I have a specific query. In my undergraduate book Physical chemistry by Engel and Reid, sample problem 5.8 calculates the change in entropy of the surroundings and system for an irreversible isothermal compression.
My question has to do with the manner in which they calculate the entropy of the surroundings.
Entropy is first operationally defined as a result of a new cyclic integral satisfying the state function criteria. This integral includes the heat transfer of all reversible parts of the carnot cycle: $$\oint\frac{\require{cancel}\cancel{d} q_{rev}}{T} = 0$$ hence the definition of entropy becomes: $\Delta{S}=\int\frac{\cancel{d} q_{rev}}{T}$ and it is from here that we derive our requirement that all subsequent calculations of entropy be done along reversible paths. This is relevant because it is my understanding (albeit one I expect to be incorrect) that sample problem 5.8 completely disregards this without explanation. I've since moved on in Chemistry, but my confusion on this point has persisted. Let me explain:
The problem has 1 mole of ideal gas at 300 K be isothermally compressed via a constant external pressure from $25.0 L$ to $10.0 L$.
The constant pressure is what causes the change to be irreversible. However, we are required to calculate a change in entropy along a reversible path, so we proceed as follows: We know the process occurs isothermally, so $\Delta U_{syst} = 0$, and $q_{reversible_{syst}} = -w = nRT\int_{V_i}^{V_f} \frac{dV}{V} = nRTln\frac{V_f}{V_i} $
After some algebra, this gives us $q_{rev_{syst}} = -2.285* 10^3 J$ and $\Delta{S}=\int\frac{\cancel{d} q_{rev_{syst}}}{T} = \Delta{S}=\frac{q_{rev_{syst}}}{T} = \frac{-2.285* 10^3 J}{3 00K} = -7.62 J*K^{-1}$
And all is well with the world. HOWEVER, this is where I run into trouble. We now turn to the surroundings, and again, there is a constant pressure, which means this process is irreversible. We are duty-bound by the definition of entropy just discussed to pursue a reversible path, considering the surroundings as our new system.
In my mind, this requires that we consider the change reversibly, ie: occurring in infinitesimal increments, which would look equal and opposite to the calculation we had just performed for the system. But when finished, this would result in us having $\Delta S_{total}=\Delta S_{syst}+\Delta S_{surr} =0$, which is not true for an irreversible process. So instead, the writers proceed as follows:
They find $P_{i}$ and $P_{external}$ using $P = \frac{nRT}{V}$
"$P_{external} = \frac{1mol*8.314Jmol^{-1}K^{-1}*300 K}{10.0L*\frac{1m^3}{10^3L}} = 2.494*10^5 Pa$ $P_{i} = \frac{1mol*8.314Jmol^{-1}K^{-1}*300 K}{25.0L*\frac{1m^3}{10^3L}} = 9.977*10^4 Pa$"
They then apply the same reasoning as in the system's case:
"Because $\Delta U = 0$, $$q = -w = P_{external}(V_f-V_f)= 2.494*10^5 Pa*(10.0*10^{-3}m^3 - 25.0*10^{-3}m^3)$$
$ = -3.741*10^{3}J $
The entropy change of the surroundings is given by:
$\Delta{S_{surr}}=\frac{q_{surr}}{T} =\frac{-q}{T}=\frac{3.741* 10^3 J}{3 00K} = 12.47 J*K^{-1}$
...$\Delta S_{total}=\Delta S_{syst}+\Delta S_{surr} = -7.62 J*K^{-1}+12.47 J*K^{-1} = 4.85 J*K^{-1}$"
I've enclosed segments borrowed directly from the book in quotes. My problem, is that although we are told time and time again that only a reversible path may be used to calculate entropy, in the case of the surroundings, $P_{external}$ is used to directly calculate heat, and subsequently that heat is used to calculate entropy. That value for heat was NOT calculated along a reversible route. How then, can we justify using it in the calculation? And if we can't, how are we to ever find anything other than $\Delta S_{total}=0$?
I know that my confusion is caused by some underlying misunderstanding, but I have not been able to pinpoint it. I'm hoping one of you will. Much of my research done on this topic, including the reading of Atkins' book, and MIT lecture notes, as well as other forums, focus on residual entropy, or worry themselves over whether the surroundings can be considered infinite or not, but none of these have seemed to answer my problem. I italicize that, because I am not confident that the answer I seek was NOT in those explanations, but I haven't seen it.
Thanks again!