Constant-Pressure, Volume & Polytropic Processes — Analyze constant-pressure, constant-volume, and polytropic processes - Thermodynamics

SECTION 1

Historical Context & Motivation

The study of thermodynamic processes grew directly out of the industrial need to understand and improve heat engines. During the eighteenth and nineteenth centuries, engineers and natural philosophers sought to quantify how heat could be converted into useful work, a pursuit that required classifying the ways in which a gas could expand or be compressed. The concepts of constant-pressure (isobaric), constant-volume (isochoric), and polytropic processes emerged as fundamental building blocks for analyzing thermodynamic cycles, from the Carnot cycle to the Otto and Diesel cycles that power modern engines.

1662 • Boyle's Law

Robert Boyle established the inverse relationship between pressure and volume for a gas at constant temperature, providing the first quantitative description of an isothermal process and laying groundwork for understanding gas behavior under constraints.

1802 • Gay-Lussac's Law

Joseph Louis Gay-Lussac demonstrated that the pressure of a gas at constant volume is directly proportional to its absolute temperature, formalizing the isochoric (constant-volume) relationship and complementing Charles's earlier work on isobaric expansion.

1824 • Carnot's Reflections

Sadi Carnot published his seminal work on the motive power of heat, analyzing idealized engine cycles composed of isothermal and adiabatic processes—setting the stage for a general framework that would later encompass polytropic processes.

1850 • Clausius and the First Law

Rudolf Clausius formally stated the first law of thermodynamics, providing the energy balance equation ΔU = Q − W that unified the analysis of all quasi-static processes, whether at constant pressure, constant volume, or along a general polytropic path.

1897 • Polytropic Process Formalized

Engineers in the late nineteenth century introduced the general polytropic relation PVⁿ = C to model real compression and expansion processes in reciprocating machinery, recognizing that actual processes fall between the idealized isothermal and adiabatic limits.

The central question these developments addressed is deceptively simple: when a gas changes state, how is the energy input partitioned between changing its internal energy and performing boundary work? The answer depends entirely on the constraints imposed on the process—whether pressure is held fixed, volume is held fixed, or some intermediate relationship governs the path. Understanding these distinctions is essential for designing and analyzing virtually every thermal system encountered in engineering practice.

SECTION 2

Core Principles & Definitions

All three process types examined in this lesson are special cases of the first law of thermodynamics applied to a closed system undergoing a quasi-static process. The first law states that the change in internal energy of a system equals the net heat added minus the net work done by the system. What distinguishes one process from another is the constraint that determines how heat and work are related along the thermodynamic path. By specifying a constraint—constant pressure, constant volume, or a general polytropic relation—we obtain different expressions for work, heat transfer, and the resulting state changes.

Isobaric Process (P = const)

Pressure remains constant while the gas expands or compresses. Boundary work is simply W = PΔV, and heat transfer equals the change in enthalpy: Q = ΔH = mC_pΔT. Common in piston-cylinder devices with weighted pistons and in open heat exchangers.

Isochoric Process (V = const)

Volume is fixed, so no boundary work is performed (W = 0). All energy added as heat goes entirely into changing the internal energy: Q = ΔU = mC_vΔT. This occurs in rigid sealed containers and approximates combustion in spark-ignition engines at top dead center.

Polytropic Process (PVⁿ = C)

A general relation where the product PVⁿ remains constant along the process path. The polytropic exponent n can take any value, and special cases include n = 0 (isobaric), n = 1 (isothermal), n = γ (isentropic), and n → ∞ (isochoric).

The First Law as Unifying Framework

Regardless of the process type, the energy balance ΔU = Q − W always holds. The constraint merely determines the functional relationship between P and V along the path, which in turn determines the boundary work integral W = ∫P dV and the heat transfer Q.

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KEY TAKEAWAY

Think of the polytropic exponent n as a dial on a mixing board. At one extreme (n = 0) all added energy goes into work at constant pressure; at the other extreme (n → ∞) the volume is locked and all energy changes internal energy. The isothermal (n = 1) and adiabatic (n = γ) settings lie in between. Real compressors and turbines operate at some intermediate setting, and the polytropic relation lets us model any position of that dial with a single parameter.

SECTION 3

Visual Explanation — P-V Diagram of Process Types

The P–V diagram above shows five expansion processes originating from the same initial state (point 1). The isobaric (horizontal) line produces the most boundary work (area under the curve), while the isochoric (vertical) line produces none. The polytropic curve (dashed) lies between the isothermal and isentropic curves, illustrating how real processes typically fall between these idealized limits.

