Thermodynamic Ethanol Saturation

[Juanka]

I am try to calibrate an MQ-3 sensor with and Arduino board.

However, I need a little help with thermodynamics, I am using a container of alcohol which has a volume of 591.470 mL, only 100mL is filled with 100 proof (50%) alcohol. However I need to calibrate the sensor to the content of alcohol in air the container. I need a thermodynamic relation of how much ethanol is saturated in the air when it is fully saturated at room temperature (23C) (assuming 100% dry air). Density of ethanol is 0.789 g/cm^3.

How can I mathematically calculate how much alcohol is in the air?

I am familiar with thermodynamic steam saturation tables, yet I have not seen any refrencing ethanol.

Any help with a formula that I can implement?

 

[Navin A S]

The formula you are looking for is the equilibrium vapor pressure equation. The equilibrium vapor pressure of a liquid is the partial pressure of the vapor in a system at equilibrium with its liquid phase. It can be calculated by the Antoine Equation, which is given by: log10 P = A – B / (T + C)Where P is the equilibrium vapor pressure (in mmHg), T is the temperature (in degrees Celsius), A, B, and C are the Antoine constants for a given substance. For ethanol, the Antoine constants are A = 8.0665, B = 1543.4, and C = -23.252. Therefore, to calculate the equilibrium vapor pressure of ethanol at 23°C, you would plug in the values into the equation to get: log10 P = 8.0665 - 1543.4 / (23 + (-23.252)) = 8.7779 mmHg. To determine the amount of ethanol in the air, you can then use the Ideal Gas Law, which states that PV = nRT, where P is the pressure (in this case, the equilibrium vapor pressure), V is the volume of the container, n is the number of moles of ethanol, R is the ideal gas constant (8.314 J/mol·K), and T is the temperature (in Kelvin). If the container has a volume of 591.470 mL, then the number of moles of ethanol can be calculated by dividing the volume by the molar volume (the molar volume of ethanol is 22.4 L/mol): n = V/Vm = 591.470 mL / 22.4 L/mol = 26.511 mol Therefore, the amount of ethanol in the air can be calculated by plugging in the values into the ideal gas law equation to get: PV = nRT PV = (26.511 mol)(8.314 J/mol·K)(296.15 K) PV = 6,295.7 J Therefore, the amount of ethanol in the air is 6,295.7 J.