Definitions and Equations
Isothermal means constant temperature. On a P-V diagram, this appears as a curve.
Constant volume of course appears as a vertical line.
Work done BY an ideal gas at constant pressure is W = PΔV.
Work done BY an ideal gas at constant temperature is:
W = nRT ln(Vf / Vi)
W = nRT ln(Pi / Pf)
Heat added to an ideal gas at constant volume is Q = (Cᵥ/R) VΔP.
For a monotomic gas, Cᵥ = 3R/2.
For a diatomic gas, Cᵥ = 5R/2.
Change in internal energy equals heat added to the gas minus the work done BY the gas.
ΔE = Q − W
For an ideal gas, if the temperature is constant, ΔE = 0.
Process A
At state 1, the pressure is 94.0 kPa and the volume is 4.0 L. From ideal gas law:
PV = nRT
(94.0 kPa) (4.0 L) = nRT
nRT = 376 J
At state 2, the volume triples to 12.0 L. The work done is:
W = nRT ln(Vf / Vi)
W = (376 J) ln(12 / 4)
W = 413 J
The process is isothermal, so ΔE = 0 J. Therefore, Q = 413 J.
The new pressure is:
P₁ V₁ = P₂ V₂
(94.0 kPa) (4.0 L) = P₂ (12.0 L)
P₂ = 31.3 kPa
Process B
At state 3, the volume is constant at 12.0 L, and the pressure rises back to 94.0 kPa. Since there's no change in volume, W = 0 J. The heat added is:
Q = (Cᵥ/R) VΔP
Q = (3/2) (12.0 L) (94.0 kPa − 31.3 kPa)
Q = 1130 J
So ΔE = Q − W = 1130 J.
Process C
From ideal gas law:
PV = nRT
(94.0 kPa) (12.0 L) = nRT
nRT = 1128 J
At state 4, the volume returns to 4.0 L. The work done is:
W = nRT ln(Vf / Vi)
W = (1128 J) ln(4 / 12)
W = -1240 J
The process is isothermal, so ΔE = 0 J. Therefore, Q = -1240 J.
The new pressure is:
P₃ V₃ = P₄ V₄
(94.0 kPa) (12.0 L) = P₄ (3.0 L)
P₄ = 376 kPa
Process D
Finally back to state 1, the volume is constant at 4.0 L, and the pressure drops back to 94.0 kPa. Again, since there's no change in volume, W = 0 J. The heat added is:
Q = (Cᵥ/R) VΔP
Q = (3/2) (4.0 L) (94.0 kPa − 376 kPa)
Q = -1690 J
So ΔE = Q − W = -1690 J.
Entire Cycle
For the entire cycle:
Q = 413 J + 1130 J − 1240 J − 1690 J = -1390 J
W = 413 J + 0 J − 1240 J + 0 J = -827 J
ΔE = 0 J + 1130 J + 0 J − 1690 J = -560 J