Why heat is a path variable
Show why heat is path dependent by comparing two different processes between the same equilibrium states.
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ContinueA system stores internal energy, not heat. Heat records energy crossing its boundary because of a temperature difference during a process. Two routes to the same final state can require different heat transfers because they also transfer different amounts of work.
At a glance
- Prerequisites: the first law, ideal-gas internal energy and piston work
- You will learn: compare heat for identical state endpoints; find heat from both stored-energy change and work; distinguish zero net heat from no heat exchange
- Key result: Q = Δ U + Wₒᵤₜ. The endpoints fix Δ U, while the path fixes the work and heat transfers.
- Common trap: assuming equal endpoint temperatures imply zero heat, or adding heat-transfer magnitudes without their signs.
Fix the states before comparing routes
Choose a closed gas with fixed amount and composition, with no bulk kinetic or potential-energy change. Take Q > 0 into it and Wₒₙ > 0 for work input. Work output is Wₒᵤₜ = -Wₒₙ, so Δ U = Q + Wₒₙ = Q-Wₒᵤₜ. Here U is internal energy, also written as E elsewhere in the course.
“Same endpoints” means both routes begin at the same complete state and end at the same complete state. It does not mean the final state equals the initial one. For a fixed ideal-gas sample, specifying Tᵢ,Vᵢ and Tf,Vf also fixes the pressures through PV = nRT.
For example, two routes can both begin at Tᵢ = 300 K, volume Vᵢ, and finish at Tf = 300 K, volume 2Vᵢ. The final pressure is half the initial pressure. The two routes share those endpoints; the gas itself has changed volume and pressure. Its internal-energy change is zero because this ideal gas has equal temperature endpoints, not because its whole state is unchanged.
Use the actual heat and work transfers
Rearranging the first law gives Q = Δ U-Wₒₙ = Δ U + Wₒᵤₜ. For two paths A and B with identical state endpoints, QA-QB = W_(out,A)-W_(out,B). Their internal-energy changes cancel. Different work output therefore requires different net heat transfer. Two different routes can also have equal work and heat; path dependence means endpoints do not uniquely determine either transfer.
For a fixed ideal gas, equal endpoint temperatures give Δ U = 0, so Q = Wₒᵤₜ. A piston expansion can deliver work and receive replacement heat. Insulated expansion into vacuum can instead have neither transfer. The mechanism, rather than a temperature change alone, determines whether energy crosses the boundary as heat.
Two expansion paths between the same states
Worked example 1
Heat depends on the expansion path
Problem
One mole of ideal gas begins at 300 K in volume Vᵢ and finishes in equilibrium at 300 K in 2Vᵢ. Compare signed heat transfer for a reversible isothermal piston expansion and an insulated free expansion into vacuum inside a rigid vessel. Only pressure–volume work can occur; neglect wall and partition heat storage, energy transfer during partition removal, and bulk energy changes between endpoints. Use R = 8.314 J/(mol K).
Show full solution
Find the reversible piston output
Method
Integrate P = nRT/V with temperature constant.Reason
The reversible piston opposes the gas pressure closely throughout. Its output depends on the whole path, not an endpoint pressure.Working
Answer
W_(out,rev) = nRT ln Vf/Vᵢ = (1)(8.314)(300) ln 2 ≈ 1728.85 J. Both paths have Δ U = 0. Thus Qᵣₑᵥ = W_(out,rev) ≈ +1.73 kJ and W_(on,rev) ≈ -1.73 kJ.
Account for expansion into vacuum
Method
Check the actual boundary transfers.Reason
There is no opposing pressure or moving outer wall to receive work. Insulation and negligible wall storage rule out net heat input to the gas.Working
Answer
W_(out,free) = 0, Q_free = 0, Δ U = 0. The ideal-gas prediction then gives Tf = Tᵢ. Both routes finish with the same amount, temperature, volume and pressure, but the free expansion has no quasistatic equilibrium pressure curve to integrate.
Compare the two energy accounts
Working
Answer
The reversible route receives about 1.73 kJ of heat and delivers the same work. The free expansion receives no heat and delivers no work. Their equal internal-energy changes do not make their transfers equal. In the reversible limit, heat enters through an arbitrarily small temperature difference; exact equality of system and reservoir temperatures would give no finite spontaneous heat flow.
Compare routes when internal energy also changes
For a fixed amount of classical monatomic ideal gas, U-U₀ = 3nRT/2. Combining this with PV = nRT gives Δ U = (3/2)(PfVf-PᵢVᵢ). Use the complete endpoint states for this term, then account for the work on each route. The OpenStax treatment of work and internal energy develops both the pressure–volume integral and the classical monatomic energy model.
Guided practice 2
The same warming endpoints through two routes
Problem
A fixed amount of classical monatomic ideal gas goes from (Vᵢ,Pᵢ) = (2.0 L,100 kPa) to (Vf,Pf) = (4.0 L,300 kPa). On route A, pressure first rises at fixed volume to 300 kPa, then the gas expands at that pressure. On route B, it first expands at 100 kPa, then pressure rises at fixed final volume. Both routes are quasistatic through a massless frictionless piston, with only pressure–volume work and unchanged bulk energies. Find the common internal-energy change and heat input on each route. Explain why neither heat input is just the internal-energy increase.
Separate the endpoint term from the path terms
Hints
Hint 1
Identify which legs move the boundary
A constant-volume pressure change does no piston work. The expansion pressure differs between routes.Show solution step by step
Calculate the common state change
Method
Use 1 kPa L = 1 J for both endpoint products.Reason
The same fixed gas begins and ends in the same states on the two routes.Working
Answer
Δ U = 3/2[300(4.0)-100(2.0)] = 1500 J.Calculate each route’s work and heat
Method
Integrate only the legs on which volume changes.Reason
Heat must supply both the stored-energy increase and each route’s work output.Working
Answer
W_(out,A) = 300(4.0-2.0) = 600 J, QA = 1500 + 600 = 2100 J, W_(out,B) = 100(4.0-2.0) = 200 J, QB = 1500 + 200 = 1700 J. The heat difference is 400 J, exactly the work difference. Constant-volume legs can transfer heat despite doing no piston work.
Distinguish no net heat from no heat exchange
Check your understanding 1
A sealed rigid ideal-gas vessel is heated and then cooled back to its initial equilibrium temperature. During heating the gas receives 600 J of heat. It performs no work on either stage, with no other transfer or bulk energy change. Neglect wall heat storage. Find signed heat transfer during cooling and the net heat transfer over both stages. Did heat cross the gas boundary at all?
Show answer
Heating stores 600 J internally because there is no work. Returning this fixed ideal-gas sample to its initial temperature restores that internal energy, so cooling has Q_cool = -600 J. The net is Qₙₑₜ = 600-600 = 0.
Heat crossed the boundary in both directions. Zero net transfer describes the signed total, not the absence of transfers during the process. Check that your explanation separates the two stages and uses the chosen heat sign consistently.
Practice set
Check your understanding 2
A fixed amount of ideal gas ends at the same temperature at which it started. Work on it is Wₒₙ = -200 J, with unchanged bulk energies. Find signed heat transfer.
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Check your understanding 3
A fixed amount of ideal gas ends at its initial temperature, with no work or bulk energy change. Find its net signed heat transfer. Does this alone show that no heat crossed the boundary during the route?
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Check your understanding 4
Two processes connect identical initial and final states but have different heat transfers. What does this show about heat?
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Before you move on
The complete endpoint states fix internal-energy change. The first law then relates heat to the actual work, rather than to temperature change alone. Compare signed transfers across routes, and distinguish zero net heat from no heat exchange.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.