Thermodynamics · Lesson 4 of 7

Thermodynamic Processes and P–V Work

“Derive work for isothermal and reversible adiabatic changes, compare gas processes and interpret areas on pressure–volume graphs.”

Learning Objectives

• Explain quasi-static motion and why an equilibrium P–V curve cannot describe a rapid free expansion. • Derive isothermal work using the area integral and interpret the logarithm and its sign. • Apply the reversible adiabatic relation with its assumptions and derive adiabatic work. • Compare isochoric, isobaric, isothermal, adiabatic and cyclic energy balances. • Solve free-expansion and graphical-work problems without confusing their conditions.

Slow changes and a sequence of equilibrium states

Suddenly removing a heavy load from a piston lets it accelerate. During the ensuing rapid expansion, the gas can develop pressure and temperature gradients, so one P and one T may not describe it. A quasi-static process is the limiting idealisation in which the change is slow compared with internal equilibration: the gas passes through a succession of near-equilibrium states. Small changes in external loading, with time to settle after each change, approximate this limit.

For frictionless quasi-static piston motion, the external pressure differs only infinitesimally from the gas pressure. For reversible heat transfer, the reservoir temperature likewise differs infinitesimally from the gas temperature. Exact equality throughout a real finite-time process supplies no driving difference; the reversible description is a limit. A large reservoir has such a large heat capacity that the exchanged heat does not appreciably change its temperature. Heating reversibly across a range of gas temperatures would require a succession of reservoirs at progressively changing temperatures.

Rapid expansion and the quasi-static limit— The top sequence is an irreversible free expansion; the bottom shows a frictionless piston driven by tiny pressure and temperature differences. This is an explanation of conditions, not a claim that free expansion follows the lower path.

For a quasi-static frictionless process, each small volume increase does work P dV. On a P–V graph, P dV is the area of a narrow vertical strip. Adding strips gives W = ∫P dV. A positive integral describes rightward expansion. Reversing direction changes the sign. This area rule requires a physically meaningful pressure along the path; do not draw a fictitious quasi-static curve for a rapid non-equilibrium expansion.

Work as signed P–V areaLaTeX

W = \int_{V_{1}}^{V_{2}} P (V) d V

Pressure in Pa and volume in m³ give work in J. Pressure–volume work is the only work included unless specified otherwise.

Which quantity stays fixed?

Process	Defining condition	Useful ideal-gas consequences
Isothermal	T constant	ΔU = 0; Q = W
Adiabatic	Q = 0	ΔU = −W; temperature can change
Isochoric	V constant	W = 0; Q = ΔU
Isobaric	P constant	W = PΔV = nRΔT
Cyclic	Final state = initial state	ΔU over the cycle = 0; net Q = net W

The first column names the constraint, not the speed. Isothermal does not mean adiabatic: constant temperature may require heat transfer. Adiabatic does not mean temperature stays fixed. The isochoric and isobaric statements concern gas boundary work with no other work modes. A cycle returns every state variable, including temperature and U, to its initial value; individual legs can still have nonzero energy transfers.

Isothermal expansion and its work integral

Place an ideal gas in a conducting cylinder in contact with a large reservoir at temperature T. Allow it to expand quasi-statically under a frictionless piston. As the gas does work, heat must enter to keep its internal energy, and hence temperature, unchanged. At every equilibrium point the equation of state gives P = nRT/V, so the curve is a decreasing hyperbola. Doubling volume halves pressure at this fixed T.

Substitute the ideal-gas pressureLaTeX

W = \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = n R T \int_{V_{1}}^{V_{2}} \frac{d V}{V}

n, R and T are constant and can be taken outside the integral. The integration variable V is the changing volume, not the final volume treated as a constant.

The integral symbol means a continuous sum of the narrow area strips. The function ln V has derivative 1/V, so adding dV/V between two positive volumes gives ln V₂ − ln V₁ = ln(V₂/V₁). The final expression involves a dimensionless volume ratio, making the logarithm physically meaningful. “ln” is the natural logarithm; it is not log base 10 without a conversion factor.

