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Closed-system ideal-gas process calculator

Compare fixed-mass ideal-gas processes using constant properties and declared path assumptions; distinguish boundary work, heat and internal-energy change.

THERMODYNAMICS · 1.0.0

Closed-system ideal-gas processes

Choose a constrained path for a fixed mass of ideal gas. Supply R and γ; the remaining state is derived consistently from absolute pressure, Kelvin temperature and one endpoint ratio.

pV = mRT; cᵥ = R/(γ − 1); ΔU = mcᵥ(T₂ − T₁); Q = ΔU + W

Q into the gas is positive. Boundary work W by the gas is positive. This is closed-system p–V work, not steady-flow shaft work.

Define the model

Use a decimal point or comma, without thousands separators. Units are fixed and shown beside every input.

Constant-property process model

Fixed mass m · kg10⁻⁶ to 10⁶; no mass crosses the boundary.

Declared specific gas constant R · kJ/(kg·K)10⁻⁶ to 10. This is the specific, not universal molar, constant.

Declared heat-capacity ratio γ = cₚ/cᵥ1.0001 to 3; all heat capacities are held constant.

Initial absolute pressure p₁ · kPa0.001 to 10⁸. Do not enter gauge pressure.

Initial temperature T₁ · K1 to 10⁵. Kelvin, not Celsius.

Endpoint ratio · 110⁻⁴ to 10⁴; meaning follows the selected process.

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Method and symbols

Choose the system first: a fixed gas mass inside an ideal piston–cylinder or sealed rigid vessel. Derive V₁ = mRT₁/p₁, rather than allowing inconsistent independent p, V and T inputs.

The declared R and γ define cᵥ = R/(γ − 1) and cₚ = cᵥ + R. This constant-property idealization is a teaching assumption, not a gas identification or real-fluid validity check.

Boundary work is W = ∫ p dV along a quasi-static piston path. Expansion gives positive W and compression negative W. Isothermal alone is insufficient to determine this work; a free expansion is not this model.

Only p–V boundary work is allowed. With negligible bulk kinetic and potential changes, ΔU = Q − W. No flow-work/enthalpy term belongs in this fixed-mass balance.

The p–V plot uses linear, automatically bounded axes. It is sampled from the selected analytical path; energy values use exact model equations, not the plotted polygon area. A sealed receiver is a closed system only while no mass enters or leaves.

Method and symbols
Quantity	Value	Unit
m	Fixed gas mass	kg
R	Specific gas constant	kJ/(kg·K)
γ	cₚ/cᵥ, constant heat-capacity ratio	1
p₁, p₂	Absolute pressure	kPa
T₁, T₂	Absolute temperature	K
V₁, V₂	Total gas volume	m³
W, Q, ΔU	Boundary work, heat, internal-energy change	kJ

Constant-property process model
Constant-property process model	State	Energy terms
Isothermal	T₂ = T₁; V₂ = rV₁; p₂ = p₁/r	W = mRT₁ ln(r); ΔU = 0; Q = W
Isobaric	p₂ = p₁; T₂ = tT₁; V₂ = tV₁	W = mR(T₂ − T₁); Q = mcₚ(T₂ − T₁)
Isochoric	V₂ = V₁; T₂ = tT₁; p₂ = tp₁	W = 0; Q = ΔU = mcᵥ(T₂ − T₁)
Reversible adiabatic	V₂ = rV₁; T₂ = T₁r^(1−γ); p₂ = p₁r^(−γ)	Q = 0; W = −ΔU

A maritime teaching example

Synthetic marine-engineering laboratory compression in a sealed, ideal piston–cylinder: m = 0.8 kg, R = 0.287 kJ/(kg·K), γ = 1.4, p₁ = 600 kPa absolute and T₁ = 300 K. These are declared constant-property teaching values, not a real compressor or diesel-cycle prediction. Select reversible adiabatic and r = V₂/V₁ = 0.5.

cᵥ = 0.287/(1.4 − 1) = 0.7175 kJ/(kg·K).
V₁ = 0.8 × 0.287 × 300/600 = 0.1148 m³; V₂ = 0.0574 m³.
T₂ = 300 × 0.5^(−0.4) = 395.8523732 K.
p₂ = 600 × 0.5^(−1.4) = 1583.409493 kPa absolute.
ΔU = 0.8 × 0.7175 × (395.8523732 − 300) = +55.0192622 kJ.
Q = 0; W = −55.0192622 kJ. Compression requires work input.

Linear axes are fitted to the current path, and may start above zero. Plot area is not used for energy calculation.

A maritime teaching example
State	Absolute pressure p · kPa	Total volume V · m³	Temperature T · K
1	600	0.1148	300
2	1,583.409493	0.0574	395.8523732

Assumptions and limits

All four models use a fixed mass of ideal gas with constant R, cᵥ, cₚ and γ. Suitability must be established independently for the gas and temperature range.
Moving-boundary paths are quasi-static; the adiabatic path is additionally reversible. Adiabatic alone does not guarantee isentropic behavior.
Only absolute pressure and Kelvin temperature are accepted. kPa × m³ = kJ; entering Pa as kPa or a molar gas constant would give the wrong model.
This is one process, not a full engine cycle. No combustion, phase change, steam/refrigerant properties, leakage, heat-transfer rate, real diesel efficiency or machinery sizing is calculated.
Initial and final derived states must have 10⁻⁶ ≤ p ≤ 10¹⁰ kPa, 1 ≤ T ≤ 10⁵ K and 10⁻¹² ≤ V ≤ 10¹² m³. Magnitudes of energy totals are limited to 10¹⁵ kJ. These broad software bounds do not validate constant properties.

These are numerical scope limits, not permissible equipment operating limits or proof that an ideal-gas approximation is suitable.

Check your understanding

Can isothermal expansion have Q = 0 in this quasi-static ideal-gas model?

Only for no volume change. Otherwise ΔU = 0 and Q = W = mRT ln(V₂/V₁).

Does a rigid vessel have zero ΔU because W = 0?

No. Heating a sealed rigid vessel gives ΔU = Q while W = 0.

Why is a real air-compressor shaft-power calculation different?

A flowing compressor is an open control volume. Its shaft balance normally uses enthalpy, not closed-system boundary work.

Primary references

Curriculum link: bounded teaching applications of first-law and energy-balance topics in İTÜ GMI 213 Thermodynamics. Official course catalogue

Related context

The method explanation and worked example are on this page. The articles below provide additional context.

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