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The ideal gas law is \(PV = nRT\) for n mole of ideal gas.

Here, you have to use the equation\(\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}\).

The initial pressure inside the balloonis \({P_1} = 1.00\;{\rm{atm}}\).

The final pressure inside the balloonis \({P_2} = 0.68\;{\rm{atm}}\).

The initial temperature is \({T_1} = {20.0^ \circ }{\rm{C}} = 293\;{\rm{K}}\).

The final temperature is \({T_2} = {5.0^ \circ }{\rm{C}} = 278\;{\rm{K}}\).

Let \({V_1}\) and \({V_2}\) be theinitial and final volume of the balloon, respectively.

Now, according to Boyle鈥檚 law, you get the following:

\(\begin{array}{c}\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}\\\frac{{{V_2}}}{{{V_1}}} = \frac{{{T_2}{P_1}}}{{{T_1}{P_2}}}\\\frac{{{V_2}}}{{{V_1}}} = \frac{{\left( {278\;{\rm{K}}} \right) \times \left( {1.00\;{\rm{atm}}} \right)}}{{\left( {293\;{\rm{K}}} \right) \times \left( {0.68\;{\rm{atm}}} \right)}}\\\frac{{{V_2}}}{{{V_1}}} = 1.4\end{array}\)

Hence, the final volume of the balloon is 1.4 times the initial volume.