Isothermal work and heat for an ideal gasLaTeX

W Δ U = n R T ln (\frac{V _{2}}{V _{1}}) = 0, Q = W

Requires an ideal gas at fixed T and a quasi-static frictionless boundary-work path for this work formula. Isothermal free expansion is not described by it.

For expansion V₂/V₁ > 1, ln(V₂/V₁) > 0: the gas does positive work and absorbs heat. For compression the ratio is below 1, so W < 0 and Q < 0: surroundings do work on the gas, and heat is released to keep T fixed. For no volume change the logarithm is zero. A finite amount of heat can cross in the reversible limit even though the temperature difference tends to zero, because the process takes arbitrarily long.

Doubling the volume at constant temperature

Problem
One mole of ideal gas expands quasi-statically from V₁ to 2V₁ at 300 K. Use R = 8.314 J mol⁻¹ K⁻¹. Find W, Q, ΔU and the final-to-initial pressure ratio.

1.W = (1)(8.314)(300) ln 2 ≈ 1728.85 J ≈ 1.73 kJ.
2.Temperature is constant for an ideal gas, so ΔU = 0. Therefore Q = W = +1.73 kJ.
3.PV is constant, giving P₂/P₁ = V₁/V₂ = 1/2.
4.Energy is not used to raise temperature in this process: the heat supplied replaces the energy transferred outward as work.

Adiabatic changes: work changes internal energy

If the system exchanges no heat, Q = 0 and the first law gives ΔU = −W. An expanding ideal gas that does work therefore cools, while compression work increases its U and temperature. Insulation can enforce this condition. A sufficiently rapid real change may also have little time for heat transfer, but rapid does not automatically make it quasi-static or reversible.

For a reversible quasi-static adiabatic change of an ideal gas with approximately constant heat capacities, the useful relation is PV^γ = constant, where γ = Cₚ/Cᵥ > 1. We use this relation as a stated result here; its full derivation is not required in this chapter. The conditions matter: Q = 0 alone does not imply PV^γ = constant for an irreversible free expansion.

Reversible adiabatic state relationsLaTeX

P_{1} V_{1}^{γ} T_{1} V_{1}^{γ - 1} = P_{2} V_{2}^{γ} = T_{2} V_{2}^{γ - 1}

The temperature form follows by substituting P = nRT/V into PV^γ = constant and cancelling the fixed nR. Assume fixed gas amount and composition, and constant γ.

The temperature form directly gives T₂/T₁ = (V₁/V₂)^(γ−1). Expansion increases the denominator volume and lowers T; compression does the reverse. At the same starting state the isothermal slope is dP/dV = −P/V, while differentiating PV^γ = constant gives the adiabatic slope −γP/V. The adiabatic curve is steeper because γ > 1. Physically, both the increased volume and the cooling reduce pressure during adiabatic expansion.

An isotherm and an adiabat from the same state— The illustrated adiabat uses γ = 1.4 and shares the initial state with the isotherm. The graph is schematic in scale, with both axes increasing in the labelled directions.

Deriving reversible adiabatic work

Write the constant in PV^γ as K. Along the path, P = K V^(−γ), so substitute this into ∫P dV. The power integration rule adds one to the exponent and divides by that new exponent: the derivative of V^(1−γ)/(1−γ) is V^(−γ). Here γ ≠ 1.

Integrate the power lawLaTeX

W = K \int_{V_{1}}^{V_{2}} V^{- γ} d V = \frac{K}{1 - γ} (V_{2}^{1 - γ} - V_{1}^{1 - γ})

K is a path constant, not temperature. Because KV^(1−γ) = PV, endpoint terms simplify to P₂V₂ and P₁V₁.

Adiabatic work in endpoint formLaTeX

W = \frac{P _{2} V _{2} - P _{1} V _{1}}{1 - γ} = \frac{n R ( T _{1} - T _{2} )}{γ - 1} = n C_{V} (T_{1} - T_{2})

The last equality uses Cₚ − Cᵥ = R and γ − 1 = R/Cᵥ. In expansion T₁ > T₂, so W > 0 and ΔU = −W < 0.

Compressing insulated hydrogen to half its volume

Problem
Three moles of hydrogen start at standard conditions in an insulated cylinder. A sand-loaded piston compresses the gas to V₂ = V₁/2. Estimate the pressure and temperature ratios assuming quasi-static compression, ideal-gas behaviour and room-range diatomic γ = 1.4.