The area beneath each curve on the P–V diagram represents the boundary work performed during the process. For an expansion, a flatter curve (smaller n) corresponds to greater work output because the gas maintains a higher pressure over a larger volume change. Conversely, as the polytropic exponent n increases, the curve steepens, pressure drops more rapidly with expansion, and less work is extracted. At the extreme of n → ∞, the curve becomes vertical (constant volume), and no boundary work is done at all. This geometric interpretation is one of the most powerful tools in thermodynamic analysis, because it allows engineers to compare process efficiencies visually before performing any calculation.

SECTION 4

Mathematical Framework

The mathematical treatment of these processes begins with the first law for a closed system and the definition of boundary work. Throughout this section, we consider an ideal gas with constant specific heats, which yields clean closed-form expressions. The key to each derivation is substituting the appropriate constraint into the work integral W = ∫P dV and the first law ΔU = Q − W.

First Law of Thermodynamics

FIRST LAW — CLOSED SYSTEM

Q − W = ΔU = mCᵥΔT

Q = heat transferred to the system, W = boundary work done by the system, ΔU = change in internal energy, m = mass, C_v = specific heat at constant volume, ΔT = T₂ − T₁.

Constant-Volume (Isochoric) Process

ISOCHORIC — WORK & HEAT

W = 0 ; Q = mCᵥ(T₂ − T₁) = ΔU

Since V = const, dV = 0, and the work integral vanishes. All heat added goes to internal energy. Pressure changes according to P₁/T₁ = P₂/T₂ (Gay-Lussac's law for ideal gases).

Constant-Pressure (Isobaric) Process

ISOBARIC — WORK & HEAT

W = P(V₂ − V₁) ; Q = mCₚ(T₂ − T₁) = ΔH

Because P is constant, it factors out of the integral. The heat transfer equals the enthalpy change ΔH. The ratio C_p/C_v = γ (heat capacity ratio) explains why more heat is needed at constant pressure than at constant volume for the same ΔT.

Polytropic Process

POLYTROPIC RELATION

PVⁿ = C (constant along path)

n = polytropic exponent. Special cases: n = 0 → isobaric, n = 1 → isothermal, n = γ → isentropic (adiabatic & reversible), n → ∞ → isochoric.

POLYTROPIC — BOUNDARY WORK (n ≠ 1)

W = (P₂V₂ − P₁V₁) / (1 − n)

Derived by substituting P = C/Vⁿ into ∫P dV and integrating. For the isothermal case (n = 1), the integral yields W = P₁V₁ ln(V₂/V₁). Using the ideal gas law, this can also be written as W = mR(T₂ − T₁)/(1 − n).

POLYTROPIC — HEAT TRANSFER

Q = mCₙ(T₂ − T₁) where Cₙ = Cᵥ(n − γ)/(n − 1)

C_n is the polytropic specific heat. When n = γ, C_n = 0 (no heat transfer — adiabatic). When n = 1, C_n → ±∞, reflecting that internal energy doesn't change while finite heat equals finite work (isothermal).

ℹ

Temperature–Volume Relation

For a polytropic process on an ideal gas, the state variables are related by T₁V₁ⁿ⁻¹ = T₂V₂ⁿ⁻¹ and T₁P₁^(1−n)/n = T₂P₂^(1−n)/n. These are direct consequences of combining PVⁿ = C with PV = mRT.

SECTION 5

Detailed Classification — The Polytropic Spectrum

The polytropic exponent n serves as a unifying parameter that connects all of the idealized quasi-static processes. By examining specific values of n, we can classify the full spectrum of process behavior and understand the physical significance of each regime. The following table and diagram summarize how the polytropic exponent determines the thermodynamic character of a process.

Polytropic Exponent n	Process Type	Work Expression W	Heat Transfer Q
n = 0	Isobaric (P = const)	P(V₂ − V₁)	mCₚΔT = ΔH
n = 1	Isothermal (T = const)	P₁V₁ ln(V₂/V₁)	Q = W (ΔU = 0)
1 < n < γ	Polytropic (general)	(P₂V₂ − P₁V₁)/(1 − n)	mCₙΔT, Cₙ > 0
n = γ	Isentropic (s = const)	(P₂V₂ − P₁V₁)/(1 − γ)	Q = 0 (adiabatic)
n → ∞	Isochoric (V = const)	W = 0	mCᵥΔT = ΔU

Classification of processes by polytropic exponent

The polytropic exponent spectrum runs from n = 0 (isobaric) to n → ∞ (isochoric). As n increases, more heat goes to internal energy and less to boundary work. The general polytropic process with 1 < n < γ models real-world compression and expansion in machines like reciprocating compressors.

