Specific gas constant for water vapour

R_{\mathrm{v}} = 461.523\ \text{J/(kg}{\cdot}\text{K)}

Value	461.523 J/(kg·K) Use in the calculator →
Status	Reference value; the source states no uncertainty
Source	Derived from CODATA 2022 R and the IAPWS molar mass of water
Categories	Thermodynamic Engineering & Trade

Specific gas constant for water vapour in every specific heat capacity unit

Unit	Value
joule per kilogram-kelvin	461.523 J/(kg·K)
kilojoule per kilogram-kelvin	0.461523 kJ/(kg·K)
BTU per pound-Fahrenheit	0.11023287 BTU/(lb·°F)
joule per kilogram-Celsius	461.523 J/(kg·°C)
foot pound-force per pound-Rankine	85.779835 ft·lbf/(lb·°R)
kilojoule per kilogram-Celsius	0.461523 kJ/(kg·°C)
joule per gram-kelvin	0.461523 J/(g·K)
calorie per gram-Celsius	0.11030664 cal/(g·°C)
kilocalorie per kilogram-Celsius	0.11030664 kcal/(kg·°C)
BTU per pound-Rankine	0.11023287 BTU/(lb·°R)

Learning zone

Water vapour's gas constant is 1.608 times air's, purely because a water molecule is lighter than the average air molecule. That ratio is the whole of psychrometrics: the humidity ratio W = 0.621945 p_w/(p − p_w) is just R_air/R_v in disguise, and it is why moist air is less dense than dry air at the same temperature and pressure, a fact that surprises people every time.

Below about 10 kPa of partial pressure, water vapour follows the ideal gas law closely enough for HVAC work, which is why this constant carries so much load in cooling-coil and cooling-tower calculations. It is also the R_v in the Clausius-Clapeyron equation used to model saturation vapour pressure in meteorology.

Used by 4 solvers

Air Latent Heat (0.68 Rule)

\dot{Q}_l = \rho_a \dot{V} h_{fg} \, \Delta W

Air Total Heat (4.5 Rule)

\dot{Q}_t = \rho_a \dot{V} \, \Delta h

Cooling Tower Heat Rejection

Q = 500 \, R \, \Delta T

Ideal Gas Law

P V = n R T

Revised 16 October 2026

Cite this page

formula.expert. (2026, October 16). Specific gas constant for water vapour. https://formula.expert/constants/specific-gas-constant-water-vapour