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Thermodynamics

Ideal-gas internal energy and enthalpy change

Change of internal energy of an ideal gas from its temperature change.

Ref: THE-01
Rev: v1
Field: Thermodynamics
Result: Δu [kJ/kg]

h = u + p v, c_{p} - c_{v} = R, Δ u = c_{v} Δ T

Lower-case properties are per unit mass; v is specific volume. The enthalpy definition is general. The temperature relations Δh = c_pΔT and Δu = c_vΔT assume an ideal gas with constant heat capacities. Here c_v = c_p − R.

Heating a gas raises its internal energy by cv times the temperature rise; the enthalpy rises by the larger cp times the rise.

Calculator

Solve for

ΔuChange in internal energyComputed

kJ/kg

c_pSpecific heat (const. pressure)

J/(kg·K)

5005,000

RSpecific gas constant

J/(kg·K)

100700

ΔTTemperature change

−2001,000

Change in enthalpy Δh100.5 kJ/kg
Heat-capacity ratio γ = cp/cv1.4

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Change in internal energy (kJ/kg) current value 71.8 kJ/kg

ΔTTemperature changeKey driver

−144718

c_pSpecific heat (const. pressure)

21.3471

RSpecific gas constant

30.590.5

Enthalpy changes exceed internal-energy changes by the flow work R ΔT.
For air γ is about 1.4.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (kJ/kg)	Local step	Change in result
ΔT Temperature change	−200 to 1,000 K	−143.6	718	861.6	+10%	+10 %
c_p Specific heat (const. pressure)	500 to 5,000 J/(kg·K)	21.3	471.3	450	+10%	+14 %
R Specific gas constant	100 to 700 J/(kg·K)	30.5	90.5	60	+10%	−4 %

Sweep of Temperature change

Temperature change (K)	Change in internal energy (kJ/kg)
−200	−143.6
−50	−35.9
100	71.8
250	179.5
400	287.2
550	394.9
700	502.6
850	610.3
1,000	718

Sweep one variable

Vary

Change in internal energy (kJ/kg)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Δu

Change in internal energy

Result · kJ/kg

What the equation gives you. Everything else feeds it.

c_p

Specific heat (const. pressure)

J/(kg·K) · typical 500 to 5,000

Specific heat at constant pressure (air 1005). Raises Δu in proportion to c_v = c_p − R.

Specific gas constant

J/(kg·K) · typical 100 to 700

Specific gas constant (air 287). Lowers c_v and so Δu.

ΔT

Temperature change

K · typical −200 to 1,000

Temperature change of the gas. Directly proportional.

Worked example with the values above

c_v = c_p − R = 1,005 − 287 = 718 J/(kg·K).
Multiply by ΔT = 100 K: 71,800 J/kg.
Δu = 71.8 kJ/kg (and Δh = 100.5 kJ/kg).

Assumptions and limits

Ideal gas with constant heat capacities.
Per unit mass.

Common mistakes

Using c_p for a closed constant-volume process.
Using °C differences incorrectly (they equal K differences).

Statistics and uncertainty

Heat capacities of real gases vary by 5–10% over wide temperature ranges, so constant-c_p results are approximate.

Where it is used

Compressor and turbine energy balances.
Heating and cooling of gases in vessels.

Maths you need

Specific heat: J/(kg K)
Ideal gas law: pv = RT

Related formulas

Practise this formula