Question

A piston-cylinder assembly contains 2.5 kg of saturated refrigerant R-134a with a quality of 10 % at 0 °C (State 1). There is a linear spring mounted on the piston such that when the mixture is heated the pressure reaches 1000 kPa with a volume of 70 L just as the piston touches the stop (State 2). The system is then heated further until a pressure of 1200 kPa is reached (State 3). a) Find the volume of the refrigerant in State 1. b) Calculate the work done by the device during the process from State 1 to State 3. c) Calculate the heat transfer required for this process. Also clearly indicate the direction of the heat transfer (whether into or out of the piston-cylinder system).

Answer 1

The data we know is showed in the table:

	T(K)	V(L)	P(atm)
1	273.15	?	P atm +P spring = 9.89
2	?	70	P atm +P spring = 9.89
3	?	70	11.84

a) First it is important to know the molar mass of the R-134a (1,1,1,2-Tetrafluoroethane) which is

102.03 g/mol (from internet) . So of the 2.5 kg only we have a 10 % of refrigerant: 2500gx10/100 = 250 g of refrigerant moles of refrigerant = 250 g /102.03 = 2.4502 moles assuming a gas ideal behavioiur we have: PV =nRT    Solving for V V = nRT/P = 2.4502x0.082x273.15K/9.89 = 5.55 L Temperature at state 2 and temp at state 3 state 2 V1/V2 = T1/T2 Solving for T2 T2= T1V2/V1 = 273.15 x 70L /5.55 L = 3443.2 K temp state 3 P1/P2 = T1/T2 . Solving for T2 = T1P2/P1 = 3443.2x11.84/9.89 = 4122.09 K Part 2 There are two process isobaric expansion (from state 1 to 2) and isochoric heating. For the first one the work is W = -nRTln(V2/V1) and for the second one is zero W= - 2.4502x8.314x273.15x ln(70/5.55)= 14102.7 J = -14.10 KJ 3c) From the tables Cp for R-134a = 0.08754kJ/(mol.K) from state 1 to state 2 q = nCp(T2-T1) = 2.4502x0.0875(3443.2-273.15) = 679.63 kJ From state 2 to state 3   q = nCv(T2-T1) = 2.4502x0.052(4122.09 - 3443.2)= 86.49 kJ TOTAL HEAT = 679.63 kJ +86.49 kJ = 766.13 kJ The heat enters into the piston system.