2}\cdot k_BT\right)^{3/2}\right]+\frac52$$

So, as we can see, for a constant pressure $P$, the entropy of the ideal gas is a monotonically decreasing function with respect to decreasing $T$; if we decrease $T$, we decrease $S$.

I suspect your confusion comes from thinking that $S$ can never decrease, but this is only the case for isolated systems. If you are forcing an ideal gas to undergo an isobaric compression, then the system is no longer isolated, and so the entropy can decrease (entropy will increase elsewhere though).

As a separate argument, the entropy is a state function, meaning its value only depends on the state, not how you got there. Now, let's consider your isobaric process, and let's say we do an isobaric expansion and then an isobaric compression back to the original state (this is the original state of the system, not the original state of the system as well as the surroundings, which is impossible to achieve). Since entropy is a state function, the entropy ends at where it started. But this means one of two things happened

1. The entropy remained constant the entire time
2. The entropy increased as well as decreased during this process.

If you showed that the change in entropy is non-zero for some part of this, then you have to conclude that it is possible to decrease the entropy of an ideal gas. Also, note that this argument is not dependent on use of an ideal gas specifically.

Answer 2

There is nothing wrong in your reasoning. The entropy of an ideal gas in a given state $(P, V, T)$ is indeed given by

$$S = nC_v \ln(T)+nR \ln (V)+S_0$$

(where $S_0$ is a constant we don't care about and this expression is meant to be used only then computing entropy differences so that the dimensions of the arguments of the logarithms work out) meaning that the entropy difference between any two states $A$ and $B$ is given by

$$\Delta S = nC_v \ln(T_B/T_A)+nR \ln (V_B/V_A)$$

which we can, using $PV=nRT$ easily rewrite as

$$\Delta S = nC_v \ln(P_B/P_A)+n C_p \ln (V_B/V_A)$$ (where we used $T_B/T_A=P_B P_B/T_AV_A$ and $R+C_v=C_p$).

So, in an isobaric process ($P_A=P_B$ so $\ln(P_B/P_A)=0$), this one is the expression we want to look at: $$\Delta S =n C_p \ln (V_B/V_A)$$

which clearly if $V_B<V_A$ is negative.

More in general, from the expression of $\Delta S$ you can see that it can be negative in a variety of ways, for example also in an isothermal process if $V_B<V_A$ (to be precise: it does not depend on the process but on the initial and starting points!)

Entropy differences can be negative if you are only looking at a subset of the system. What can not decrease is the entropy of the universe. But we are in this case only looking at the gas! If we get a negative entropy difference in the gas it means the rest of the world has increased its entropy.

Indeed, to go from $A$ to $B$ using an isothermal compression, you have given the system an amount of work $p(V_A-V_B)$ to decrease its entropy but this means that the system has released an amount of heat $$Q=n C_p (T_B-T_A)$$ "in the atmosphere", thus heating it up and producing an amount $\Delta S$ of entropy in the universe which (because we are dealing with a reversible process) perfectly balances the decrease in the gas, meaning $$\Delta S_{universe}=\Delta S_{gas}+\Delta S_{environment} = 0$$

If you want to compute this analytically, you need to find a transformation from $A$ to $B$ from which entropy is easy to compute (meaning it is either isothermal, with $\delta S = \delta Q/T$, or adiabatic, with $\delta S =0$) and you will see that the "emitted entroyp", i.e. the "emitted heat divided by the temperature" will perfectly balance the value we computed of $$\Delta S =n C_p \ln (V_B/V_A)$$

In the easiest case of an isothermal compression, the entropy difference is, using $\Delta S = nC_v \ln(T_B/T_A)+nR \ln (V_B/V_A)$ with $T_B=T_A$, given by

$$\Delta S_{gas} = nR \ln (V_B/V_A) <0$$

so it is negative also in this case! However, the "heat emitted (at temperature $T$)" is, using the formula for an isothermal process $$Q=-nRT\ln(V_B/V_A)<0$$ with the minus sign indicating it is heat released so that - as we are at constant temperature - the produced entropy in the environment is

$$\Delta S_{env} = Q/T = -nR\ln(V_B/V_A)$$

so that $$\Delta S_{env}+\Delta S_{gas}=0$$ as expected.