AkCalculators

Thermodynamic state variables — ideal gas relations and practical guidance

Thermodynamic analysis commonly begins by specifying a set of state variables that fully describe a system in equilibrium. For a simple compressible substance the primary intensive properties include pressure (P), temperature (T) and specific volume (v). For many gases at low-to-moderate pressures, the ideal gas relation provides a convenient and accurate first-order model:

Ideal gas law (specific form):
P · v = R · T
where P is absolute pressure (Pa), v is specific volume (m³/kg), T is absolute temperature (K), and R is the specific gas constant (J/kg·K = universal R / M).

What this calculator does

Given any two independent quantities among P, T and v, this tool uses the ideal gas relation to compute the third. It also computes derived intensive properties:

• Density ρ = 1 / v (kg/m³)
• Specific internal energy u ≈ cv · T (J/kg) for ideal gases (cv constant)
• Specific enthalpy h ≈ cp · T (J/kg)

Where cp and cv are specific heat capacities. For a chosen gas the calculator uses default cp values (approximate, temperature-independent) and computes cv = cp − R when necessary. The tool is intended for teaching, hand-calcs and preliminary design; use real-gas property tables or equations of state for high accuracy near phase changes or at high pressure.

Units and conversions

This page accepts several pressure unit inputs (Pa, kPa, bar, atm) and temperature as Kelvin or Celsius (converted to K internally). Specific volume is in m³/kg. Always use absolute temperature (K) for thermodynamic calculations.

When the ideal gas model is ok — and when it isn't

The ideal gas assumption is reasonable when the gas is dilute and not close to condensation or critical points (e.g., many air applications up to a few MPa and moderate temperatures). It fails for high-pressure compressors, near-critical steam/water, cryogenic and very dense gases. For steam/water, use steam tables; for high accuracy use real-gas models (e.g., Peng–Robinson) or software libraries.

Practical examples

Example 1 — Standard conditions: Given P = 101325 Pa and T = 300 K, for air (R=287.058) compute v = R·T/P = 287.058×300/101325 ≈ 0.850 m³/kg. Density ρ = 1/v ≈ 1.176 kg/m³. u ≈ cv·T (cv≈1005−287.058 ≈ 717.9) → u ≈ 717.9×300 ≈ 215,370 J/kg. h ≈ cp·T ≈ 1005×300 ≈ 301,500 J/kg.

Example 2 — Given P and v: If P = 2e5 Pa and v = 0.1 m³/kg, then T = P·v / R. For air, T = 2e5 × 0.1 / 287.058 ≈ 69.7 K (very low — indicates refrigeration/cryogenic conditions or nonphysical input for air; check units and assumptions).

Limitations & best practices

• Always confirm units and that temperatures are absolute (K).
• For substances like steam at saturation use specialized steam property tables.
• Specific heats (cp, cv) are temperature-dependent; constant cp/cv is an approximation adequate for many engineering estimates at moderate temperature ranges.
• If you need higher fidelity, pair this calculator with real-gas libraries or published property tables.

This calculator is designed to help students and engineers quickly check states, perform hand calculations, and produce printable results and CSV outputs for lab notebooks.