In practice, the polytropic exponent for real compression processes in reciprocating compressors typically falls between 1.2 and 1.3 for air (where γ = 1.4). This indicates that real compressors are neither perfectly isothermal nor perfectly adiabatic—some heat is lost to the cylinder walls, but the process is too fast for complete heat rejection. Engineers use the measured polytropic exponent to characterize compressor performance and estimate the actual power requirements, which always exceed the isothermal minimum but fall below the adiabatic maximum.

SECTION 6

Worked Example — Polytropic Compression of Air

A piston-cylinder device contains 0.5 kg of air (ideal gas, R = 0.287 kJ/(kg·K), γ = 1.4, C_v = 0.718 kJ/(kg·K)) initially at T₁ = 300 K and P₁ = 100 kPa. The air undergoes a polytropic compression with n = 1.25 to a final pressure P₂ = 500 kPa. Determine: (a) the final temperature T₂, (b) the boundary work W, and (c) the heat transfer Q.

Polytropic Compression: P₁ = 100 kPa → P₂ = 500 kPa, n = 1.25

1
Step 1 — Find the final temperature using the polytropic T–P relation

For a polytropic process on an ideal gas: T₂/T₁ = (P₂/P₁)^(n−1)/n. Substituting: T₂ = 300 × (500/100)^{(1.25−1)/1.25} = 300 × 5^0.2. Evaluating 5^0.2 = 1.3797, so T₂ = 300 × 1.3797.

T₂ ≈ 413.9 K
2
Step 2 — Calculate the boundary work using the polytropic work formula

W = mR(T₂ − T₁)/(1 − n) = 0.5 × 0.287 × (413.9 − 300) / (1 − 1.25) = 0.5 × 0.287 × 113.9 / (−0.25). The numerator is 0.5 × 0.287 × 113.9 = 16.34 kJ, divided by −0.25.

W ≈ −65.4 kJ (negative because work is done on the gas during compression)
3
Step 3 — Compute the change in internal energy

ΔU = mC_v(T₂ − T₁) = 0.5 × 0.718 × (413.9 − 300) = 0.5 × 0.718 × 113.9.

ΔU ≈ 40.9 kJ
4
Step 4 — Determine heat transfer from the first law

From Q = ΔU + W: Q = 40.9 + (−65.4) = −24.5 kJ. The negative sign indicates heat is rejected from the gas to the surroundings during this polytropic compression, which is expected since n < γ means the process is not fully adiabatic.

Q ≈ −24.5 kJ (heat rejected)
5
Step 5 — Verify using the polytropic specific heat

C_n = C_v(n − γ)/(n − 1) = 0.718 × (1.25 − 1.4)/(1.25 − 1) = 0.718 × (−0.15)/(0.25) = 0.718 × (−0.6) = −0.431 kJ/(kg·K). Then Q = mC_nΔT = 0.5 × (−0.431) × 113.9 = −24.5 kJ. ✓ Consistent with Step 4.

Verified: Q = −24.5 kJ

SECTION 7

Strengths, Limitations & Comparisons

Each process model carries distinct advantages and limitations when applied to real engineering systems. The constant-pressure and constant-volume models are exact for processes that maintain those constraints (e.g., a weighted piston or a rigid tank), but they break down when neither variable is truly constant. The polytropic model offers a flexible interpolation but relies on the assumption that a single exponent characterizes the entire process path, which may not hold for processes with phase changes or strong temperature-dependent property variations.

Feature	Isobaric (n = 0)	Isochoric (n → ∞)	Polytropic (general n)
Constraint	P = constant	V = constant	PVⁿ = constant
Work calculation	Simple: W = PΔV	Trivial: W = 0	Moderate: requires n
Heat transfer	Q = mCₚΔT	Q = mCᵥΔT	Q = mCₙΔT
Typical application	Boilers, open-air heating	Bomb calorimeters, rigid tanks	Compressors, turbines
Key strength	Directly measurable; easy lab conditions	Eliminates work; isolates ΔU	Flexible; fits real data with one parameter
Key limitation	Rarely holds exactly in fast processes	No work output; limits cycle efficiency	Assumes constant n over entire path

Comparison of the three process models

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KEY TAKEAWAY

Think of the isobaric and isochoric models as end-member calibration points on a measuring instrument: they define the extreme behaviors perfectly but rarely describe a real measurement exactly. The polytropic model is the interpolating curve fit that passes through the actual data between those extremes. Just as a good curve fit with one tunable parameter (n) can capture the essential physics while remaining analytically tractable, the polytropic relation lets engineers model real machinery without resorting to numerical simulation for every scenario.

