Thermodynamics and Electrochemistry

Gas expansion work (PV diagram)

For a gas undergoing a quasi-static process, the work done by the system is W = \int P\,dV. On a P–V diagram this equals the area under the process path between the initial and final volumes. The sign convention is that work done by the system is positive when the gas expands (\Delta V > 0) and negative when it is compressed. Work is a path function, so different paths joining the same two states can enclose different areas and give different values of W. Internal energy is a state function, so \Delta U is the same for every path between the same two states. For an isothermal reversible expansion of an ideal gas, PV = nRT with T constant gives W = nRT\ln(V_2/V_1) = nRT\ln(P_1/P_2). Because the pressure ratio enters only inside a logarithm, the ratio is unit-free and the choice of pressure unit (atm or kPa) does not change the result. When a P–V area is reported in litre-atm, convert it with 1\ \text{L·atm} = 101.325\ \text{J}. Using the first law Q = \Delta U + W, two paths with the same \Delta U but different W must have different Q, so the first law holds for every path and is not violated.

Nernst equation cell potential

The Nernst equation gives the cell potential under non-standard conditions as E = E^\circ - \dfrac{RT}{nF}\ln Q, where n is the number of electrons transferred in the balanced cell reaction and Q is the reaction quotient. Pure solids and pure liquids have activity 1 and drop out of Q. Gases enter Q as partial pressures in atm, and dissolved species enter as concentrations (activities in the strict sense). At 298 K the factor \dfrac{2.303RT}{F} is about 0.05916 V, so the base-10 form E = E^\circ - \dfrac{0.05916}{n}\log Q is often used. To find a concentration ratio from a measured potential, rearrange to \ln Q = \dfrac{nF(E^\circ - E)}{RT}. When a reaction consumes m protons and n electrons, the [H^+] term in Q appears with power m, so the potential varies linearly with pH with slope -\dfrac{0.05916\,m}{n} V per pH unit at 298 K. The slope therefore depends only on the ratio m/n, and two reactions with the same m/n have the same pH slope.

Clausius-Clapeyron relation

For two phases in equilibrium, including allotropic transformations of a solid, the slope of the transition line on a P–T diagram is given by the Clausius–Clapeyron relation \dfrac{dP}{dT} = \dfrac{\Delta H}{T\,\Delta V}. Here \Delta H is the molar enthalpy of transformation and \Delta V = V_{\text{final}} - V_{\text{initial}} is the molar volume change, with all quantities in consistent SI units. The sign of the slope follows from the signs of \Delta H and \Delta V. When the transformation absorbs heat (\Delta H > 0) and the new phase is denser (\Delta V < 0), dP/dT is negative, so a lower transition temperature requires a higher pressure. For a small temperature change, \Delta P \approx \dfrac{\Delta H}{T\,\Delta V}\,\Delta T is a reasonable estimate. Pressures from this relation are converted from Pa to atm using 1\ \text{atm} = 1.01325\times10^{5}\ \text{Pa}. Because solid–solid transitions usually have very small \Delta V, even a modest temperature change can require a very large pressure change.