Answer:
The answer is 11.7 ft
Explanation:
You can use the combined gas law from Boyle's law, Charles's law, and Gay-Lussac's Law. Only because hydrogen behaves like an ideal gas for this conditions.
[tex]\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}[/tex]
where the subscripts denote the pressure "p", volume "V" and the temperature "T" (in Kelvin) at two different times. Let's consider [tex]t_1[/tex] as the balloom at 150,000 ft so
[tex]p_1 = 0.14 \ lb/in^2[/tex]
[tex]V_1 = \frac{4}{3} \pi R_1^3 = 523598.77 \ ft^3[/tex]
and [tex]T_1 = -67^\circ F = 218.15\ K[/tex].
Then, [tex]t_2[/tex] is the moment when the balloon is on the ground.
[tex]p_2 = 14.7 \ lb/in^2[/tex] and [tex]T_2 = 68^\circ F = 293.15\ K[/tex].
From the first equation,
[tex]V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}[/tex], then
[tex]V_2 = 6701.07 ft^3 [/tex] and the radius would be
[tex]R_2 = \sqrt[3]{\frac{3 V_2}{4 \pi}} = 11.7 \ ft [/tex].