The heat required to sustain animals while they hibernate comes from the biochemical combustion of fatty acids, such as arachidonic acid, \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) What volume of air, measured at \(298 \mathrm{K}\) and \(1.00 \mathrm{atm}\) is required to burn \(2.00 \mathrm{kg} \mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2} ?\) Air is approximately \(78.1 \% \mathrm{N}_{2}\) and \(20.9 \% \mathrm{O}_{2},\) by volume. Other gases make up the remaining \(1.0 \%\)

To solve the problem, one needs to first write down the combustion reaction equation to see how arachidonic acid reacts with oxygen to produce carbon dioxide and water: \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2} + 31\mathrm{O}_{2} \rightarrow 20\mathrm{CO}_{2} + 16\mathrm{H}_{2}\mathrm{O} \) Note that the coefficients in front of reactants tell us the stoichiometric amounts needed (or produced) per each molecule (or mole) of arachidonic acid.

First, convert the mass of \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) to moles by dividing it with its molecular weight. The molecular weight of \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) is \(304 \mathrm{g/mole}\). Therefore, for \(2.00 \mathrm{kg} = 2000 \mathrm{g}\) of \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\), we have \(2000 \mathrm{g} / 304 \mathrm{g/mole} = 6.58 \mathrm{moles}\) of arachidonic acid. According to the stoichiometry of the combustion reaction, for each mole of \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) we need \(31 \mathrm{moles}\) of \(\mathrm{O}_{2}\). Therefore, we need \(6.58 \mathrm{moles} * 31 = 204 \mathrm{moles}\) of \(\mathrm{O}_{2}\) to burn \(2.00 \mathrm{kg}\) of arachidonic acid.

Knowing that air consists of \(20.9 \%\ \mathrm{O}_{2}\) by volume, we can use this percentage to calculate the total volume of air needed to provide \204 \mathrm{moles}\ \mathrm{O}_{2}\. First, convert the moles of \(\mathrm{O}_{2}\) to volume at the given conditions using the Ideal Gas Law: \(V = nRT/P\), where \(n\) is the number of moles, \(R\) is the ideal gas constant (0.0821 \(\bm{L·atm/mol·K}\)), \(T\) is the temperature in Kelvin, and \(P\) is the pressure in atm. Substituting these values, we get \(V = 204 \mathrm{moles} * 0.0821 \bm{L·atm/mol·K} * 298 \bm{K} / 1.00 \bm{atm} = 5004 \bm{L}\). As this represents \(20.9 \%\ of the total air volume, if we let \(x\) be the total volume of air, we get: \(0.209x = 5004 \bm{L}\), solving for \(x\) we get an approximation of \(x = 23950 \bm{L}\) of air.