## How to calculate Mixing Ratio and Specific Humidity

Posted: June 1, 2020 in physical chemistry, physics, thermodynamics
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If your weather station displays barometric pressure, temperature and relative humidity like the one pictured above, you can calculate the amount of water vapor in the air expressed either as grams of water vapor per kilogram of dry air (known as Mixing Ratio) or as grams of water vapor per kilogram of vapor-containing air (known as Specific Humidity). The two measures are very similar for cooler air; differences only become apparent for warmer air.

Formulas for calculating Mixing Ratio and Specific Humidity

In the formulas below, barometric pressure P is expressed in hectopascals (hPa), temperature T is expressed in degrees Celsius, relative humidity rh is expressed in %, and e is the Euler number 2.71828 [raised to the power of the contents of the square brackets]. In developing these formulas, the following textbook was consulted: Atmospheric Thermodynamics by Grant W. Petty, Sundog Publishing, Madison Wisconsin. ISBN-10: 0-9729033-2-1

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Derivation of Mixing Ratio formula

Mixing Ratio is defined as the mass of water vapor in grams mixed into a kilogram of dry air.
Both water vapor and dry air exert pressures, so by the law of partial pressures
Barometric pressure (P) = Pressure of dry air (Pdry) + Pressure of water vapor (Pvap)
and therefore
Pdry = P – Pvap (1)
The ratio Pvap/Pdry also expresses the molar ratio of water vapor to dry air since by the ideal gas law PV = nRT
Pvap/Pdry = nvapRTV^-1 / ndryRTV^-1 = moles vapor / moles dry air (RTV^-1 cancels)
The molar mass of water is 0.0180153 kg/mol (ref: Petty, see above)
The molar mass of dry air is 0.0289655 kg/mol (ref: Petty, see above)
Therefore
Pvap × 0.0180153 / Pdry x 0.0289655 = kg vapor / kg dry air
Pvap × 0.622 / Pdry = kg vapor / kg dry air
Substituting Pdry by equation 1
Pvap × 0.622 / (P – Pvap) = kg vapor / kg dry air
Therefore
Mixing Ratio = Pvap × 622 / (P – Pvap) units: g vapor / kg dry air (2)
Pvap can be calculated from temperature T and relativity humidity rh using the formula originated and published by CarnotCycle blog
Pvap = 6.112 × e^[(17.67 × T)/(T+243.5)] × (rh/100)
where T is expressed in degrees Celsius and rh is expressed in %.
Substituting the formula for Pvap into equation 2 completes the derivation.

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Derivation of Specific Humidity formula

Specific Humidity is defined as the mass of water vapor in grams per kilogram of moist air.
By the law of partial pressures
Barometric pressure (Pmoist) = Pressure of dry air (Pdry) + Pressure of water vapor (Pvap)
The ratio Pvap/(Pdry + Pvap) expresses the molar ratio of water vapor to moist air since by the ideal gas law PV = nRT
Pvap/(Pdry + Pvap) = nvapRTV^-1 / (ndryRTV^-1 + nvapRTV^-1) = moles vapor / moles moist air
Pvap/(Pdry + Pvap) = nvap / (ndry + nvap) = moles vapor / moles moist air (RTV^-1 cancels)
Let M1 = molar mass of water vapor = 0.0180153 kg
Let M2 = molar mass of dry air = 0.0289655 kg
M1Pvap / (M2Pdry + M1Pvap) = kg vapor / kg moist air (3)
Substitute Pdry = Pmoist – Pvap in equation 3
M1Pvap / (M2Pmoist – M2Pvap + M1Pvap) = kg vapor / kg moist air
M1Pvap / (M2Pmoist – (M2 – M1)Pvap) = kg vapor / kg moist air
Divide by M2
(M1/M2)Pvap / (Pmoist – (1 – M1/M2)Pvap) = kg vapor / kg moist air
0.622 × Pvap / (Pmoist – 0.378 × Pvap) = kg vapor / kg moist air
Therefore
Specific Humidity = Pvap × 622/ (Pmoist – 0.378 × Pvap) units: g vapor / kg moist air (4)
Pvap can be calculated from temperature T and relativity humidity rh using the formula originated and published by CarnotCycle blog
Pvap = 6.112 × e^[(17.67 × T)/(T+243.5)] × (rh/100)
where T is expressed in degrees Celsius and rh is expressed in %.
Substituting the formula for Pvap into equation 4 completes the derivation.

