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?
Best Answer
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
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.