Lesson 4 - Gas Transfer

Unlike the unit of dissolved gas concentration which we have previously been dealing with, which is known as the molar fraction and comes from the field of chemistry and the gas laws, the preferred unit of CO₂ dissolved concentration within the brewing industry is the volume per volume (at STP). A typical dissolved CO₂ range for lagers and ales would be between 1.5 and 3.5 v/v (this range is merely to illustrate a point and is not definitive). So what does this mean, and how can we compare it to the units we’ve used before?

Well, at first the idea of a CO₂ concentration of 3.5 volumes/volume (at STP) might sound impossible? After all, it suggests that there are more volumes of gas dissolved in beer than the volume of beer! There is a saying in English about trying to put a ‘quart into a pint pot’ i.e. trying to put more into a glass (or pot) than it will hold. This seems relevant in our discussion. However, to avoid ‘Old English’, we will talk in litres to illustrate this in the example below.   

It is important to know that more CO₂ can dissolve in beer than the volume of beer, over the range of temperatures and pressure we are considering in the brewing process. We just need to know the densities of the gas and beer, and for that, we consider them both to be at STP, namely 273.15 K and 101,325 Pa a. Let’s consider this in an example:

Consider the situation of a litre of CO₂ dissolved in a half litre of beer. What would be the concentration of dissolved CO₂ expressed in volumes/volume (at STP)? (We may assume that the molar weight of beer is 18 kg/kmol and its density is 1,000 kg/m³.)

So, we take our volumes for this illustration as 1.0 L (CO₂ gas to dissolve) and 0.5 L (beer in glass).

Assuming the gas and beer are at STP, then the volumes/volume concentration would be:

To get the equivalent g/L and molar fraction, we need to head back to the ideal gas law and the concept of molar volume, where 1 kmol of a gas occupies 22.4 m³ at STP (review the earlier parts of the lesson if this isn’t clear). To determine the density of CO₂ gas (molecular weight 44) at STP conditions we use the following equation:

Hence, the mass of 1.0 L of CO₂ at STP would be:

Hence, the equivalent g/L would be 0.00196/0.5 L = 0.00392 kg/L= 3.92 g/L

So, the molar fraction would be given as the ratio of number of kmoles of CO₂ equivalent to a mass of 0.00392 g, to the number of kmoles of beer equivalent to 1 L (as the density is 1,000 kg/m³).

Hence, the molar fraction would be:

Before we move on to look at the technology used for gas transfer, let’s have a brief look at how gas transfers with time i.e. gas transfer dynamics.