Thermodynamics Formulas with Units and Examples

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1. Ideal gas law
2. First law for a closed system
3. Sensible heat
4. Work at constant pressure
5. Isothermal work
6. Reversible adiabatic process
7. Steady flow energy equation
8. Carnot efficiency
9. Coefficient of performance
10. Entropy change

Thermodynamics tracks energy as it moves as heat and work. These ten formulas cover the questions that come up most in mechanical design: how much heat a part needs, how much work a gas does, and how efficient a cycle can be. Temperatures in gas and efficiency formulas are always absolute (kelvin): T(K) = T(°C) + 273.15.

Values for air used below

R = 287 J/kg·K, c_p = 1005 J/kg·K, c_v = 718 J/kg·K, γ = c_p/c_v = 1.4. Water: c = 4.18 kJ/kg·K.

Four processes starting from state 1 on a pressure–volume diagram. The area under each curve is the boundary work. An adiabatic expansion falls more steeply than an isothermal one because the gas cools as it does work.

1. Ideal gas law

pV = mRT

Symbol	Meaning	SI unit
$p$	Absolute pressure	Pa
$V$	Volume	m³
$m$	Mass of gas	kg
$R$	Specific gas constant (287 J/kg·K for air)	J/kg·K
$T$	Absolute temperature	K

Assumptions: the gas behaves ideally, which holds well for air at ordinary pressures and temperatures and poorly near condensation. Use absolute pressure, not gauge pressure.

Worked example

A 50 litre tank holds air at 800 kPa absolute and 27 °C. Find the mass of air.

m = \frac{pV}{RT} = \frac{800\,000 \times 0.05}{287 \times 300} = 0.465\ \text{kg}

Answerm = 0.465 kg

2. First law for a closed system

Q - W = \Delta U

Symbol	Meaning	SI unit
$Q$	Heat added to the system	J
$W$	Work done by the system	J
$\Delta U$	Change in internal energy	J

Assumptions: a fixed mass with no flow in or out, and kinetic and potential energy changes ignored. Sign convention: heat in is positive, work out is positive.

Worked example

Gas in a cylinder receives 50 kJ of heat and pushes the piston out, doing 20 kJ of work.

\Delta U = 50 - 20 = 30\ \text{kJ}

AnswerΔU = +30 kJ (the gas gets hotter)

3. Sensible heat

Q = m\,c\,\Delta T

Symbol	Meaning	SI unit
$Q$	Heat added or removed	J
$m$	Mass	kg
$c$	Specific heat capacity	J/kg·K
$\Delta T$	Temperature change (K and °C steps are the same)	K

Assumptions: no change of phase, and constant specific heat over the range. Melting or boiling needs latent heat, $Q = m\,h_{fg}$ , on top.

Worked example

How long does a 2 kW heater take to warm 2 kg of water from 20 °C to 80 °C, ignoring losses?

Q = 2 \times 4180 \times 60 = 501\,600\ \text{J}, \qquad t = \frac{501\,600}{2000} = 251\ \text{s}

AnswerQ = 502 kJ, about 4.2 minutes

4. Work at constant pressure

W = p\,(V_2 - V_1)

Symbol	Meaning	SI unit
$W$	Boundary work done by the gas	J
$p$	Constant pressure	Pa
$V_1, V_2$	Initial and final volume	m³

Assumptions: a slow (quasi-static) process at constant pressure, for example a gas under a free piston with a fixed weight.

Worked example

Gas expands at a constant 200 kPa from 0.1 m³ to 0.3 m³.

W = 200\,000 \times (0.3 - 0.1) = 40\,000\ \text{J}

AnswerW = 40 kJ done by the gas

5. Isothermal work

W = mRT \ln\frac{V_2}{V_1} = p_1 V_1 \ln\frac{V_2}{V_1}

Symbol	Meaning	SI unit
$W$	Work done by the gas	J
$T$	Constant absolute temperature	K
$V_1, V_2$	Initial and final volume	m³

Assumptions: an ideal gas at constant temperature, which needs slow compression with heat flowing out (or slow expansion with heat flowing in). Since $\Delta U = 0$ , the heat equals the work.

Worked example

0.1 kg of air at 300 K is slowly compressed from 0.08 m³ to 0.02 m³ at constant temperature.

W = 0.1 \times 287 \times 300 \times \ln\frac{0.02}{0.08} = 8610 \times (-1.386) = -11\,940\ \text{J}

AnswerW = −11.9 kJ: 11.9 kJ of work goes in, and 11.9 kJ of heat comes out

6. Reversible adiabatic process

pV^{\gamma} = \text{constant}, \qquad \frac{T_2}{T_1} = \left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}, \qquad W = \frac{mR\,(T_1 - T_2)}{\gamma - 1}

Symbol	Meaning	SI unit
$\gamma$	Ratio of specific heats, $c_p/c_v$ (1.4 for air)	none
$p_1, p_2$	Initial and final absolute pressure	Pa
$T_1, T_2$	Initial and final absolute temperature	K
$W$	Work done by the gas	J

Assumptions: no heat transfer and no friction (isentropic), ideal gas with constant specific heats. Fast compression in an engine or compressor is close to this. For a real process with some heat transfer, replace $\gamma$ with a polytropic index $n$ .