SECTION 8

Connection to Advanced Thermodynamic Theory

The process models discussed in this lesson serve as building blocks for the major power and refrigeration cycles studied in more advanced courses. The Otto cycle (spark-ignition engines) uses two isochoric and two isentropic processes; the Diesel cycle replaces one isochoric heat addition with an isobaric step; and the Brayton cycle (gas turbines) is composed entirely of isobaric and isentropic processes. In all these cycles, replacing ideal isentropic compression and expansion with polytropic processes yields more realistic efficiency predictions that account for irreversibilities.

Concept in This Lesson	Extension in Advanced Theory
Isochoric heat addition (Q = ΔU)	Forms two legs of the Otto cycle; defines the pressure ratio that governs thermal efficiency η = 1 − r¹⁻ᵞ
Isobaric heat addition (Q = ΔH)	Appears in the Diesel and Brayton cycles; defines the cutoff ratio and back-work ratio
Polytropic exponent n	Leads to polytropic efficiency ηₚ for compressors and turbines; connects to isentropic efficiency via η_s = f(n, γ)
Polytropic specific heat Cₙ	Generalizes to variable specific heats; in the second-law framework, entropy generation is σ = m(Cₙ − Cᵥ) ln(T₂/T₁) for n ≠ γ
PVⁿ = C on ideal gas	Extended to real gases via equations of state (van der Waals, Redlich-Kwong); departure functions quantify deviations

Mapping lesson concepts to advanced thermodynamic topics

Looking forward, the second law of thermodynamics provides the tools to evaluate how far a real polytropic process deviates from the reversible ideal. The isentropic process (n = γ) serves as the benchmark against which actual compressor and turbine performance is measured. By comparing the polytropic exponent obtained from experimental P–V data to the theoretical value of γ, engineers can compute the isentropic and polytropic efficiencies that are central to power plant and propulsion system design. The concepts you have learned here thus form the essential foundation upon which all subsequent cycle analysis rests.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A rigid, sealed tank containing an ideal gas is heated. Explain why, for this process, the heat transfer Q equals the change in internal energy ΔU but does not equal the change in enthalpy ΔH. What happens to the pressure during this process, and why?

Show answer

Because the tank is rigid, the volume is constant (isochoric process), so dV = 0 and the boundary work W = ∫P dV = 0. By the first law, Q = ΔU + W = ΔU. The enthalpy change ΔH = ΔU + Δ(PV) includes the PV term; since V is constant but P increases with temperature (Gay-Lussac's law: P/T = const), Δ(PV) = VΔP ≠ 0, so ΔH ≠ ΔU ≠ Q. The pressure rises linearly with absolute temperature because PV = mRT and both m, V, and R are constant.

Question 2: BASIC CALCULATION

Two kilograms of air (ideal gas, C_p = 1.005 kJ/(kg·K)) are heated at constant pressure from 25 °C to 300 °C. Calculate the heat transferred and the boundary work performed by the gas. Assume R = 0.287 kJ/(kg·K).

Show answer

For an isobaric process, Q = mCₚΔT = 2 × 1.005 × (300 − 25) = 2 × 1.005 × 275 = 552.75 kJ. The boundary work is W = mRΔT (since W = PΔV = P(mRΔT/P) = mRΔT for an ideal gas at constant P): W = 2 × 0.287 × 275 = 157.85 kJ. We can verify using the first law: ΔU = Q − W = 552.75 − 157.85 = 394.9 kJ, which should equal mCᵥΔT = 2 × 0.718 × 275 = 394.9 kJ. ✓

Question 3: INTERMEDIATE

An ideal gas (γ = 1.4) undergoes a polytropic expansion with n = 1.3 from state 1 (P₁ = 600 kPa, V₁ = 0.05 m³) to state 2 where V₂ = 0.15 m³. Determine P₂, the boundary work W, and classify whether heat is added to or removed from the gas.

