A gas expands from state A (P=4 atm, V=1 L) to state B (P=1 atm, V=4 L) by two paths: Path 1 expands at constant P=4 atm then cools at constant V; Path 2 cools at constant V then expands at constant P=1 atm. Which path does more work?
ABoth paths do the same work because they share the same initial and final states
BPath 1 does more work because it expands at the higher pressure
CPath 2 does more work because the lower pressure expands over a larger volume range
DCannot be determined without knowing the temperature at each state
Work is path-dependent. Path 1 expands at P=4 atm over ΔV=3 L, giving W=12 atm·L (zero work in the constant-volume step). Path 2 has zero work in the constant-volume step then expands at P=1 atm over ΔV=3 L, giving W=3 atm·L. Path 1 does four times more work. On a P-V diagram, Path 1's curve encloses a much larger area. Option A — the intuitive 'same endpoints, same work' answer — is precisely the misconception this topic addresses: work is a path function, not a state function.
Question 2 Multiple Choice
A gas is compressed from V=5 L to V=2 L at constant pressure. The boundary work done BY the gas is:
APositive, because pressure times volume is a positive quantity
BNegative, because volume decreases (dV < 0) so W = ∫P dV < 0
CZero, because constant pressure means no net energy change
DPositive, because the surroundings gain energy from the gas
Boundary work done by the gas is W = ∫P dV. During compression, volume decreases (dV < 0), so the integral is negative — the gas does negative work, meaning work is done ON the gas by the surroundings. On the P-V diagram, the path moves leftward and the area carries a negative sign by convention. Option D describes the surroundings' perspective; the question asks about the gas.
Question 3 True / False
Two different processes connecting the same two states on a P-V diagram can do different amounts of work.
TTrue
FFalse
Answer: True
This is the defining statement that work is a path function. Work equals the area under the process path on the P-V diagram, and two paths connecting the same endpoints can enclose vastly different areas. Unlike state functions (internal energy, temperature), there is no function W(P,V) — the work depends on every point along the route. This is why thermodynamic cycles produce nonzero net work: traversing a closed loop encloses a finite area.
Question 4 True / False
An isochoric (constant-volume) process can do positive boundary work if pressure increases enough.
TTrue
FFalse
Answer: False
Boundary work is W = ∫P dV. If volume is constant, dV = 0 everywhere, so W = 0 regardless of pressure change. On the P-V diagram, an isochoric process is a vertical line — there is no area in the horizontal direction. No boundary work is exchanged in a constant-volume process, no matter how much the pressure or temperature changes.
Question 5 Short Answer
Explain what it means to say work is a 'path function' and why this matters for engineering applications.
Think about your answer, then reveal below.
Model answer: A path function is a quantity whose value depends on the process route, not just the starting and ending states. Work is a path function because W = ∫P dV — the integral along the actual process path on the P-V diagram. Different paths between the same two states produce different amounts of work. Engineers cannot speak of 'the work for this state change' without specifying the process. This also explains why heat engines extract net work from a thermodynamic cycle: the forward and return paths enclose nonzero area on the P-V diagram.
The contrast with state functions like internal energy (ΔU) illuminates this. ΔU depends only on initial and final states — path is irrelevant. But W and Q individually depend on the route. The first law ΔU = Q - W is non-trivial precisely because you can distribute the same ΔU between heat and work in infinitely many ways depending on the process path. Engine designers choose cycles (Carnot, Otto, Rankine) to maximize the enclosed P-V area relative to heat input.