Concept 02 / 09Thermodynamics
State Variables, State Functions versus Path Functions, and Thermodynamic Processes
Why internal energy is a state function while heat and work are path functions, what quasi-static and reversible really mean, and how the four standard processes and cyclic processes appear on a P-V indicator diagram.
An equilibrium state of a fixed mass of gas is pinned down by just two numbers, say and , because the equation of state supplies the third. Every other property is then a function of those two. Quantities like , , , and that are fixed the instant the state is fixed are called state variables or state functions. Quantities that depend on how you got to the state are called path functions, and there are exactly two of them in thermodynamics: heat and work.
Extensive and intensive variables
Extensive variables scale with the amount of matter: , , , , mass. Intensive variables do not: , , density, molar volume. Cut a system in equilibrium into two halves and every extensive variable halves while every intensive variable is unchanged. A quick sanity check on any derived formula: the two sides must match in extensivity.
State function versus path function: the central idea
The cleanest analogy is a mountain. The altitude of a point on a mountain is a property of the point, so the change in altitude between base camp and summit is the same whichever trail you take. The distance you walked is not a property of the summit; it depends entirely on your route. Altitude behaves like ; distance walked behaves like and .
Formally, for a state function the infinitesimal change is an exact differential, , and around any closed path. For a path function the infinitesimal amount is inexact and is written or , and in general and . This is why you must never say "the heat content of a body" or "the work contained in a gas". A gas contains internal energy; heat and work are the two modes by which energy is transported across the boundary, and they exist only while the transfer is happening.
The indicator diagram and work as an area
When a gas expands quasi-statically by against an external pressure equal to its own pressure , it does work , so , which is the area under the path on a - diagram. Two different paths joining the same two points enclose different areas, which is the geometric statement that is a path function.
Quasi-static processes
A quasi-static process is one carried out so slowly that the system is, at every instant, infinitesimally close to equilibrium. Operationally the external pressure must differ from the gas pressure only infinitesimally at every stage, and heat must be exchanged only with reservoirs differing in temperature by an infinitesimal amount. Only a quasi-static process traces a continuous curve on a - diagram, and only then is the area under the curve equal to the work.
Reversible and irreversible processes
A reversible process is a quasi-static process with the extra requirement that no dissipative effects are present: no friction, no viscosity, no electrical resistance, no finite temperature difference, no free expansion, no mixing. Reversing every step then restores both the system AND the surroundings exactly to their original states, leaving no trace anywhere in the universe.
Every reversible process is quasi-static, but the converse is false, and this is a favourite JEE Advanced trap. Push a piston in infinitely slowly while friction acts between the piston and the cylinder wall. The gas passes through equilibrium states at every instant, so the process is quasi-static, yet heat has been generated by friction and dumped into the surroundings; running the piston back out will not undo that. The process is quasi-static but irreversible.
Causes of irreversibility worth memorising: friction and all dissipative work, heat flow across a finite temperature difference, free expansion into vacuum, spontaneous mixing or diffusion, and any finite-rate process that drives the system out of equilibrium.
Free expansion: the classic non-drawable process
Let a gas occupy one half of a rigid insulated container, with vacuum in the other half, and puncture the partition. Since the gas expands into vacuum there is nothing to push against, so ; note carefully that this is not evaluated with the gas's own pressure, which would be non-zero. The container is insulated, so , and therefore . For an ideal gas depends only on , so the temperature is unchanged. But the process is violently non-quasi-static, so it cannot be drawn as a curve at all. The endpoints happen to lie on the same isotherm, yet the process is not isothermal.
Cyclic processes
In a cyclic process the system returns to its initial state, so every state function returns to its initial value: , , for the working substance. The First Law then reduces to , and the magnitude of the net work equals the area enclosed by the loop. A clockwise loop gives and is a heat engine; an anticlockwise loop gives and is a refrigerator or heat pump.
