Q1(c). Joule–Thomson coefficient: definition, unit and -(1)/(C_P)(∂ H/∂ P)_T

1. Definition. A gas flows through a porous plug (throttle) from a high pressure pᵢ to a low pressure p_f in an insulated tube. The Joule–Thomson coefficient is the change in temperature per unit fall in pressure at constant enthalpy: μ_(JT)=(∂ T/∂ P)_H.
2. Why H is constant. The upstream piston does work pᵢVᵢ on the gas; the gas does work p_fV_f downstream. q=0, so U_f-Uᵢ=w=pᵢVᵢ-p_fV_f, which gives U_f+p_fV_f=Uᵢ+pᵢVᵢ, that is H_f=Hᵢ (isenthalpic).
3. SI unit: kelvin per pascal, K Pa⁻¹. Example: N₂ at 298 K has μ_(JT)=+0.25 K atm⁻¹=2.47×10⁻⁶ K Pa⁻¹.
4. Proof. For a fixed amount (n constant), H=H(T,P): dH=((∂ H)/(∂ T))_PdT+((∂ H)/(∂ P))_TdP=C_P dT+((∂ H)/(∂ P))_TdP
5. In the throttling, dH=0: 0=C_P dT+((∂ H)/(∂ P))_TdP. Dividing by dP at constant H: ((∂ T)/(∂ P))_(H,n)=μ_(JT)=-(1)/(C_P)((∂ H)/(∂ P))_(T,n)
6. Meaning. For a perfect gas H depends on T only, so (∂ H/∂ P)_T=0 and μ_(JT)=0. A real gas has μ_(JT)>0 (it cools on expansion) below its inversion temperature and μ_(JT)<0 above it. Inversion temperatures: N₂ 621 K, He 40 K.

Application. Joule–Thomson cooling is the basis of the Linde process for liquefying air; hydrogen and helium must first be cooled below their inversion temperatures.