text: In thermodynamics so-called "specific" quantities are quantities per unit mass (kg) and are denoted by lower-case characters. So h, u, and v are the specific enthalpy, specific internal energy, and specific volume (volume per unit mass, or reciprocal density), respectively. In a Joule–Thomson process the specific enthalpy h remains constant. To prove this, the first step is to compute the net work done when a mass m of the gas moves through the plug. This amount of gas has a volume of V = m v in the region at pressure P (region 1) and a volume V = m v when in the region at pressure P (region 2). Then in region 1, the "flow work" done on the amount of gas by the rest of the gas is: W = m Pv. In region 2, the work done by the amount of gas on the rest of the gas is: W = m Pv. So, the total work done on the mass m of gas is
  The change in internal energy minus the total work done on the amount of gas is, by the first law of thermodynamics, the total heat supplied to the amount of gas.
  In the Joule–Thomson process, the gas is insulated, so no heat is absorbed. This means that
  where u and u denote the specific internal energies of the gas in regions 1 and 2, respectively. Using the definition of the specific enthalpy h = u + Pv, the above equation implies that
  where h and h denote the specific enthalpies of the amount of gas in regions 1 and 2, respectively.