A hydrogen-filled balloon to be used in high altitude atmosphere studies will eventually be 100 ft in diameter. At 150,000 ft, the pressure is 0.14 lb/in2 and the temperature is - 67°F. Assume the balloon is spherical in shape. What is the diameter of the hydrogen balloon at the ground at 14.7 lb/in2 and 68°F

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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].

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