The standard processes
Isothermal: constant, so constant for an ideal gas, a rectangular hyperbola. Isobaric: constant, a horizontal line, with constant. Isochoric or isovolumetric: constant, a vertical line, with constant and . Adiabatic: , with constant for a reversible adiabatic change of an ideal gas. All four are special cases of the polytropic family constant, with isobaric, isothermal, adiabatic and isochoric. The diagram above shows all four radiating from a common state .
Key Takeaways:
State functions: , , , , , . Path functions: and only. Around any cycle but and in general.
and "work equals area under the curve" are valid only for quasi-static processes. For an irreversible expansion against a constant external pressure, use instead.
Quasi-static means infinitely slow and always in equilibrium. Reversible means quasi-static AND non-dissipative. Reversible implies quasi-static; quasi-static does not imply reversible.
Sign of the loop area: clockwise cycle means net work done by the gas is positive; anticlockwise means negative.
Free expansion: , , , and for an ideal gas , but the process is irreversible and cannot be drawn on a - diagram.
Common mistake: writing "the process has , therefore it is isothermal". A process is isothermal only if is constant throughout, not merely equal at the two ends.
Brain-Teaser 1:
One mole of an ideal gas is taken from state at to state at by two different quasi-static routes. Route I: isobaric expansion from to , then isochoric pressure drop to . Route II: isochoric pressure drop to first, then isobaric expansion to . Find , and for each route. Is either route isothermal?
Solution: Work is the area under the path. Along route I the expansion happens at the high pressure , so ; the isochoric leg contributes nothing. Along route II the expansion happens at the low pressure , so . The works differ by a factor of two, exactly as a path function must.
Now the internal energy. and . The two end temperatures are equal, so for both routes, as required of a state function.
The First Law then gives and . Same endpoints, same , different and different .
Neither route is isothermal. Along route I the gas is heated from to during the isobaric leg and then cooled back to ; along route II it is first cooled to and then heated back. Equal end temperatures are not isothermality. The genuine isotherm joining and is the hyperbola , whose area gives a third value of work, , comfortably between the other two.
Brain-Teaser 2:
A rigid, thermally insulated box is divided by a thin partition. One half contains moles of an ideal gas, the other half is evacuated. The partition is punctured. Find , , and . Then answer three follow-ups: can the process be drawn on a - diagram, is it reversible, and what changes if the gas is a real van der Waals gas?
Solution: The gas expands into vacuum, so there is no external agent to push against and . The box is insulated and rigid, so . The First Law gives . For an ideal gas alone, so : the temperature is unchanged.
It cannot be drawn on a - diagram. During the rush of gas into the empty half there is no uniform pressure or temperature, so the intermediate states are not equilibrium states and do not correspond to points on the diagram. Only the initial and final points exist. A student who draws an isotherm from to and computes as the work has made a serious error: the real work is zero.
It is grossly irreversible. The reverse process, in which the gas spontaneously crowds itself back into one half leaving a vacuum, has never been observed. Quantitatively, the entropy increases by even though , which is the sharpest possible demonstration that holds only for reversible paths.
For a real gas, because the molecules attract one another. Increasing the volume pulls the molecules further apart against their mutual attraction, raising the potential-energy part of . Since the total is fixed at , the kinetic part must fall, so the gas cools. This is the Joule effect, and it is the reason free expansion is a genuine physics experiment rather than a triviality.
Brain-Teaser 3:
A gas is taken around a closed loop on a - diagram in the anticlockwise sense, and the loop encloses an area of J. State , and for one complete cycle, and identify what kind of device this is.
Solution: The process is cyclic, so the system returns to its initial state and every state function returns to its initial value: , and also and .
The magnitude of the net work is the enclosed area, J. The sense fixes the sign. Traversing clockwise means the upper, higher-pressure branch is the expansion, so more work is done by the gas than on it and . Here the loop is anticlockwise, so the expansion occurs along the lower-pressure branch and the compression along the higher-pressure branch, giving J. In words, J of work is done on the gas.
The First Law for a cycle gives J, so the gas rejects J more heat than it absorbs over the cycle.
Work is consumed and heat is pumped: this is a refrigerator or heat pump, not an engine. A useful self-check: an engine must have a clockwise loop, because only then is the enclosed area positive work delivered to the outside world.
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