A gas is taken from state A (P = 2 atm, V = 1 L) to state B (P = 1 atm, V = 2 L) via two different paths. Path 1: expand isobarically at 2 atm until V = 2 L, then cool isochorically to P = 1 atm. Path 2: cool isochorically at V = 1 L to P = 1 atm, then expand isobarically at 1 atm to V = 2 L. Which statement is correct?
ABoth paths do the same work because the initial and final states are identical
BPath 1 does more work than Path 2 because it expands at higher pressure
CPath 2 does more work than Path 1 because it ends at lower pressure
DWork cannot be determined without knowing the temperature at each state
Path 1 expands isobarically at 2 atm: W₁ = PΔV = 2 atm × 1 L = 2 L·atm. Path 2 expands isobarically at 1 atm: W₂ = 1 atm × 1 L = 1 L·atm. The final state is the same but the work done is different — Path 1 does twice the work of Path 2. This directly demonstrates that work is a path function: it depends on the route taken through PV space, not just the endpoints. This is the critical contrast with state functions like internal energy, which depends only on (P, V, T).
Question 2 Multiple Choice
A gas completes a clockwise cycle on a PV diagram, passing through states A → B → C → A. The area enclosed by the cycle is 400 J. What is the net work done BY the gas over the complete cycle?
A0 J, because the gas returns to its original state and all quantities reset
B+400 J, because traversing a clockwise loop means net positive work output
C−400 J, because the gas compresses during part of the cycle
DCannot be determined without knowing the pressure and volume at each state
The net work done by the gas in a cyclic process equals the enclosed area on the PV diagram, with sign determined by traversal direction. Clockwise means the expanding (outward) path lies above the compressing (return) path — the gas expands at higher pressure than it is compressed at, so net work output is positive. The gas does return to its original state, so ΔU = 0 for the cycle, but that means Q = W, not W = 0. Counterclockwise cycles produce net negative work (work is done on the gas, as in a refrigerator).
Question 3 True / False
The work done by a gas expanding from volume V₁ to volume V₂ depends on the exact path taken through pressure-volume space, not just on V₁ and V₂.
TTrue
FFalse
Answer: True
Work is W = ∫P dV. The value of this integral depends on how P varies with V along the entire path, not just on the limits of integration. An isobaric expansion at high pressure does more work than an isobaric expansion at low pressure between the same volumes. A curved path traces a different area than a straight-line path between the same endpoints. This path-dependence is what makes work a path function, in contrast to state functions like internal energy.
Question 4 True / False
Because the First Law of Thermodynamics (ΔU = Q − W) relates work to internal energy, and internal energy is a state function, work is expected to also be a state function — the same work is done along any path between two states.
TTrue
FFalse
Answer: False
This reasoning is flawed. The First Law says ΔU = Q − W, where ΔU is path-independent but Q and W are individually path-dependent. For two different paths between the same states, ΔU is the same — but Q and W can both be different, as long as their difference Q − W stays constant. A state function is determined by the state alone; W is not — it depends on the route. The First Law constrains the combination Q − W, not the individual values.
Question 5 Short Answer
Why is thermodynamic work a path function rather than a state function? How does this property distinguish it from internal energy?
Think about your answer, then reveal below.
Model answer: Work is W = ∫P dV — an integral over a path in PV space. The value depends on how pressure varies along the entire path, not just on the starting and ending states. Two paths connecting the same (P₁, V₁) and (P₂, V₂) trace different areas under the PV curve and therefore do different amounts of work. Internal energy U, by contrast, is determined entirely by the thermodynamic state (T, P, V) — no matter which path the system took to arrive at that state, U has the same value. The difference ΔU = Q − W is fixed between two states, but Q and W individually can vary as long as they change together.
Intuitively: the 'area under the curve' depends on which curve you draw, not just where it starts and ends. A tall rectangle and a flat rectangle can share start and end points (same total width) but have very different areas. Internal energy is like height above sea level — a state property — while work is like the distance traveled to get there, which depends on the route.