A large scuba tank (see Figure \({\rm{9}}{\rm{.16}}\)) with a volume of \({\rm{18\;L}}\) is rated for a pressure of \({\rm{220 bar}}\). The tank is filled at \({\rm{2}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C}}\) and contains enough air to supply \({\rm{1860\;L}}\) of air to a diver at a pressure of \({\rm{2}}{\rm{.37\;atm}}\) (a depth of \({\rm{45}}\)feet). Was the tank filled to capacity at \({\rm{2}}{{\rm{0}}^{\rm{^\circ }}}{\rm{C}}\)?

The ideal gas law, or the universal gas equation, is a state equation for a hypothetical ideal gas. Despite its shortcomings, the ideal gas law approximates the behaviour of many gases in a number of settings rather well.

The ideal gas law connects pressure, volume, temperature, and the number of moles in a gas.It is expressed as

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

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(\(^{\rm{o}}{\rm{C}}\)) andnot kelvins (\({\rm{K}}\)) (absolute scale).

As a result, we need toconvert temperatures from degrees Celsius to kelvins.

We have

\({0^\circ } = (0) + 273.15K\)

So,

\(\begin{aligned}{}{20^\circ } = (20) + 273.15K\\ = 283.15K\end{aligned}\)

Also, we need to convert the pressure given in bar to atm.

We have

\({\rm{1 bar = 0}}{\rm{.9869 atm}}\)

So,

\(\begin{aligned}{}{\rm{220bar = (220) \times (0}}{\rm{.9869) atm}}\\{\rm{ = 217}}{\rm{.118\;atm}}{\rm{.}}\end{aligned}\)

So, we have

\(\begin{aligned}{}{\rm{Initial Volume }}\left( {{{\rm{V}}_{\rm{1}}}} \right){\rm{ = 18\;L}}\\{\rm{Initial Temperature }}\left( {{{\rm{T}}_{\rm{1}}}} \right){\rm{ = 283}}{\rm{.15K}}{\rm{.}}\\{\rm{Initial Pressure }}\left( {{{\rm{P}}_{\rm{1}}}} \right){\rm{ = 217}}{\rm{.118\;atm}}\\{\rm{Final Volume }}\left( {{{\rm{V}}_{\rm{2}}}} \right){\rm{ = ?L}}{\rm{.}}\\{\rm{Final Temperature }}\left( {{{\rm{T}}_{\rm{2}}}} \right){\rm{ = 283}}{\rm{.15\;K}}{\rm{.}}\\{\rm{Final Pressure }}\left( {{{\rm{P}}_{\rm{2}}}} \right){\rm{ = 2}}{\rm{.37\;atm}}\end{aligned}\)

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

Here, \({{\rm{T}}_{\rm{1}}}{\rm{ = }}{{\rm{T}}_{\rm{2}}}\)

So,

\(\begin{aligned}{}{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{ = }}{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}\\{\rm{(217}}{\rm{.118) \times (18) = (2}}{\rm{.37) \times }}\left( {{{\rm{V}}_{\rm{2}}}} \right)\\{{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{(217}}{\rm{.118) \times (18)}}}}{{{\rm{2}}{\rm{.37}}}}\\{{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{3908}}{\rm{.124}}}}{{{\rm{2}}{\rm{.37}}}}\\{\rm{ = 1649\;L}}\end{aligned}\)

Therefore, at \({\rm{2}}{{\rm{0}}^{\rm{^\circ }}}\), the tank is not completely filled as \({{\rm{V}}_{\rm{2}}}{\rm{ = 1649\;L}}\).