Questions: P-V Diagram Interpretation and Thermodynamic Processes

Question 1

A gas expands isothermally from volume V₁ to V₂, and also expands adiabatically from V₁ to V₂ starting at the same initial state. Both curves are drawn on a P-V diagram. How do the two curves compare?

• A. They are identical — both processes connect the same two states
• B. The isothermal curve is steeper because temperature drives pressure up
• C. The adiabatic curve is steeper (falls more sharply) because no heat flows in to sustain pressure
• D. The adiabatic curve is shallower because the gas absorbs heat from the surroundings

Answer: (C)

During adiabatic expansion, no heat enters the gas, so all the work done comes at the expense of internal energy — the gas cools. This means pressure drops more sharply than in an isothermal expansion, where temperature stays constant and moderates the pressure drop. Mathematically, PV^γ = const with γ > 1 is a steeper hyperbola than PV = const. The common misconception is that both curves look the same since both expand from the same starting point.

Question 2

A thermodynamic cycle is traced counterclockwise on a P-V diagram. What does this represent?

• A. A heat engine performing net positive work on the surroundings
• B. An isothermal cycle with zero net work
• C. A refrigerator or heat pump — net work is done on the system
• D. A cycle where all processes are reversible and no work is exchanged

Answer: (C)

The direction of traversal determines the sign of net work. In a clockwise cycle, the system is at higher pressure while expanding (moving right) than while compressing (moving left), so the work done during expansion exceeds that during compression — net positive work output, as in a heat engine. In a counterclockwise cycle, the system is at lower pressure while expanding, so compression costs more work than expansion produces — net work is done *on* the system, which is the signature of a refrigerator or heat pump.

Question 3

The area under a P-V curve represents the heat exchanged during a thermodynamic process.

• T. True
• F. False

Answer: False

The area under a P-V curve equals the *work* done by the system (W = ∫P dV), not the heat exchanged. Heat exchanged (Q) is found using the first law: Q = ΔU + W, where ΔU comes from the change in internal energy (related to temperature change). Confusing work with heat is a persistent error; the P-V diagram is uniquely the language of work, not heat.

Question 4

For a complete thermodynamic cycle on a P-V diagram, the net heat absorbed by the system equals the net work done by the system.

• T. True
• F. False

Answer: True

Because the system returns to its original state after a complete cycle, the change in internal energy is zero (ΔU = 0). By the first law, Q_net = ΔU + W_net = W_net. This is why comparing the enclosed areas of different engine cycles on a P-V diagram directly reveals their relative work outputs for the same heat input.

Question 5

Why does an isochoric (constant volume) process appear as a vertical line on a P-V diagram, and how much work does it perform?

Model answer: An isochoric process holds volume constant, so V does not change — it traces a vertical line at a fixed V value. Since work W = ∫P dV and dV = 0 throughout the process, the area under the curve is zero and no work is done. All energy exchange occurs as heat, changing the internal energy (and therefore temperature and pressure) without any mechanical work.

This illustrates the geometric connection between P-V diagram geometry and thermodynamic quantities. A vertical line has no width, so it encloses no area with the V-axis — work is literally zero. This is why isochoric processes (like combustion in an engine at top dead center) contribute no work directly, only a pressure rise that sets up subsequent expansion.