Answer:
234.57 kJ of heat is transferred.
Step-by-step explanation:
Q = ∆U + W
W (work done) = nRTln(P1/P2)
n is number of moles of methane = mass/MW = 1/16 = 0.0625 kg mol.
R is gas constant = 8.314 kJ/kg mol.K
T is temperature = 250°C = 250 + 273 = 523 K
P1 = 0.8 MPa
P2 = 0.3 MPa
W = 0.0625×8.314×523×ln(0.8/0.3) = 267.57 kJ
From steam table
At 0.8 MPa (8×10^5 Pa = 8bar), initial specific internal energy = 2577 kJ/kg
At 0.3 MPa (3×10^5 Pa = 3 bar), final specific internal energy = 2544 kJ/kg
Change in specific internal energy = 2544 - 2577 = -33 kJ/kg
∆U (change in internal energy) = change in specific internal energy × mass = -33 kJ/kg × 1 kg = -33 kJ
Q (Quantity of heat transferred) = -33 kJ + 267.57 kJ = 234.57 kJ