A cylinder of medical oxygen has a volume of \({\rm{35}}{\rm{.4\;L}}\) and contains \({{\rm{O}}_{\rm{2}}}\)at a pressure of \({\rm{151 atm}}\) and a temperature of \({\rm{2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{C}}\). What volume of \({{\rm{O}}_{\rm{2}}}\) does this correspond to at normal body conditions, that is, \({\rm{1 atm}}\)and\({\rm{3}}{{\rm{7}}^{\rm{^\circ }}}{\rm{C}}\)?

The general gas equationis known to be an ideal gas law. Despite its flaws, the ideal gas law provides a goodapproximation of the behaviour of various gases under a variety of situations.

Pressure, volume, temperature, and the number of moles of a gas are all related by the ideal gas law.It is given by

\({\rm{PV = nRT}}\)

where,

\(\begin{aligned}{}{\rm{P = pressure of the gas}}{\rm{. }}\\{\rm {V = volume of the gas}} {\rm{.}}\\{\rm{ n = number of moles of the gas}}{\rm{.}}\\{\rm{ T = temperature of the gas}}{\rm{. }}\end{aligned}\)

and

\(\begin{aligned}{}{\rm{R = Ideal gas constant }}\\{\rm{ = 0}}{\rm{.08206\;L atm mo}}{{\rm{l}}^{{\rm{ - 1}}}}{{\rm{K}}^{{\rm{ - 1}}}}\end{aligned}\)

ifthe amount of the gas \({\rm{(n)}}\)is in moles, temperature \({\rm{(T)}}\) is in kelvin \(\left( {\rm{K}} \right){\rm{,}}\)and pressure \({\rm{(P)}}\)is in atm.

So, we have the Combined Gas Law with a constant number of moles of an ideal gas under two distinct circumstances.

\({\rm{PV = nRT}}\)

\(\begin{aligned}{}\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = nR}}\\\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{ = nR}}\end{aligned}\)

i.e.\(\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\)

where,

\({{\rm{V}}_{\rm{1}}}{\rm{,}}{{\rm{V}}_{\rm{2}}}{\rm{ = }}\)Initial and final volumes,

\({{\rm{P}}_{\rm{1}}}{\rm{,}}{{\rm{P}}_{\rm{2}}}{\rm{ = }}\)Initial and final pressures, and

\({{\rm{T}}_{\rm{1}}}{\rm{,}}{{\rm{T}}_{\rm{2}}}{\rm{ = }}\)Initial and final absolute temperatures.

However, the temperatures provided are in degrees Celsius andnot kelvins (absolute scale).As a result, we must convert temperatures from degrees Celsius to kelvins.

We have\({{\rm{0}}^{\rm{^\circ }}}{\rm{ = (0) + 273}}{\rm{.15K}}\)

So,

\(\begin{aligned}{}{\rm{2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{ = (25) + 273}}{\rm{.15K}}\\{\rm{ = 298}}{\rm{.15K3}}{{\rm{7}}^{\rm{^\circ }}}\\{\rm{ = (37) + 273}}{\rm{.15K}}\\{\rm{ = 310}}{\rm{.15K}}\end{aligned}\)

So,

Initial volume \(\left( {{{\rm{V}}_{\rm{1}}}} \right){\rm{ = 35}}{\rm{.4\;L}}{\rm{.}}\),

Final volume\(\left( {{{\rm{V}}_{\rm{2}}}} \right){\rm{ = ?}}\;\;\;{\rm{L}}\),

Initial temperature \(\left( {{{\rm{T}}_{\rm{1}}}} \right){\rm{ = 298}}{\rm{.15K}}{\rm{.}}\;\;\;\)

Final temperature \(\left( {{{\rm{T}}_{\rm{2}}}} \right){\rm{ = 310}}{\rm{.15K}}\),

Initial pressure \(\left( {{{\rm{P}}_{\rm{1}}}} \right){\rm{ = 151\;atm}}\), and

Final pressure\(\left( {{{\rm{P}}_{\rm{2}}}} \right){\rm{ = 1 }}atm\).

\(\begin{aligned}{l}{\rm{PV = nRT}}\\\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\\\frac{{{\rm{(151) \times (35}}{\rm{.4)}}}}{{{\rm{298}}{\rm{.15}}}}{\rm{ = }}\frac{{{\rm{(1) \times }}\left( {{{\rm{V}}_{\rm{2}}}} \right)}}{{{\rm{310}}{\rm{.15}}}}\\\frac{{{\rm{5345}}{\rm{.4}}}}{{{\rm{298}}{\rm{.15}}}}{\rm{ = }}\frac{{{\rm{(1) \times }}\left( {{{\rm{V}}_{\rm{2}}}} \right)}}{{{\rm{310}}{\rm{.15}}}}\\{\rm{17}}{\rm{.93 = }}\frac{{{\rm{(1) \times }}\left( {{{\rm{V}}_{\rm{2}}}} \right)}}{{{\rm{310}}{\rm{.15}}}}\\{{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{(17}}{\rm{.93) \times (310}}{\rm{.15)}}}}{{\rm{1}}}\\{{\rm{V}}_{\rm{2}}}{\rm{ = 5560}}{\rm{.54\;L}}{\rm{.}}\end{aligned}\)

Therefore, at normal body temperatures, the volume of an oxygen cylinder is \({\rm{5560}}{\rm{.54\;L}}{\rm{.}}\)