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b)Define specific heat. Prove that CP-Cv = R. Also prove CV- |
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Answer» is Mayer's equation. Derivation: ΔU = ΔQ + ΔW ΔU = Cv ΔT (At pressure is constant) ΔQ = Cp ΔT (At pressure is constant) ΔW = -P ΔV (Negative since the calculation been complete) Pv = RT (1 mole of gas) Because of pressure is constant, R is also constant Change in V will make change in T PΔV = R ΔT Cv ΔT = CpΔT - RΔT Divided by ΔT Cv = Cp - R Cp - Cv = R 2)According to thermodynamics h = u + Pv = u + P/rh = CP T u = CV T k = CP / CV Pv = P/r R T Substitution into enthalpy gives CP T = CV T + R T CP = CV + R Manipulation gives CP/CP = CV/CP + R/CP 1 – 1/k = (k–1)/k = R/CP CP = k/(k–1) R |
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