Question

A weather balloon contains 8.80 moles of helium at a pressure of 0.992 atm and a temperature of 25⁰C at ground level. What is the volume of the balloon under these conditions?

Short answer

The volume of the balloon is \(\left( {\rm{V}} \right){\rm{ = 217}}{\rm{.03 L}}{\rm{.}}\)

Step-by-step solution

Definition of the ideal gas law

The general gas equation, commonly known as the ideal gas law, is a state equation for a hypothetical ideal gas. Despite its flaws, the ideal gas law provides a decent approximation of the behavior of various gases under a variety of situations.

Explanation of the ideal gas law

The pressure, volume, temperature, and the number of moles of a gas are all related by the ideal gas law.

It is given as

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

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

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

The 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.

Finding the volume of the balloon

Given:

\(\begin{aligned}{}{\rm{Volume (V) = ?L}}{\rm{. }}\\{\rm{Pressure (P) = 0}}{\rm{.992 atm}}{\rm{. }}\\{\rm{Temperature (T) = 2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{.}}\end{aligned}\))

The number of moles of the gas \({\rm{(n) = 8}}{\rm{.8}}\))moles.

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

So, we need to convert the temperature in degreesCelsiusto kelvins.

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

So,

\(\begin{aligned}{}{\rm{2}}{{\rm{5}}^{\rm{^\circ }}}{\rm{ = (25) + 273}}{\rm{.15K}}\\{\rm{ = 298}}{\rm{.15K}}\\{\rm{PV = nRT }}\\{\rm{Volume (V) = }}\frac{{{\rm{nRT}}}}{{\rm{P}}}\\{\rm{ = }}\frac{{{\rm{(8}}{\rm{.8) \times (0}}{\rm{.08206) \times (298}}{\rm{.15)}}}}{{{\rm{(0}}{\rm{.992)}}}}\\{\rm{ = }}\frac{{{\rm{215}}{\rm{.30}}}}{{{\rm{0}}{\rm{.992}}}}\\{\rm{ = 217}}{\rm{.03\;L}}{\rm{.}}\end{aligned}\))

Hence, the volume of the balloon is \(\left( {\rm{V}} \right){\rm{ = 217}}{\rm{.03 L}}{\rm{.}}\))