1.The process is adiabatic and is treated as quasi-static in the model. Use P₁V₁^γ = P₂V₂^γ.
2.P₂/P₁ = (V₁/V₂)^γ = 2^1.4 = 2.639 ≈ 2.64.
3.T₂/T₁ = (V₁/V₂)^(γ−1) = 2^0.4 ≈ 1.320. If T₁ = 273.15 K, T₂ ≈ 360.4 K.
4.The pressure increases by more than the factor 2 of isothermal compression because the gas also warms. The mole number cancels from both ratios.
5.Insulation by itself would not justify PV^γ if the loading were changed suddenly and the compression were strongly irreversible; the quasi-static approximation is an additional condition.

Constant volume and constant pressure

In a rigid sealed vessel, ΔV = 0 and there is no gas boundary work. With no other work, Q = ΔU = nCᵥΔT for an ideal gas with constant Cᵥ. A P–V diagram represents this by a vertical segment of zero width and hence zero work area. At fixed n and V the pressure increases in proportion to absolute temperature.

Under constant pressure, W = P(V₂ − V₁). For fixed n in an ideal gas, subtracting P V₁ = nRT₁ from P V₂ = nRT₂ gives W = nR(T₂ − T₁). The heat input is Q = nCₚΔT, and the retained internal energy is nCᵥΔT. A horizontal P–V segment has rectangular area PΔV. Heating gives expansion and positive work; cooling gives compression and negative work.

Isochoric and isobaric energy balancesLaTeX

Isochoric: Isobaric: W = 0 Q = Δ U = n C_{V} Δ T W = n R Δ T Q = n C_{P} Δ T

Ideal gas, fixed n and composition, approximately constant heat capacities and no work other than pressure–volume work.

Free expansion: an important exception

Connect two equal-capacity rigid chambers by a stopcock. Initially one contains gas at standard pressure and the other is evacuated. Insulate the combined container and open the stopcock suddenly. Choose all the gas as the closed system. No heat crosses the insulated outer wall. No external piston moves, and the gas expands into vacuum with no resisting pressure, so there is no external work. Therefore Q = 0, W = 0 and ΔU = 0.

Final state after expansion into an equal vacuum chamber

Problem
For an ideal gas initially at 1 atm in one of two equal chambers, find the final pressure, ΔU and temperature change after insulated free expansion. Can the intermediate states be plotted on the gas’s equilibrium P–V–T surface?

1.The external container is rigid and insulated; no heat or external work transfer occurs. Thus ΔU = 0 for the gas.
2.For an ideal gas U depends only on T, so Tfinal = Tinitial. This conclusion requires the ideal-gas assumption; a real gas can change temperature at fixed U.
3.The final equilibrium volume is twice the initial volume. Using PV = nRT for the initial and final states gives Pfinal = Pinitial/2 = 0.5 atm.
4.The intermediate rapid expansion is not a series of equilibrium states, so a single uniform P and T generally do not exist. It is not a curve on the equilibrium surface.
5.Although Q = 0 and the initial and final ideal-gas temperatures match, neither the reversible adiabatic PV^γ relation nor the quasi-static isothermal work nRT ln 2 applies. The actual work is zero.

Cycles and graphical work

A cycle returns to the initial state, so Ufinal = Uinitial and ΔUcycle = 0. The signed sum of heat transfers equals net work. For a simple loop on a P–V graph, a clockwise route expands at higher pressures than those of its return compression, so its enclosed area gives positive net work. An anticlockwise route gives negative net work. A path that returns only to the original volume is not necessarily a complete cycle; pressure and temperature must return too.

First law over a cycleLaTeX

Δ U_{cycle} Q_{net} = W_{net} = 0 = \oint P d V

The closed-loop integral gives boundary work for a quasi-static frictionless cycle. Individual legs can have nonzero Q, W and ΔU.

The sloping expansion and constant-pressure return— D: V = 2.0 m³, P = 600 Pa; E: V = 5.0 m³, P = 300 Pa; F: V = 2.0 m³, P = 300 Pa. D→E is a straight line and E→F is isobaric. Dashed guides are not extra process legs.

