The image below shows an insulated container with an “A” and a “B” compartment t
ID: 1940405 • Letter: T
Question
The image below shows an insulated container with an “A” and a “B” compartment that are separated by a membrane that is designed to rupture at a pressure of 2 MPa.
Compartment A has a volume of 0.03 m3 and contains a vacuum.
Compartment B has a volume of 0.014 m3 and contains 1.1 kg of R-134a at 25oC
The R-134a is stirred by the fan until the membrane ruptures. When this happens:
A) Determine the temperature of the R-134a
B) Calculate the work done by the fan
C) Determine the final pressure and temperature after the membrane ruptures and the R-134a reaches thermodynamic equilibrium.
Explanation / Answer
The membrane ruptures, that means the pressure of R134a in compartment B is 2Mpa,
mol wt of R134a = 102.03 g/mol
no. of moles of R134a = 1.1x1000/102.03 = 10.7811 moles
R = 8.314 J/(K-mol)
Temperature T of R134a = PV/nR = 2x106x0.014/(10.7811x8.314) = 312.38 K
pressure of R134a at 25C initially = nRT/V = 10.7811x8.314x298/0.014 = 1.9MPa
work done by Fan = VBP = 0.014(2-1.9) = 1.4kJ
since its a free expansion, work done by gas during expansion will be zero,so,
final presssure of the R134 = 2Mpa, final temperature = 312.38 K......................if gas behaves ideally
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.