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P Mander  June 2020, derivations added February 2023

## Compute dew point temperature from RH&T

Posted: August 1, 2017 in physical chemistry, physics, thermodynamics
Tags: , , , , , , , , , , , , Relative humidity (RH) and temperature (T) data from an RH&T sensor like the DHT22 can be used to compute not only absolute humidity AH but also dew point temperature TD There has been a fair amount of interest in my formula which computes absolute humidity (AH) from measured relative humidity (RH) and temperature (T) since it adds value to the output of RH&T sensors. To further extend this value, I have developed another formula which computes dew point temperature TD from measured RH and T.

Formula for computing dew point temperature TD

In this formula (P Mander 2017) the measured temperature T and the computed dew point temperature TD are expressed in degrees Celsius, and the measured relative humidity RH is expressed in %

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Strategy for computing TD from RH and T

1. The dew point temperature TD is defined in the following relation where RH is expressed in % 2. To obtain values for Psat, we can use the Bolton formula[REF, eq.10] which generates saturation vapor pressure Psat (hectopascals) as a function of temperature T (Celsius)  These formulas are stated to be accurate to within 0.1% over the temperature range –30°C to +35°C

3. Substituting in the first equation yields Taking natural logarithms Rearranging Separating TD terms on one side yields – – – –

P Mander August 2017 If the man who almost single-handedly invented chemical thermodynamics – the American mathematical physicist Josiah Willard Gibbs – had owned an automobile, he would have had no trouble figuring out the action of antifreeze.

“The problem reduces to consideration of a binary solution in equilibrium with solid solvent,” I can hear old Josiah saying. “Such a thermodynamic system has two degrees of freedom, so at constant pressure there must be a relation between temperature and composition.”

And indeed there is. The relation corresponds to the observed depression of the freezing point of a solvent by a solute. What’s more, its exact form confirms how antifreeze really works.

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Computing chemical potential

We have Josiah Willard Gibbs to thank for introducing the concept of chemical potential (μ) as a sort of generalized force driving the flow of chemical components between coexistent phases.

When the phases are in equilibrium at constant temperature and pressure, the chemical potential of any component has the same value in each phase The key point to note here is that μi is the chemical potential of component i in an arbitrary state, i.e. in a mixture of components. In order to compute this potential we need to know two things: the chemical potential of the pure substance μi0 at a pressure p (such as that of the atmosphere), and the mole fraction (xi) of the component in the mixture. Assuming an ideal solution, use can then be made of the textbook formula …(1)

With pressure and temperature fixed, this equation has a single variable (xi), from which we can draw the conclusion that the variation in chemical potential of a component in an ideal solution is determined solely by its own mole fraction.

The significance of this fact can be appreciated by considering the following diagrams Here is water in equilibrium with ice at 273K. The chemical potentials of the solid and liquid phases are equal; there is no net driving force in either direction. Now consider the effect of adding an antifreeze agent to the liquid phase Assuming the temperature held constant at 273K, the addition of antifreeze reduces the mole fraction of water, lowering its chemical potential in accordance with equation 1. The coexistent solid phase now has a higher potential, providing the driving force to transform ice into water. Since the temperature is held constant, this equates to the lowering of the freezing point of water in the mixture.

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To obtain a formula for the freezing point of water in a solution containing antifreeze, we start with the equilibrium relation where the zero superscript indicates a standard potential, i.e. that the solid phase consists of pure ice whose mole fraction x is unity. Substituting the left hand side with we obtain which after differentiation with respect to temperature at constant pressure and subsequent integration yields the formula for the freezing point of water in a solution containing antifreeze at 1 atmosphere pressure: The terms on the right are the molar enthalpy of fusion of water (ΔHf0), the freezing point of pure water (Tf0), the gas constant R and the mole fraction of water (xH2O) in the solution containing antifreeze.

The latter is the only variable, confirming that the freezing point of water in a solution containing antifreeze is determined solely by the mole fraction of water in the mixture – in other words the extent to which the water is diluted by the antifreeze agent.

This is how antifreeze works. There is nothing active about its action. It exerts its effect passively by being miscible and thereby reducing the mole fraction of water in the liquid mixture. There’s really nothing more to it than that.

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Using the formula Values for constants

Enthalpy of fusion of water ΔHf0 = 6.02 kJmol-1
Freezing point of pure water Tf0 = 273.15 K
Gas constant R = 0.008314 kJmol-1K-1

Example

651 grams of the antifreeze agent ethylene glycol (molecular weight 62.07) are added to 1.5 kg of water (molecular weight 18.02). What is the freezing point of water in this solution?

Strategy

1. Calculate the mole fraction of water in the solution Number of moles of water = 1500/18.02 = 83.2
Number of moles of ethylene glycol = 651/62.07 = 10.5
Mole fraction of water = 83.2/(83.2 + 10.5) = 0.89

2. Calculate the freezing point of water in the solution The solution will give antifreeze protection down to 261.65K or –11.5°C

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P Mander March 2015