Library / Thermodynamics / THE-04

Thermodynamics

Ideal-gas entropy change

Entropy change of an ideal gas between two states.

Ref: THE-04
Rev: v1
Field: Thermodynamics
Result: Δs [kJ/(kg·K)]

s_{2} - s_{1} = c_{p} \ln \frac{T_{2}}{T_{1}} - R \ln \frac{p_{2}}{p_{1}}

Constant c_p; absolute temperature and pressure. Setting the entropy change to zero gives T₂/T₁ = (p₂/p₁)^(γ−1)/γ with γ = c_p/c_v.

Entropy goes up when a gas is heated and down when it is compressed; the formula balances those two effects.

Calculator

Solve for

ΔsEntropy changeComputed

kJ/(kg·K)

c_pSpecific heat

kJ/(kg·K)

0.55

RGas constant

kJ/(kg·K)

0.11

T₁Initial temperature

150800

T₂Final temperature

1501,500

p₁Initial pressure

kPa

101,000

p₂Final pressure

kPa

105,000

Isentropic outlet temperature T₂s445.7 K
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.

Entropy change (kJ/(kg·K)) current value 0.1155 kJ/(kg·K)

T₂Final temperatureKey driver

−1.091.22

c_pSpecific heat

−0.1422.16

p₂Final pressure

−0.6091.17

T₁Initial temperature

−0.870.812

p₁Initial pressure

−0.5450.776

RGas constant

−0.8730.375

Compressing air from 100 to 400 kPa and heating it from 300 K to 500 K ends hotter than the isentropic outlet of about 446 K, so the entropy rises.
A positive Δs for an adiabatic process indicates irreversibility; a negative one is impossible.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (kJ/(kg·K))	Local step	Change in result
T₂ Final temperature	150 to 1,500 K	−1.094	1.22	2.314	+10%	+82.9 %
c_p Specific heat	0.5 to 5 kJ/(kg·K)	−0.1425	2.156	2.299	+10%	+44.4 %
p₂ Final pressure	10 to 5,000 kPa	−0.6094	1.174	1.784	+10%	−23.7 %
T₁ Initial temperature	150 to 800 K	−0.8702	0.8121	1.682	+10%	−82.9 %
p₁ Initial pressure	10 to 1,000 kPa	−0.5453	0.7764	1.322	+10%	+23.7 %
R Gas constant	0.1 to 1 kJ/(kg·K)	−0.8729	0.3748	1.248	+10%	−34.4 %

Sweep of Final temperature

Final temperature (K)	Entropy change (kJ/(kg·K))
150	−1.094
318.8	−0.3369
487.5	0.09007
656.3	0.3888
825	0.6188
993.8	0.8058
1,163	0.9635
1,331	1.1
1,500	1.22

Sweep one variable

Vary

Entropy change (kJ/(kg·K))Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Δs

Entropy change

Result · kJ/(kg·K)

What the equation gives you. Everything else feeds it.

c_p

Specific heat

kJ/(kg·K) · typical 0.5 to 5

Specific heat at constant pressure (air 1.005). Scales the temperature term.

Gas constant

kJ/(kg·K) · typical 0.1 to 1

Specific gas constant (air 0.287). Scales the pressure term.

T₁

Initial temperature

K · typical 150 to 800

Initial absolute temperature. A hotter start gives a smaller temperature ratio, which lowers Δs.

T₂

Final temperature

K · typical 150 to 1,500

Final absolute temperature. Raises Δs.

p₁

Initial pressure

kPa · typical 10 to 1,000

Initial absolute pressure. Raises Δs.

p₂

Final pressure

kPa · typical 10 to 5,000

Final absolute pressure. Lowers Δs.

Worked example with the values above

Temperature term: c_p ln(T₂/T₁) = 1.005 × 0.5108 = 0.5134.
Pressure term: R ln(p₂/p₁) = 0.287 × 1.386 = 0.3979.
Δs = 0.1155 kJ/(kg·K).

Assumptions and limits

Ideal gas with constant c_p.
Absolute temperatures and pressures.

Common mistakes

Using °C or gauge pressure.
Forgetting the second term.

Statistics and uncertainty

Δs is a small difference of two larger terms, so 1% errors in temperature can change Δs by tens of per cent near the isentropic point.

Where it is used

Compressor and turbine irreversibility.
Second-law checks on cycles.

Maths you need

Natural logarithm: ln
Ideal gas: pv = RT

Related formulas

Practise this formula