Ideal Gases with Variable Specific Heats

For room-temperature air, cp ≈ 1.005 kJ/kg·K and cv ≈ 0.718 kJ/kg·K are fine. Heat that same air to 1500 K and cp climbs past 1.2. The constant-cp shortcut quietly drifts, and the error compounds across a cycle.

The fix: tabulate u(T) and h(T)

For an ideal gas, u and h depend on temperature only. So textbooks tabulate them directly against temperature:

T (K)	u (kJ/kg)	h (kJ/kg)
300	210.49	300.19
350	250.02	350.49
400	286.16	400.98
500	359.49	503.02
600	435.11	607.02

Read u and h straight off the table at each state’s temperature — no cp needed. The difference h2 − h1 is the true enthalpy change, variable-cp included.

Worked example

Air is heated from 300 K to 500 K at constant pressure. Heat added per kg:

q = h2 − h1 = 503.02 − 300.19 = 202.83 kJ/kg

The constant-cp estimate (1.005·200 = 201.0) is close here. Push to 300→1000 K and the gap widens: the table gives h(1000) − h(300) = 1046.04 − 300.19 = 745.85, while cp·ΔT = 1.005·700 = 703.5 — a 6% error.

The gas constant still closes P·v = R·T

Variable specific heats change u and h, not the equation of state. Air’s R = 0.287 kJ/kg·K still gives v = R·T/P. At 100 kPa and 300 K:

v = 0.287 · 300 / 100 = 0.861 m³/kg

The property lookup calculator handles this: pick “Ideal gas” and the gas (air, CO₂, CO, O₂, NO₂), enter any two of P, T, v, and it closes the state, pulling h, u, s from the variable-cp table.

When constant cp is fine

Short temperature spans near 300 K, cold-air-standard cycles, and quick estimates. When you’re spanning hundreds of degrees, use the tables — your grader’s solution manual does.