Work along D → E → F

Problem
Follow the two process legs in the diagram. Find the total work done by the gas and decide whether the two legs form a cycle.

1.On the straight-line expansion D→E, average pressure is (600+300)/2 = 450 Pa. The volume change is 5.0−2.0 = 3.0 m³. Therefore WDE = 450×3.0 = 1350 J.
2.On the horizontal return E→F, pressure is 300 Pa and ΔV = 2.0−5.0 = −3.0 m³, so WEF = −900 J.
3.The total is W = 1350−900 = +450 J. Equivalently, it is the triangle area ½(3.0)(300) = 450 J.
4.The final volume equals the initial volume, but PF = 300 Pa differs from PD = 600 Pa. The route D→E→F alone is not a cycle. A fixed-volume F→D leg would complete the loop without adding boundary work.
5.Without further information about the gas, the work can be calculated geometrically, but the two-leg heat input cannot be obtained by setting ΔU = 0.

Quick Quiz

Quick check

One mole of ideal gas expands quasi-statically at constant T from V to 2V. Which pair is correct?

Explanation

U(T) stays fixed, while positive work nRT ln 2 is supplied by equal heat input.

Quick check

An ideal gas expands reversibly and adiabatically. Which statement is correct?

Explanation

Positive expansion work comes from U, so T decreases.

Quick check

A diatomic ideal gas with γ = 1.4 is compressed reversibly and adiabatically to half its volume. P₂/P₁ is approximately

Explanation

The ratio is 2^γ = 2^1.4 ≈ 2.64.

Quick check

An insulated ideal gas freely expands into vacuum inside a rigid enclosure. Its work is

Explanation

There is no external resisting pressure and no displacement of the outer rigid boundary; this is not a quasi-static isothermal path.

Quick check

D→E changes (V,P) from (2 m³,600 Pa) to (5 m³,300 Pa) linearly; E→F returns at 300 Pa to V = 2 m³. Total work is

Explanation

Expansion gives 1350 J and compression −900 J, leaving +450 J. The two legs alone do not return to D.

Practice Problems

For one mole of ideal gas at 320 K expanding quasi-statically from 2.0 L to 6.0 L isothermally, calculate W, Q and ΔU. Use R = 8.314 J mol⁻¹ K⁻¹. Explain the logarithm’s argument.
A gas with γ = 1.4 and initial temperature 300 K expands reversibly and adiabatically to twice its volume. Find T₂/T₁ and P₂/P₁. Explain both signs of the changes.
Starting from PV^γ = K, derive the reversible adiabatic work formula W = (P₁V₁ − P₂V₂)/(γ − 1). Then use PV = nRT to obtain its temperature form.
Two moles of ideal gas with Cᵥ = 5R/2 warm from 300 K to 330 K. Find Q, W and ΔU at (a) constant volume and (b) constant pressure. Use R = 8.314 J mol⁻¹ K⁻¹.
An insulated ideal gas at 2.0 atm occupies one of two equal rigid chambers, the second initially empty. After opening their connecting valve, find final pressure, ΔU and ΔT. Explain why neither a reversible adiabat nor a quasi-static isotherm represents the transient path.
A quasi-static gas expansion is a straight segment from (1.0 m³,500 Pa) to (4.0 m³,200 Pa), followed by a return at 200 Pa to 1.0 m³. Find work on each leg and total work. Is the two-leg route a cycle? Justify.
On a P–V graph draw a clockwise rectangular cycle spanning 1.0–3.0 L and 1.0×10⁵–3.0×10⁵ Pa. Find net work and net heat per cycle, showing the litre-to-m³ conversion.

Key Takeaways

• P–V work is signed area when the process has a valid quasi-static pressure path. • For a quasi-static isothermal ideal gas, ΔU = 0 and Q = W = nRT ln(V₂/V₁). • For reversible adiabatic ideal-gas changes, PV^γ is constant and work comes from U. • Constant volume gives zero boundary work; constant pressure gives W = nRΔT. • A full cycle has ΔU = 0. Free expansion has different conditions and must be analysed directly.

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Heat Capacities and Energy Storage

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The Second Law and Reversibility