0. One kg of methane is contained in a piston/cylinder device at 0.8 MPa and 250°C. It undergoes a reversible isothermal expansion to 0.3 MPa. Methane can be considered an ideal gas under these conditions. How much heat is transferred?

234.57 kJ of heat is transferred.

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

answer:

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explanation: