Heat flow In case of a heat flow rate \dot Q from
T1 to
T2 (with T_1 \geq T_2) the rate of entropy production is given by : \dot S_\text{i}=\dot Q\left(\frac{1}{T_2}-\frac{1}{T_1}\right). If the heat flow is in a bar with length
L, cross-sectional area
A, and thermal conductivity
κ, and the temperature difference is small : \dot Q=\kappa \frac{A}{L}(T_1-T_2) the entropy production rate is : \dot S_\text{i}=\kappa \frac{A}{L}\frac{(T_1-T_2)^2}{T_1 T_2}.
Flow of mass In case of a volume flow rate \dot V from a pressure
p1 to
p2 : \dot S_\text{i}=-\int _{p_1} ^{p_2} \frac{\dot V}{T} \mathrm{d}p. For small pressure drops and defining the flow conductance
C by \dot V=C(p_1-p_2) we get : \dot S_\text{i} = C\frac{(p_1 - p_2)^2}{T}. The dependences of \dot S_\text{i} on and on are quadratic. This is typical for expressions of the entropy production rates in general. They guarantee that the entropy production is positive.
Entropy of mixing In this Section we will calculate the
entropy of mixing when two ideal gases diffuse into each other. Consider a volume
Vt divided in two volumes
Va and
Vb so that . The volume
Va contains
amount of substance na of an
ideal gas a and
Vb contains amount of substance
nb of gas b. The total amount of substance is . The temperature and pressure in the two volumes is the same. The entropy at the start is given by :S_\text{t1}=S_\text{a1}+S_\text{b1}. When the division between the two gases is removed the two gases expand, comparable to a Joule–Thomson expansion. In the final state the temperature is the same as initially but the two gases now both take the volume
Vt. The relation of the entropy of an amount of substance
n of an ideal gas is :S=nC_\text{V}\ln\frac{T}{T_0}+nR\ln\frac{V}{V_0} where
CV is the molar heat capacity at constant volume and
R is the molar
gas constant. The system is an adiabatic closed system, so the entropy increase during the mixing of the two gases is equal to the entropy production. It is given by :S_\Delta=S_\text{t2}-S_\text{t1}. As the initial and final temperature are the same, the temperature terms cancel, leaving only the volume terms. The result is :S_\Delta=n_\text{a}R\ln\frac{V_\text{t}}{V_\text{a}}+n_\text{b}R\ln\frac{V_\text{t}}{V_\text{b}}. Introducing the concentration
x =
na/
nt =
Va/
Vt we arrive at the well-known expression :S_\Delta=-n_\text{t}R[x\ln x+(1-x)\ln(1-x)].
Joule expansion The
Joule expansion is similar to the mixing described above. It takes place in an adiabatic system consisting of a gas and two rigid vessels a and b of equal volume, connected by a valve. Initially, the valve is closed. Vessel a contains the gas while the other vessel b is empty. When the valve is opened, the gas flows from into b until the pressures in the two vessels are equal. The volume, taken by the gas, is doubled while the internal energy of the system is constant (adiabatic and no work done). Assuming that the gas is ideal, the molar internal energy is given by . As
CV is constant, constant
U means constant
T. The molar entropy of an ideal gas, as function of the molar volume
Vm and
T, is given by : S_\text{m}=C_\text{V}\ln\frac{T}{T_0}+R\ln\frac{V_\text{m}}{V_0}. The system consisting of the two vessels and the gas is closed and adiabatic, so the entropy production during the process is equal to the increase of the entropy of the gas. So, doubling the volume with
T constant gives that the molar entropy produced is : S_\text{mi}=R\ln 2.
Microscopic interpretation The Joule expansion provides an opportunity to explain the entropy production in statistical mechanical (i.e., microscopic) terms. At the expansion, the volume that the gas can occupy is doubled. This means that, for every molecule there are now two possibilities: it can be placed in container a or b. If the gas has amount of substance
n, the number of molecules is equal to
n⋅
NA, where
NA is the
Avogadro constant. The number of microscopic possibilities increases by a factor of 2 per molecule due to the doubling of volume, so in total the factor is 2
n⋅
NA. Using the well-known Boltzmann expression for the
entropy : S=k\ln \Omega, where
k is the
Boltzmann constant and Ω is the number of microscopic possibilities to realize the macroscopic state. This gives change in molar entropy of :S_{\text{m}\Delta} = S_\Delta / n = k \ln(2^{n \cdot N_\text{A}}) / n = k N_\text{A} \ln 2 = R \ln 2 . So, in an irreversible process, the number of microscopic possibilities to realize the macroscopic state is increased by a certain factor. ==Basic inequalities and stability conditions==