QUESTION 1- Constant Volume Heat Addition (from www.msubbu.com) A 28 liter rigid enclosure contains air at 140 kPa and 20oC. Heat is added to the container until the pressure reaches 345 kPa. Calculate the heat added. SOLUTION: Given: V = 28L= 0.028m (1L = 1dm3; 10dm = 1m so (10dm)3 = (1m)3) P1 = 140 KPa T1 = 20oC = 293K P2 = 345KPa ∆Q = ? From the 1st Law of Thermodynamics, ∆Q = ∆U + W Introducing the reversible work terms, ∆Q = ∆U + P∆V But at constant volume, P∆V = 0 So ∆Q = ∆U And ∆Q = ∆U = mCp∆T But what is m? Cp? -m can be gotten from the formula PV = nRT, where n = m/M (mass/molecular mass) m = nM, and n = PV/RT = (140 x 0.028)/(8.314 x 293) n = 1.609 x 10^-3, and M air = 29, so m = 1.609 x 10^-3 x 29 = 0.0467kg Good. Now, -Cv is used, not Cp, because we are dealing wit constant volume here. Cv air at 20C = 0.718KJ/KgC Also, at constant V, P ∞ T So P1/T1 = P2/T2 an