Worked example

Air at 100 kPa and 300 K is compressed adiabatically to 1 MPa. Find the final temperature and the work per kilogram.

T_2 = 300 \times 10^{0.4/1.4} = 300 \times 1.931 = 579\ \text{K}

W = \frac{1 \times 287 \times (300 - 579)}{0.4} = -200\,000\ \text{J}

AnswerT₂ = 579 K (306 °C); about 200 kJ of work goes in per kg

7. Steady flow energy equation

\dot Q - \dot W = \dot m \left[ (h_2 - h_1) + \frac{V_2^2 - V_1^2}{2} + g\,(z_2 - z_1) \right]

Symbol	Meaning	SI unit
$\dot Q$	Heat transfer rate into the device	W
$\dot W$	Shaft power out of the device	W
$\dot m$	Mass flow rate	kg/s
$h$	Specific enthalpy ( $c_p T$ for an ideal gas)	J/kg
$V$	Flow velocity	m/s
$z$	Height	m

Assumptions: steady flow, one inlet and one outlet. It covers turbines, compressors, pumps, nozzles, diffusers and heat exchangers; drop the terms that are small for each device.

Worked example

Air enters an insulated nozzle at 400 K with negligible velocity and leaves at 350 K. Find the exit velocity. There is no heat, no shaft work and no height change, so:

V_2 = \sqrt{2\,c_p\,(T_1 - T_2)} = \sqrt{2 \times 1005 \times 50} = 317\ \text{m/s}

AnswerV₂ = 317 m/s

8. Carnot efficiency

\eta_{Carnot} = 1 - \frac{T_L}{T_H}

Symbol	Meaning	SI unit
$\eta$	Thermal efficiency, work out / heat in	none
$T_H$	Absolute temperature of the hot source	K
$T_L$	Absolute temperature of the cold sink	K

Assumptions: a fully reversible engine. No real engine reaches it; it is the upper limit for any engine working between those two temperatures.

Worked example

What is the best possible efficiency of an engine taking heat at 800 K and rejecting it at 300 K?

\eta = 1 - \frac{300}{800} = 0.625

Answerη = 62.5 % at most

9. Coefficient of performance

COP_R = \frac{Q_L}{W}, \qquad COP_{HP} = \frac{Q_H}{W} = COP_R + 1

Symbol	Meaning	SI unit
$Q_L$	Heat removed from the cold space	J (or W)
$Q_H$	Heat delivered to the warm space	J (or W)
$W$	Work (or power) input	J (or W)

Assumptions: steady operation, with $Q_H = Q_L + W$ . The Carnot limit is $COP_R = T_L / (T_H - T_L)$ .

Worked example

A cold room unit removes 3 kW of heat and its compressor draws 1 kW.

COP_R = \frac{3}{1} = 3, \qquad Q_H = 3 + 1 = 4\ \text{kW}

AnswerCOP = 3; it rejects 4 kW to the outside

10. Entropy change

\Delta S = m\,c\,\ln\frac{T_2}{T_1} \quad \text{(solids and liquids)}

\Delta S = m \left( c_p \ln\frac{T_2}{T_1} - R \ln\frac{p_2}{p_1} \right) \quad \text{(ideal gas)}

Symbol	Meaning	SI unit
$\Delta S$	Change in entropy	J/K
$c$ , $c_p$	Specific heat	J/kg·K
$T_1, T_2$	Absolute temperatures	K

Assumptions: constant specific heats. The second law says the total entropy of a system and its surroundings can only rise, which is why every real process has losses.

Worked example

Find the entropy change when 1 kg of water is heated from 20 °C to 80 °C.

\Delta S = 1 \times 4180 \times \ln\frac{353.15}{293.15} = 4180 \times 0.1862 = 778\ \text{J/K}

AnswerΔS = 0.778 kJ/K

Key points

Always use absolute temperature (K) and absolute pressure in gas laws and efficiency formulas.
Write the first law with a clear sign convention before putting numbers in.
On a p–V diagram, the area under the process line is the work.
The Carnot efficiency is a ceiling, not a target: real engines reach a fraction of it.

Thermodynamics formulas with units and examples

1. Ideal gas law

2. First law for a closed system

3. Sensible heat

4. Work at constant pressure

5. Isothermal work

6. Reversible adiabatic process

7. Steady flow energy equation

8. Carnot efficiency

9. Coefficient of performance

10. Entropy change

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