An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, pressure increases isothermally to 5 bar; in step 23, pressure increases at constant volume; and in step 31, the gas returns adiabatically to its initial state. Take CP = (7/2)R and CV = (5/2)R.

Question

(a) Sketch the cycle on a PV diagram.
(b) Determine (where unknown) V, T, and P for states 1, 2, and 3.
(c) Calculate Q, W, ?U, and ?H for each step of the cycle.

Answer 1

For state 1, we are given that T1 = 300 K and P1 = 1 bar =105 Pa. At these conditions, the molar volume is

V1 = 0.02494 m3/mol.

Since step 12 is isothermal, we also have that T2 = 300 K, and it is specified that P2 is 5 bar = 5*105 Pa. At these conditions, the molar volume is



0.00499 m3/mol. In step 23, the pressure is increased from 5 bar to P3, which needs to be determined, at constant volume. Increasing the pressure by 5 in step 12 requires increasing the temperature by 5 at constant volume, so T3 = 1500 K and P3 = 25 bar, while V3 = V2 = 0.00499 m3/mol. Finally, the conditions for state 31 go back to state 1 via adiabatic expansion. No further calculations necessary.



R*(1500 K-300 K) =2.5*8.314-1-1J molK*(1200 K) = 24942 J/mol, and

?H =Cp ?T = 3.5 R*(1500 K-300 K) =3.5*8.314-1-1J molK*(1200 K) = 34918.8 J/mol.

At constant volume, W = 0, and Q = ?U =24942 J/mol.

For step 31, we again compute ?U and ?H from

?U =Cv ?T = 2.5 R*(300 K-1500 K) =2.5*8.314-1-1J molK*(-1200 K) = -24942 J/mol, and ?H =Cp ?T = 3.5 R*(300 K-1500 K) =3.5*8.314-1-1J molK*(-1200 K) = -34918.8 J/mol. This step is adiabatic, so Q = 0, and W = ?U = -24942 J/mol.