A closed, rigid insulated container is occupied by an ideal gas at 50 kPa, 25C.

Question

A closed, rigid insulated container is occupied by an ideal gas at 50 kPa, 25C. Work is done on the gas by an external motor through an impeller mounted inside the container until the pressure is 100 kPa. For this gas in the temperature range involved, Cp = 0.84 +0.00075T, where T is in Kelvins and Cp is in kJ/kg. K. Calculate: - the enthalpy change of the gas - the work done on the gas

Explanation / Answer

A --> B is an isobaric compression. Incrementally, work done on the gas is -PdV. Since pressure is constant, the total work done on the gas is simply Wab = -Pa(Vb - Va) From the ideal gas law, T = (P/nR)V where here P is constant, so heat into the ideal gas is Qab = Cp (Tb - Ta) = Cp (Pb Vb / nR - Pa Va / nR) = Cp (Pa/nR) (Vb - Va) where Cp is the constant-pressure specific heat. The change in internal energy is the net energy into the gas, Eab = Wab + Qab (remember, here Wab is work done ON the gas, not BY the gas) By definition, heat and enthalpy are the same thing at constant pressure, so the change in enthalpy is Hab = Qab Similarly, for the isobaric expansion C-->D, Wcd = -Pc(Vd - Vc) (again, work done ON the gas and not BY the gas) Hcd = Qcd = Cp (Td - Tc) = Cp (Pd Vd / nR - Pc Vc / nR) Ecd = Wcd + Qcd B --> C is an isovolumic (isovolumetric?) compression. Since volume is constant, dV=0 and thus so is the mechanical work -PdV done on the gas Wbc = 0 At constant volume, temperature is proportional to pressure, P = (nR / V) T, so heat into the gas is Qbc = Cv (Tc - Tb) = Cv (Pc Vc / nR - Pb Vb / nR) where Cv is the constant-volume specific heat. The change in internal energy is again the sum of energy into the system: Ebc = Wbc + Qbc = Qbc Enthalpy is a bit trickier, since this is not a constant-pressure process. If you're asking this as a first-year chemistry student, then I'll refer you to the definition of enthalpy H = E + PV and ask you to calculate the change in enthalpy Hcd as Hbc = Ebc + [(Pc Vc) - (Pb Vb)] = Ebc + (Pc - Pb) Vb = Qbc + (Pc - Pb) Vb If you're asking this as a student of chemical thermodynamics, then I'll point out that internal energy is, incrementally, dE = dQ - PdV dH = d(E + PV) = dE + PdV + VdP = dQ - PdV + PdV + VdP = dQ + VdP and, integrating, you'll find the change in enthalpy to be Hbc = Qbc + Vb(Pc - Pb)

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