Entropy for Open Systems

We would like to derive a thermodynamic relation for entropy (as opposed to the statistical definition given above) in general systems which are not necessarily closed, that is which allow an exchange of heat with the surroundings. Firstly note that for systems which are not closed, the defintion of reversibility is more general : It means that the system retraces its path in time when the external conditions are reversed in time. So consider now an infinitesimal reversible transformations. We can then write (recall the kinetic theory discussion again, work done $=$ force $\times$ distance = $P \times A \times$ distance.)

$$dW = PdV .$$ (4.16)

Thus for a general infinitesimal reversible transformation, we have from the first law

$$dQ= dU + PdV$$ (4.17)

On the other hand, the entropy $S$ can be treated as a function of $U$ and $V$ (see the statistical definition given earlier), which under an infinitesimal reversible transformation changes by

$$dS = \left({\partial S \over \partial U}\right)_{V} dU + \left({\partial S \over \partial V}\right)_{U} dV$$ (4.18)

Define (this will be checked for consistency later!),

$$\left( \partial S \over \partial U \right)_{V} = {1 \over T}$$ (4.19)

and

$$\left( \partial S \over \partial V \right)_{U} = {P \over T}$$ (4.20)

Then Eq.(4.18) becomes

$$TdS = dU + PdV$$ (4.21)

Which then implies on comparing with the first law (4.17) that for an infinitesimal reversible transformation (in a general, not necessarily closed system), the change in entropy is given by

$$dS = {dQ \over T}$$ (4.22)

One can use (4.22) to compute the the difference in entropy between two states $A$ and $B$. Choose a reversible path joining the two states and integrate the above equation to get

$$S(A)-S(B) = \int_{B}^{A} {dQ \over T}$$ (4.23)

Note: $S$ is a state function, so its value is fixed once the independent parameters, say $U$ and $V$ are fixed. On the other hand, to calculate it using the formula above, one must choose a reversible path since for a general irreversible path the quantity calculated depends on the path taken and is not the entropy difference between the two states. (It is an assumption of classical thermodynamics that one can always find a reversible transformation between any two states connected by an irreversible one). To illustrate this point, consider two different isothermal expansions of an ideal gas from volume $V_1$ to volume $V_2$.
1. Reversible isothermal expansion.

$$\Delta S_{gas} =\int {dQ \over T} = {1 \over T} \int dQ .$$ (4.24)

Now for an ideal gas $U=U(T)$ only (see section(1)), hence in the expansion $\Delta U =0$ and so from the first law the heat absorbed by the gas equals the work done. So

$$\Delta S_{gas}= {1 \over T} \int p dV = N k \int { dV \over V} = N k \log \left({V2 \over V1}\right) .$$ (4.25)

The entropy change of the heat reservoir is

$$\Delta S_{reservoir} = -\Delta \left( { Q \over T} \right) = -\Delta S_{gas}$$ (4.26)

Is it surprising that the entropy of the reservoir has decreased ? No, there is nothing forbidding it since it is not an isolated system. Indeed if one looks at the change of entropy of the combined larger system of gas plus reservoir, it is zero, in agreement with the second law (for isolated systems). Note that reversibility in the above system can be achieved by storing the work done by the gas on expansion in a spring attached to the piston, so that this can be used to compress the gas thus reversing the expansion.
2. Free Expansion. Since the intial and final states (volume and temperature) of the gas are the same as before for the reversible case, the change in entropy of the gas is as before (because $S$ is a state function, and anyway its calculation requires choosing a reversible path joining the initial and final states). Since as before $\Delta U =0$ for the gas and as no work is done in a free expansion, $\Delta W=0$ also, implying that no heat is exchanged between the gas and its reservoir. Hence the entropy change of the reservoir is zero. Thus there is a net increase of entropy of the combined gas plus reservoir system. Furthermore the expansion is clearly irreversible (Why?)