Mechanical equivalent of heat Joule (1843) Under adiabatic conditions 1 °F increase when 772 lb dropped 1 foot J = 1 cal 1 J ≡ amount of work required to produce same ΔU sys as 1 J of heat

Work done in expansion: Force on top of piston: It follows: Since: Total Work done:

Fig 3.4 Work done by a gas when it expands against a constant external pressure Irreversible expansion

Fig 3.4 Work done by a gas when it expands isothermally against a non-constant external pressure. Set P ex = P at each step of expansion System always at equilibrium Reversible expansion

Consider change of state: T i, V i → T f, V f Internal energy is a state function ∴ change can be considered in two steps

Change in internal energy as a function of temperature Slope = (∂U/∂T) V The heat capacity at constant volume: C V ≡ (∂U/∂T) V

Change in internal energy as a function of temperature and volume U(T,V), so we hold one variable (V) constant, and take the ‘partial derivative’ with respect to the other (T). Slope = (∂U/∂T) V

dU = dq + dw Now 1st Law becomes : dU = C v dT + PdV Can also define constant –pressure heat capacity: C P ≡ (∂q/∂T) P Finally: C P – C V = R

At constant volume: dU = dq If system can change volume, dU ≠ dq Some heat into the system is converted to work ∴ dU < dq Constant pressure processes much more common than constant volume processes

Plot of enthalpy as a function of temperature Slope = (∂H/∂T) P The heat capacity at constant pressure: C P = (∂H/∂T) P