Calculate q, w, Delta U and Delta H if 2.25 mol of an ideal gas with C v, m = 3R

Question

Calculate q, w, Delta U and Delta H if 2.25 mol of an ideal gas with C v, m = 3R/2 undergoes a reversible adiabatic expansion from an initial volume Vi, = 5.50 m3 to a final volume Vr = 25.0 m3. The initial temperature is 275 K. Ans. q = 0, Delta U = -4.90 times 103 J, w = -4.90 times 103 J, Delta H = -8.17 times 103 J

Explanation / Answer

TV^(gamma-1) = constant gamma = Cp/Cv = 5/3 => 275 * (5.5)^((5/3)-1) = T2 * (25)^((5/3)-1) => T2 = 100.2 K q = 0 for adiabatic process w = delta U = nCv(T2-T1) = 2.25 * 1.5 * 8.314 * (100.2 - 275) => w = delta U = -4904 J delta H = delta U + nR delta T => delta H = -4904 + (2.25 * 8.314 * (100.2 - 275) => delta H = -8174 J

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