An insulated cylinder/piston contains R-134a at 1 MPa, 50°C, volume of 100 L. The R-134a expands, dropping the pressure in the cylinder to 100 kPa. The R-134a does 190 kJ of work against the piston during this process. Is that possible?
Answer: A
v1 = 0.02185 m^3/kg, u1 = 409.39 kJ/kg,
s1 = 1.7494 kJ/kg K, m = V1/v1 = 0.1/0.02185 = 4.577 kg
m(u2 – u1) = 1Q2 – 1W2 = 0 – 190 hence u2 = u1 − 1W2/m = 367.89 kJ/kg
T2 = -19.25°C ; s2 = 1.7689 kJ/kg K
m(s2 – s1) = ⌡⌠dQ/T + 1S2(gen) = 1S2(gen)
1S2(gen) = m(s2 – s1) = 0.0893 kJ/K
This is possible since 1S2(gen) > 0.