Please give an explanation.

Thank you.

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.

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

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.

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

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.

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

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