a weather balloon is inflated to a volume of 28.5 L at a pressure of 748 mmHg and a temperature of 28.0 degrees C. The balloon rises in the atmosphere to an altitude of 25,000 feet, where the pressure is 385 mmHg and the temperature is -15.0 degrees C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

We are given with the conditions of the weather balloon and required to determine the volume of the balloon when the altitude is raised to 25,000 feet. We determine the number of moles through ideal gas law PV = nRT. Plugging to this equation, the number of moles is 1.1351 moles. We substitute this with another set of conditions, volume is 47.46 liters.

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Tringa0 Tringa0 · Answer 2 · 2019-10-15T08:32:40+02:00

Answer:

The volume of the balloon at this altitude is 47.46 Liters.

Explanation:

Initial pressure of the gas = [tex]P_1=748 mmHg[/tex]

Initial volume of balloon = [tex]V_1=28.5 L[/tex]

Initial temperature of the gas in the balloon = [tex]T_1=28^oC=301.15 K[/tex]

Moles of gas in balloon = n

[tex]P_1V_1=nRT_1[/tex]...[1]

Moles of gas will remain the same when the temperature pressure and volume conditions are altered.

Final pressure of the gas at given altitude= [tex]P_2=385 mmHg[/tex]

Final volume of balloon at given altitude = [tex]V_2=?[/tex]

Final temperature of the gas in the balloon at given altitude = [tex]T_2=-15^oC=258.15 K[/tex]

[tex]P_2V_2=nRT_2[/tex]...[2]

From [1] and [2]

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

[tex]V_2=\frac{P_1V_1\times T_2}{T_1\times P_2}[/tex]

[tex]=\frac{748 mmHg\times 28.5L\times 258.15 K}{301.15 K\times 385 mmHg}=47.46 L[/tex]

The volume of the balloon at this altitude is 47.46 Liters.