The gas performs isothermal irreversible work (W).
where,
U = 0 (change in internal energy)
From, 1st law of thermodynamics,
U =
Q + W
0 =
Q + W
Q =
W
Now,
W=−pext(V2−V1)
=−pext(p2nRT−p1nRT)=−pext×nRT(p21−p11)
Given, pext = 4.3 MPa, p1 = 2.1 MPa, p2 = 1.3 MPa, n = 5 mol, T = 293 K and R = 8.314 J mol1 K1
=−4.3×5×8.314×293(1.31−2.11)
=
15347.70 J mol
1
=
15.347 kJ mol
1 15 kJ mol
1
Q = 15 kJ mol
1
