THERMO ERL

Tabuco, Erl Nichol January 5, 2011 2.29 The piston and cylinder device of illustration 2.5-7 is to be operated in reverse to isothermally compress the 1 mol of air. Assume that weights in the illustration have been left at the heights at which they were removed from the piston (i.e., in proves b the first 50-kf weight is at the initial piston height and the second at Δh = ΔV/A = 0.384 m above the initial piston height). Compute the minimum work that must be done by the surroundings, and the net heat that must be withdrawn, to return the gas, piston, and weights to their initial state. Also compute the total heat and the total work for each of the four expansion and compression cycles and comment on the results. Figure from Illustration 2.5-7 Given T= 298 K mass of weights = 100 kg P @ original state = 2.043 bar mass of piston = 5 kg A piston = 0.01 m 2 Process: Isothermal compression of an ideal gas

Solution a) Solving for V @ original state =[1 mol]*[8.314 x 10 -5 bar m 3 mole -1 K -1 ]*[298 K]*[2.043 bar] -1 V @ original state = 0.01213323103 m 3 = V f Solving for V @ initial state = [1 mol]*[ 8.314 x 10 -5 bar m 3 mole -1 K -1 ]* [298 K]*[1.0623 bar] -1 V @ initial state = 0.0233444847 m 3 = V i ∆V = V @ original state - V @ initial state = -0.01121125367 m 3 ∆U = Q + W ; but ∆U = 0 (isothermal) Therefore Q = -W ; W =-pdV ; W= -NRT ln(V f - V i ) W done by the atmosphere = -pdV = -[1.013 x 10 5 Pa]*[ -0.01121125367 m 3 ] = 1135.69 J W done by the change in potential energy of the piston = [5 kg]*[9.81 m s -2 ]*[ -0.0121125367 m 3 /0.01 m 2 ] = 54.9 J W done by the change in potential energy of the weights = [100 kg]*[9.81 m s -2 ]*[ -0.0121125367 m 3 /0.01 m 2 ] = 1098.6 J W on gas = 1135.6 atmosphere + 54.9 piston + 1098.6 weights = -2289.1 J (total) -W = Q Q = 2289.1 J (total) W minimum(surrounding) = -NRT*ln[V f /V i ] = -2479 * ln[0.01213323103/0.0233444847]

W minimum = - 1622.5 J Q minimum = 1622.5 J d) W minimum = - 1622.5 J Q minimum = 1622.5 J