Suggested Response - Q1

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a)    Describe the uses and sources of nitrogen in a brewery.

b)    A cylinder of compressed nitrogen (N2) has a volume of 60 L. If the cylinder gas pressure and temperature are 100 bar g and 20°C respectively, what mass of nitrogen is present in the cylinder? (Atmospheric pressure can be assumed to be 100 kPa.)

c)    If nitrogen is to be used to nitrogenate ales and stouts, determine a conversion factor to convert volumes of nitrogen per volume of beer (at STP) to g/L.

Response:

a)

Uses include: nitrogenation of beer; purging of air/oxygen from vessels and containers, although this is less common than the use of CO2; and draught dispense.

Sources are:

  • Purchased in bulk, having been produced by air liquefaction (essentially cooling air at elevated pressures and low temperatures to selectively separate nitrogen from air).

  • Produced in smaller quantities on-site using a nitrogen generator, which typically operates on the principle of pressure swing adsorption (use of a selective adsorbent material which removes oxygen from pressurised air. The adsorbent is contained in two columns where each column is pressurised and depressurised during operation).

b)

The solution involves an application of the universal gas law:

P V space equals space n R T

Substituting for the specific gas constant RSpecific:

P V space equals space m R subscript S p e c i f i c end subscript T

Where RSpecific is given by:

R subscript S p e c i f i c end subscript space equals space fraction numerator R over denominator M. m a s s end fraction space equals space fraction numerator 8 comma 314.5 over denominator 2 space cross times space 14 end fraction space equals space 297 space J divided by k g space K

Remember that we should use absolute pressure and temperature, hence re-arranging the universal gas law for mass m:

m space equals space fraction numerator P V over denominator R subscript S p e c i f i c end subscript T end fraction space equals space fraction numerator left parenthesis 100 space cross times space 10 to the power of 5 space plus space 100 space cross times space 10 cubed right parenthesis space cross times space 0.06 over denominator 297 space cross times space left parenthesis 20 space plus space 273 right parenthesis end fraction

m space equals space 6.97 space k g

c)

Let’s consider a dissolved N2 concentration of 1 v/v at STP in 1 L of water as a basis for the determination of our conversion factor.

1 v/v at STP in 1 L of water is a concentration of 1 L of dissolved N2 in 1 L of water at STP.

Method 1:

Using the ideal gas law, we can determine the mass of 1 L of N2:

P V space equals space m R subscript S p e c i f i c end subscript T

m space equals space fraction numerator P V over denominator R subscript S p e c i f i c end subscript T end fraction space equals space fraction numerator 101 comma 325 space cross times space 0.001 over denominator open parentheses begin display style fraction numerator 8 comma 314.5 over denominator 28 end fraction end style close parentheses space cross times space 273.15 end fraction space equals space 1.25 space g

Hence a dissolved N2 concentration of 1 v/v at STP is equivalent to 1.25 g/L.

Or simpler method 2:

There is an alternative way of obtaining this answer by simply using the relationship that 1 kmol of any gas occupies 22.4 m3 at STP.

That is, 1 kmol, being 28 kg of N2 gas, would occupy 22.4 m3 = 22,400 L at STP. Therefore, 1 L of N2 gas (in our 1 L of beer) will weigh 28 x (1/22,400) = 0.00125 kg = 1.25 g.

Hence a dissolved N2 concentration of 1 v/v (1 volume N2 / 1 L water) at STP is equivalent to 1.25 g/L.

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