91Ó°ÊÓ

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 provided 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, not kelvins (absolute scale).

As a result, we must convert temperatures from degrees Celsius to kelvins. We have

\({0^\circ }{\rm{C}} = \left( 0 \right) + 273.15K\)

So,

\(\begin{aligned}{}{\rm{2}}{{\rm{1}}^{\rm{^\circ }}}{\rm{ = (21) + 273}}{\rm{.15K}}\\{\rm{ = 294}}{\rm{.15K - 4}}{{\rm{8}}^{\rm{^\circ }}}\\{\rm{ = ( - 48) + 273}}{\rm{.15K}}\\{\rm{ = 225}}{\rm{.15K}}\end{aligned}\)

So,

\(\begin{aligned}{}{\rm{Initial Volume }}\left( {{{\rm{V}}_{\rm{1}}}} \right){\rm{ = 11}}{\rm{.2\;L}}\\{\rm{Initial Temperature }}\left( {{{\rm{T}}_{\rm{1}}}} \right){\rm{ = 294}}{\rm{.15\;K}}\\{\rm{Initial Pressure }}\left( {{{\rm{P}}_{\rm{1}}}} \right){\rm{ = 745 torr}}\\{\rm{Final Volume }}\left( {{{\rm{V}}_{\rm{2}}}} \right){\rm{ = ?}}\;\;\;{\rm{L}}\\{\rm{Final Temperature }}\left( {{{\rm{T}}_{\rm{2}}}} \right){\rm{ = 225}}{\rm{.15\;K}}\\{\rm{Final Pressure }}\left( {{{\rm{P}}_{\rm{2}}}} \right){\rm{ = 63}}{\rm{.1 torr}}{\rm{.}}\end{aligned}\)

\(\begin{aligned}{}{\rm{PV = nRT}}\\\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\\\frac{{{\rm{(745) \times }}\left( {{\rm{1}}{\rm{.41 \times 1}}{{\rm{0}}^{\rm{4}}}} \right)}}{{{\rm{294}}{\rm{.15}}}}{\rm{ = }}\frac{{{\rm{(63}}{\rm{.1) \times }}\left( {{{\rm{V}}_{\rm{2}}}} \right)}}{{{\rm{225}}{\rm{.15}}}}\\\frac{{{\rm{1050}}{\rm{.45 \times 1}}{{\rm{0}}^{\rm{4}}}}}{{{\rm{294}}{\rm{.15}}}}{\rm{ = }}\frac{{{\rm{(63}}{\rm{.1) \times }}\left( {{{\rm{V}}_{\rm{2}}}} \right)}}{{{\rm{225}}{\rm{.15}}}}\\{\rm{35711}}{\rm{.37 = }}\frac{{{\rm{(63}}{\rm{.1) \times }}\left( {{{\rm{V}}_{\rm{2}}}} \right)}}{{{\rm{225}}{\rm{.15}}}}\\{{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{(35711}}{\rm{.37) \times (225}}{\rm{.15)}}}}{{{\rm{63}}{\rm{.1}}}}\\{{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{804}}{\rm{.04 \times 1}}{{\rm{0}}^{\rm{4}}}}}{{{\rm{63}}{\rm{.1}}}}\\{\rm{ = 127423}}{\rm{.381\;L}}{\rm{.}}\\{\rm{ = 12}}{\rm{.74 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{\;L}}{\rm{.}}\end{aligned}\)

Therefore, the final volume of balloon at\({\rm{20Km}}\)is\({\rm{12}}{\rm{.74 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{\;L}}\).