How to calculate the specific heat capacity of gases

Question

Could someone explain why the Specific Heat Capacity (SHC) of dry air is 1.005kJ/kg.K whereas the accepted SHC for ventilation air is quoted 1.300kJ/kg.K? Assuming the air for the ventilation calc’s is humid say 60RH @20°C it is estimated that there would be about 8.8g/kg of moisture in the air. [Using the psychrometric chart from the British Standards Institute] The SHC of water is 4.200kJ/kg.K = 4.2J/g The SHC of water vapour is 1.86kJ/kg.K = 1.86J/g If there are about 9g of moisture (water vapour) in the air; 9 x 1.86 = 16.74J of energy will be required. The water molecules in vapour form will displace the dry air molecules therefore the dry air SHC will be less than 1.005kJ/kg.K. If dry air is about 1kg/m3 and the water vapour has displaced some of the dry air molecules then the SHC of the humid air (dry air + water vapour) is going to be less than 1.300kJ/kg.K. To convert H2O from liquid to gas (water to water vapour) requires 2.3kJ/g of latent heat which is more energy than is required to raise 1kg of water 1K.

Also is there any relationship between atomic mass and SHC? As air is mainly Nitrogen (14 atomic number) @ 78%, and Oxygen (16) 21% [+ 1% of various other gases]. Both the main gases are diatomic: Dry Air = (0.78 x (2x14)) +( 0.21 x (2x16)) = 28.56 + Argon and CO2 = 28.946 atomic mass per mole. Water = Hydrogen (2x1) + Oxygen (16) = 18.00 or there about due to impurities per mole. The argument that humid air is less dense than dry air can be deduced from the above.

If a mole of a gas at standard conditions has a volume of 22.4ltrs and 1000 ltrs = 1m3 therefore 1000/22.4 = 44.64. (moles in 1m3) 44.64 x 28.95 = 1292 which is close to the 1.300kJ/kg.K [ventilation air however it would be dry and not 60RH]. Deduct 1 mole from dry air (1000 – 22.4 = 978ltrs) 978/22.4 = 43.66 moles. (43.66 x 28.95) + (1 x 18.00) = 1.281kJ/kg.K. The SHC of humid air approx’ 1m3? Virtually the same as dry air.

Or is it coincidental? Any help would be appreciated - thanks

Answer 1

The heat capacity of moist air follows the simplest prescription you can think of, $$ c_s = 1.005+1.82 H\,\frac{\rm kJ}{\rm kg\cdot {}^\circ C}$$ where 1.005 is the figure for dry air and $H$ is the number of kilograms of water vapor per kilogram of air.

At 20 Celsius degrees, the saturated vapor pressure for water vapor is 2.3 kPa. Sixty percent of it is 1,380 Pa, about 0.014 of the total pressure. This partial pressure doesn't represent the percentage of the mass. Dry air has 29 g/mole or so, water has 16+1+1=18 g/mole, so the mass ratio is 18/29 times 0.014 = 0.0087. Even when multiplied by 1.82, we get 0.016 or so: we may raise 1.005 to 1.02 at most. The humidity doesn't affect the heat capacity too much.

The actual reason for your wildly different number 1,300 is different and simple: you have confused kilograms and cubic meters.

Just to be sure, the latent heat plays no role here because no phases are being changed (condensed/vaporized) during the heating of a decently moist air. The heat capacity of liquid water plays no role, either: no water is actually liquid (in big chunks) in this problem.

The value 1,300 of something is actually the volumetric specific heat capacity of air, so the right units are $\rm kJ/ m^3\cdot {}^\circ C$. The word "volumetric" means that it's computed per unit volume. Divide it by 1.275 kilograms per cubic meters, the density of air, and you will get 1.02 kJ/kg per degree, in agreement with the result two paragraphs above.

The heat capacity of individual gases is simple to compute when one considers the molar heat capacity or, equivalently (up to multiplication by Avogadro's constant), heat capacity of a single molecule. A spherically symmetric atom would have $3k/2$, a diatomic one has $5k/2$, more general molecules that remember the precise orientation of everything have $3k$ etc. Multiply it by $N_A$ to get the molar heat capacities $3R/2$ etc. This may also be divided by the molar mass to get the specific heat capacity etc. When you consider the appropriate weighted average of the molecules/atoms that make up air, you may calculate the heat capacity of the air, too. Thermodynamics of diluted gases is simple because the interactions are so weak so we're really summing simple properties of individual molecules only.