# Calculation of irreversible surrounding entropy

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!

What they are doing is modeling the surroundings as something like a weight sitting on top of the massless frictionless piston plus an ideal infinite reservoir at 300 K. Lifting a weight is a mechanically reversible process that does nothing more than increase the (conservative) potential energy of the weight. So this part doesn't figure in the entropy change.

Now for the reservoir. An ideal infinite reservoir is one that can absorb or reject any amount of heat without its temperature changing because of its infinite thermal capacity. Plus, its change in entropy is always $\frac{Q_{res}}{T}$, where $Q_{res}$ is the heat absorbed by the reservoir.

So, in this problem, the surroundings are regarded as an ideal reversible entity.

What they are doing in the calculation in parenthesis is determining the actual amount of heat transferred to the ideal infinite reservoir in the process. This is equal to minus the heat transferred to the system. So the entropy change of the reservoir is $\frac{-Q_{syst}}{T}$

• Thank you so much for responding Chester! Maybe I can clarify my confusion even further. When you say "What they are doing in the calculation in parenthesis..." I assume you're referring to step with "P_external(Vf−Vf)." If the corresponding entropy were indeed equal to -(that of the system), then when considered together, \D S_{syst} + \D S_{surr} would equal zero no? The argument I see being made in the book is that the entropy changes of the syst and surr are different, via that calculation with P_ext, and I am unsure why we are allowed to use P_ext in a reversible calculation for entropy. – Yajibromine Aug 12 '17 at 23:15
• I am unsure about this because P_{ext} is being used in the calculation of irreversible heat transfer. By the logic of the text, this produces q_{irrev}, which I would have thought could not be used to calculate entropy change on the grounds that entropy is defined explicitly using reversible heat. Furthermore, if the q for the surroundings were to be calculated using a reversible path that traverses the same change, it would result in dS_surr = -dS_syst, and total S would then be zero, which cannot be true due to the process being irreversible. Thanks again, I only hope to clarify! – Yajibromine Aug 12 '17 at 23:18
• In looking at reversible paths to get entropy change, the reversible path for the surroundings does not have to be the same as the reversible path for the system. Each entity can be treated separately in determining its entropy change (which is really a material property which depends only on the initial and final states of that entity). However, in the present irreversible process, even though the system experiences an irreversible path, the actual path for the surroundings is reversible. Here is a simple primer: physicsforums.com/insights/grandpa-chets-entropy-recipe – Chet Miller Aug 13 '17 at 0:07
• Here are the initial and final states of the surroundings: Initial: A pair or weights totaling (W+w) at elevation z = 0 and an ideal reservoir at temperature T, with no heat added yet. Final: A weight W at elevation $z=z_f >0$, a weight w at elevation z = 0 (this weight w was removed from the piston at time zero), and an ideal reservoir at temperature T that has received Q heat added. Irrespective of what happened to the system, what is the entropy change for the surroundings? – Chet Miller Aug 13 '17 at 0:34
• In the above example, $P_{ext}=W/A$, where A is the cross sectional area of the piston and cylinder and $\Delta V=Az_f$. – Chet Miller Aug 13 '17 at 0:41

The gas pressure equals to the external pressure during isothermal irreversible expansion. To maintain P(gas)=P(ext), heat is supplied reversibly. By this reason delta S(surr) must be calculated above. Am I clear?

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