Derivation of the Chemical Potential Equation

X - 8 DELGADO-BONAL ET AL.: DISEQUILIBRIUM ON MARS

T Appendix A: Derivation of the chemical −CpT ln( ) + RT ln(p/p0) (A7) potential equation T0 When the temperature is T=T0, the above expression is The expression that is commonly used in planetary at- reduced to the more familiar equation: mospheres is usually written as (Kodepudi and Prigogine [1998], Eq. 5.3.6): µ(p, T0) = µ(p0,T0) + RT0 ln(p/p0) (A8) However, as has been explained in Section 2, this expres- µ(p, T ) = µ(p0,T ) + RT ln(p/p0) (A1) sion is only useful for situations where the temperature is

where µ0 is the chemical potential at unit pressure (1 close to the standard temperature of reference (usually 298 atm), p0 is the pressure at standard conditions and R the K), i.e., the Earth environment. When the temperatures are gas constant. diﬀerent, the complete expression for the chemical potential In order to calculate the chemical potential for arbitrary is needed. p and T, the knowledge of the chemical potential at (p0,T) The temperature dependence of the heat capacity is usu- is needed. This previous step, usually omitted, can be per- ally expressed as a polynomial function with tabulated con- formed as (Eq. (5.3.3) in (Kodepudi and Prigogine [1998])): stants:

2 Cp(T ) = a + bT + cT (A9) T Z T −H (p ,T 0) µ(p ,T ) = µ(p ,T ) + T · m 0 dT 0 (A2) If the coefficients b and c are much smaller than a, as usu- 0 T 0 0 T 02 0 T0 ally happens, Equation A7 represents an useful approximation, useful for those environments where the temperature where T0 is the temperature at standard conditions and variations are not large and the heat capacity can be con- Hm is the molar enthalpy of the compound. sidered as constant. For Venus and hot exoplanets a more Therefore, the most complete equation to determine the detailed expression might be needed. chemical potential for a compound at a particular pressure The molar enthalpy at an arbitrary temperature reads and temperature is the expression: now: Z T T 0 2 T Z −H (p ,T ) H(p0,T ) = H(p0,T0) + a + bT + cT dT µ(p, T ) = µ(p ,T )+T · m 0 dT 0+RT ln(p/p ) T 0 0 T 02 0 T0 0 T0 (A3) b c b c The molar enthalpy Hm(p0,T ) at an arbitrary tempera- H(p ,T ) = H(p ,T )−(aT + T 2+ T 3)+(aT 0+ T 02+ T 03) ture T can be obtained from the values of the heat capacity 0 0 0 0 2 0 3 0 2 3 (A10) at constant pressure, Cp(T ), and the tabulated values of The enthalpy integral in Eq. A3 becomes: enthalpy at a standard temperature H(p0,T0): Z T T 0 Hm(p0,T ) = Hm(p0,T0) + Cp(T )dT (A4) Z −H (p ,T ) m 0 dT 0 = T0 T 02 Using this last relation, the enthalpy integral can be sim- T0 plified to: b c Z T 1 − H(p ,T ) − (aT + T 2 + T 3) dT 0 0 0 0 2 0 3 0 T 02 T0 0 Z T 0 b 02 c 03 R T ∗ ∗ (aT + T + T ) 0 Z T −H (p ,T 0) Z T Hm(p0,T0) + Cp(T )dT − 2 3 dT = (A11) m 0 dT 0 = − T0 dT 0 T 02 T 02 T 02 T0 T0 T0 (A5) b 2 c 3 1 1 + H(p0,T0) − (aT0 + T0 + T0 ) − For the case of Mars, we can consider the heat capacity 2 3 T T0 as a constant, Cp = Cp(T ), being able to extract it from the T b c 2 2 integral and obtaining (see the discussion for hot exoplanets − a ln − (T − T0) − (T − T ) T 2 6 0 later): 0 Z T H (p ,T ) + C (T 0 − T ) Note that when we consider the coefficients b and c = − m 0 0 p 0 dT 0 = zero, i.e., C is not temperature dependent, then C = a T 02 p p T0 and the expression A11 is reduced to A6. Z T H (p ,T ) − C T + C T 0 We can use A11 into the chemical potential formula de- − m 0 0 p 0 p dT 0 = T 02 tailed in A3 and obtain a general equation for the chemical T0 potential for an arbitrary pressure and temperature, useful for any planetary atmosphere: Z T Z T 1 0 1 0 −(Hm(p0,T0) − CpT0) 02 dT − Cp 0 dT = T T T T 0 0 T Z T −H (p ,T 0) µ(p, T ) = µ(p ,T )+T · m 0 dT 0+RT ln(p/p ) T 0 0 T 02 0 0 T0 1 1 T +(Hm(p0,T0) − CpT0) − − Cp ln (A6) T T0 T0 T µ(p, T ) = µ(p0,T0) + RT ln(p/p0)+ Using (A6) and (A3), we obtain an useful approximation T0 for the calculation of the chemical potential at the Martian b 2 c 3 1 1 environment − T H(p0,T0) − (aT0 + T0 + T0 ) − 2 3 T T0 T b c 2 2 − aT ln − T (T − T0) − T (T − T0 ) T 1 1 T0 2 6 µ(p, T ) = µ(p0,T0) + T · (Hm(p0,T0) − CpT0) − T0 T T0 (A12) DELGADO-BONAL ET AL.: DISEQUILIBRIUM ON MARS X - 9

Although these equations have been obtained by direct and α is the volumetric thermal expansion coefficient, integration of the chemical potential equation, it is impor- 1 ∂V ; for an ideal gas is easy to prove that α = 1 , tant to remark that it is possible to obtain the same equa- V ∂T p T and therefore µJT = 0. tions by Legendre transformations. By definition, the Gibbs If we consider the first step in Figure 1, i.e., the variation potential function is given as: in temperature without any changes in the pressure, and G = H − TS (A13) knowing the enthalpy and entropy variations on tempera- The G function is called chemical potential in the case of ture (considering Cp as constant): one mol, and including the proper dependence of pressure and temperature in the equation, we can write: T µ(p0,T ) = h(p0,T0)+Cp(T −T0)−T ·s(p0,T0)−CpT ln( ) µ(p, T ) = h(p, T ) − T · s(p, T ) (A14) T0 (A15) The determination of the enthalpy and entropy at differ- We consider now the second step in Figure 1, the pres- ent (p,T) can be done using: sure variation. The Joule-Thompson coefficient is zero for an ideal gas and therefore the pressure term in the enthalpy Z T Z p becomes zero, being the pressure dependence due to the en- h(p, T ) = h(p ,T ) + C dT + −µ C dp 0 0 p JT p tropy expression. The final equation for the chemical poten- T0 p p0 T 0 tial reads:

" # Z T C Z p ∂V s(p ,T ) = s(p ,T ) + p dT − dp T p 0 0 0 µ(p, T ) = h(p0,T0)−T ·s(p0,T0)+Cp(T −T0)−CpT ln( )+RT ln( ) T T p ∂T 0 p0 0 p T T0 p0 (A16) where µJT is the Joule-Thompson coeﬃcient It is trivial to prove that Equation A8 and Equation A16 are in fact the same equation where µ(p ,T ) = h(p ,T ) − ∂T V 0 0 0 0 µJT = = (αT − 1) T0 · s(p0,T0). ∂P H Cp