Equations of state

1. Equations of state Work is a function of the (change in) pressure and volume: dw = –pexdV Heat is a function of the (change in) temperature: dq = CdT Energy is therefore a function of all three: pressure, volume and temperature. Under reversible conditions ... dU = dq + dw = CdT – pdV The equation of state relates these three variables with an equation. Equations of state are specific for particular chemical systems. Ideal gas equation of state pV =nRT V and n are extensive variables: they depend on the size of the system. Intensive ideal gas equation of state pVm =RT Vm =V/n Virial equation of state pVm =RT(1 + B/Vm + C/Vm2 + …) CHEM 471: Physical Chemistry B is the second virial coefficient

2. Equations of state Solid/liquid equation of state Vm =Vm°[1 + α(T – T°)][1 – κ(p – p°)] α = coefficient of expansion κ = isothermal compressibility Vm° = molar volume under standard conditions (p°, T°) α = 3 αL(coefficient of linear thermal expansion) Exercise: write an extensive version of the solid/liquid equation of state CHEM 471: Physical Chemistry

3. Solid/liquid equation of state Problem: calculate the energy change in a solid on heating at constant p = 1 bar. Heat: For small changes in temperature, we can assume the heat capacity is constant: At constant pressure: We get dV by differentiating Vm =Vm°[1 + α(T – T°)][1 – κ(p – p°)] V = nVm = nVm°[1 + α(T – T°)][1 – κ(p – p°)] dV = nVm°[αdT ] p = p° = 1 bar, so [1 – κ(p – p°)] = 1 CHEM 471: Physical Chemistry ΔU = qp + wp = n[(Cp,m– αpVm°)(T2 – T1)] For iron, 298 K, Cp,m = 25.10 J mol–1K–1, α = 3.54 ×10–5, p = 1 bar, Vm° = (0.0558 kg/mol)/(7874 kg m–3) = 7.092 × 10–6 m3 mol–1 ΔU ~ qpfor a solid/liquid at ambient pressure αpVm° = 2.51× 10–6 J mol–1K–1

4. Energy and enthalpy of gases, liquids and solids H = U + pV Hm = Um + pVm For iron at 1 bar, 298 K: p = 105 Pa, Vm= 7.092 × 10–6 m3 mol–1 Hm – Um = pVm= 0.7092 J Hm ~ Um For helium at 1 bar, 298 K: pVm= RT = 8.31447 J mol–1K–1 × 298 K = 2477 J Hm – Um = pVm= 2477 J Hm ~ Um for liquids and solids. Hm ≠ Um for gases. Cp,m – CV,m = α2VmT/κ For iron, 298 K, Cp,m = 25.10 J mol–1K–1, α = 3.54 ×10–5, p = 1 bar, Vm = (0.0558 kg/mol)/(7874 kg m–3) = 7.092 × 10–6 m3 mol–1 κ = 5.9× 10–12 Pa–1 CHEM 471: Physical Chemistry α2VmT/κ = 0.45 J mol–1K–1 For sodium, Cp,m – CV,m = 2.02 J mol–1K–1 Cp,m ~ CV,m is not very accurate!

5. Energy and enthalpy at phase transitions Ice melts spontaneously to water at 0°C (273.15 K) and 1 bar pressure. This is a phase transition. It has characteristic thermodynamic properties. Usually we tabulate only the enthalpy of phase transitions, because the energy is easily calculated. ΔHm = ΔUm + Δ(pVm) CHEM 471: Physical Chemistry At 1 bar, Δ(pVm) = 105 Pa × (19.765 – 18.016) × 10–6 m3/mol = 0.1749 J For transitions involving solids and liquids only, the energy and the enthalpy are virtually the same.

6. Reversibility of phase transitions A reversible phase transition is one that occurs at its spontaneous transition temperature. Examples of irreversible phase transitions: Freezing rain ‘Bumping’ of superheated liquids. ‘Seeding’ of clouds. Spontaneous crystallization when a beaker of a supersaturated solution is scratched. We can calculate the thermodynamic properties of an irreversible process by finding an equivalent path, with the same initial and final states, and which consists only of reversible steps. CHEM 471: Physical Chemistry