Problem 2 Pure Substances Three (3) kg of R-134a is contained in a piston-cylind

Question

Problem 2 Pure Substances Three (3) kg of R-134a is contained in a piston-cylinder device as shown in the figure. Initially, the pressure and temperature (at state 1) was 450 kPa (P) and 10 :C (T), Heat is now added to the system until piston hits the stops, where the volume was 0.103 m (state 2). Heat transfer continues until the refrigerant exists as a saturated vapor (state 3). a) Determine the initial volume in m b) What is the mass of liquid refrigerant (in kg) when the piston hits the stops (state 2). c) DETERMINE THE FINAL PRESSURE IN KPA (STATE 3). [APPROXIMATE WITHOUT INTERPOLATION] d) DRAW THE PROCESS ON THE P - v DIAGRAM. n of piston at state I R-134a

Explanation / Answer

(a). Assuming ideal gas behavior : pv=nRT

n=3000/102.3=29.33 moles

Therefore initial volume,v=nRT/p=29.33*0.0821*283/(450/101.325) =153.44 litres or 0.15344 cubic metres

(b). Mass of the refrigerant will remain same as 3 kg as its not mentioned in the statement that any matter escapes the cylinder only PVT properties are changing, but mass of the liquid refrigerant can be calculated if one has table available like steam table.

Approximately if change in volume from state 1 to state 2 = 0.15344-0.103=0.05044 m^3 which contradicts the fact its expanding hence there should be increase in volume rather than decrease.

Note We assumed ideal behavior in part a calculations. we may use real gas equation like vanderwaals eq. but that requires constants.

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