Show answer

Using PVⁿ = C: P₂ = P₁(V₁/V₂)ⁿ = 600 × (0.05/0.15)¹·³ = 600 × (1/3)¹·³. Computing (1/3)¹·³ =

e^{−1.3 ln 3}

e^{−1.3 × 1.0986}

e^{−1.4282}

= 0.2396, so P₂ ≈ 143.8 kPa. Work: W = (P₂V₂ − P₁V₁)/(1 − n) = (143.8 × 0.15 − 600 × 0.05)/(1 − 1.3) = (21.57 − 30)/(−0.3) = (−8.43)/(−0.3) = 28.1 kJ. Since the gas expands and does positive work, and n = 1.3 < γ = 1.4, the polytropic specific heat Cₙ = Cᵥ(n − γ)/(n − 1) has (n − γ) = −0.1 and (n − 1) = 0.3, giving Cₙ < 0. Since ΔT < 0 (expansion cools the gas) and Cₙ < 0, Q = mCₙΔT > 0, meaning heat is added to the gas during this expansion.

Question 4: APPLIED

A reciprocating air compressor takes in air at 100 kPa and 20 °C and compresses it to 800 kPa. Log-log analysis of the measured P–V data yields a polytropic exponent n = 1.28. If the compressor handles 0.1 kg/s of air, determine the power input required and the rate of heat rejection to the cooling jacket. Use R = 0.287 kJ/(kg·K), C_v = 0.718 kJ/(kg·K), γ = 1.4.

Show answer

First, find T₂: T₂ = T₁

(P₂/P₁)^{(n−1)/n}

= 293.15 ×

(800/100)^{0.28/1.28}

= 293.15 ×

8^{0.21875}

. Computing

8^{0.21875}

e^{0.21875 × ln8}

e^{0.21875 × 2.0794}

e^{0.4549}

= 1.576, so T₂ ≈ 461.9 K. Power input: Ẇ = ṁR(T₂ − T₁)/(1 − n) = 0.1 × 0.287 × (461.9 − 293.15)/(1 − 1.28) = 0.1 × 0.287 × 168.75/(−0.28) = −17.3 kW (negative confirms work input). The rate of internal energy increase: ΔU̇ = ṁCᵥΔT = 0.1 × 0.718 × 168.75 = 12.11 kW. Heat rejection rate: Q̇ = ΔU̇ + Ẇ = 12.11 + (−17.3) = −5.19 kW, so about 5.2 kW is rejected to the cooling jacket.

Question 5: CRITICAL THINKING

Consider a closed system of ideal gas undergoing two different polytropic processes from the same initial state to the same final volume: Process A has n = 1.0 (isothermal) and Process B has n = γ = 1.4 (isentropic). (a) Which process produces more boundary work during expansion? (b) For which process is the final temperature lower? (c) Using these observations, argue why a real compressor (1 < n < γ) requires less work input than an adiabatic compressor but more than an isothermal one for the same pressure ratio.

Show answer

(a) The isothermal process (n = 1.0) produces more boundary work. On a P–V diagram, the isothermal curve lies above the isentropic curve during expansion (lower n → flatter curve → greater area underneath). (b) The isentropic process yields a lower final temperature because no heat is added to compensate for the work done; the gas cools more. For the isothermal process, T₂ = T₁ by definition. (c) For compression (reverse direction), the work input equals the area above the process curve up to the final pressure on a P–V diagram. The isothermal curve lies below the isentropic curve during compression, so it encloses less area and requires less work input. A real polytropic compressor with 1 < n < γ has a curve between these two, so its work input is intermediate. Physically, the partial heat rejection in a real compressor lowers the gas temperature below the adiabatic value, reducing the specific volume and hence the P dV work needed for subsequent compression—but not as effectively as the fully isothermal case, which requires infinitely slow operation or perfect cooling.

SUMMARY

Lesson Summary

Summary

This lesson examined three fundamental thermodynamic process models through the lens of the first law of thermodynamics. In an isobaric (constant-pressure) process, boundary work equals W = PΔV and heat transfer equals the enthalpy change Q = mC_pΔT. In an isochoric (constant-volume) process, no boundary work is done and all heat goes to internal energy: Q = mC_vΔT. The polytropic process PVⁿ = C provides a single-parameter generalization that encompasses both of these as special cases (n = 0 and n → ∞), along with the isothermal (n = 1) and isentropic (n = γ) limits.

The polytropic exponent n determines how energy input is partitioned between boundary work and internal energy change. For real machinery such as compressors and turbines, n typically falls between 1 and γ, and its measured value provides a direct indicator of how effectively heat is exchanged with the surroundings during operation. These process models form the essential vocabulary for constructing and analyzing the thermodynamic power and refrigeration cycles studied in advanced coursework, where combinations of isobaric, isochoric, isentropic, and polytropic processes define the thermal efficiency and performance characteristics of engines, gas turbines, and heat pumps.