Final answer:
The volume of the air in the balloon when cooled to 32 °F with its pressure reduced to 14.7 lb/in² is approximately 0.255 m³, calculated using the combined gas law after converting temperatures to Kelvin and pressures to atmospheres.
Step-by-step explanation:
To find the volume of the air in the balloon after it is cooled to 32 °F and the pressure is reduced to 14.7 lb/in², we can use the combined gas law, which states that the pressure, volume, and temperature of a gas are related by the formula (P1*V1)/T1 = (P2*V2)/T2, where P stands for pressure, V for volume, and T for temperature in Kelvin. To work with these values, we must first convert all units to a consistent set, particularly temperature to Kelvin and pressure to a common unit.
First, we convert the temperatures from Fahrenheit to Kelvin. The formula to convert from Fahrenheit to Kelvin is K = (F - 32) * 5/9 + 273.15. Therefore:
80 °F = (80 - 32) * 5/9 + 273.15 ≈ 299.82 K,
32 °F = (32 - 32) * 5/9 + 273.15 ≈ 273.15 K.
We also need to convert the pressure from lb/in² to atm, with 1 atm = 14.696 lb/in². Thus:
19.0 lb/in² ≈ 19.0 / 14.696 atm,
14.7 lb/in² ≈ 14.7 / 14.696 atm.
Now we can solve for V2 using the combined gas law:
(19.0 / 14.696 atm * 0.300 m³) / 299.82 K = (14.7 / 14.696 atm * V2) / 273.15 K.
After calculating and solving for V2, we find that the volume of the air in the container under the new conditions is approximately 0.255 m³.
Remember that in an actual calculation, you would need to retain the significant figures as per the measurements' precision.