Question

A child has a toy balloon with a volume of 1.80 liters. The temperature of the balloon when it was filled was 20° C and the pressure was 1.00 atm. If the child were to let go of the balloon and it rose 3 kilometers into the sky where the pressure is 0.667 atm and the temperature is -10° C, what would the new volume of the balloon be? (Don't forget to convert the temperature to K)

Answer 1

Answer:

The new volume of the balloon is 2.422 liters.

Explanation:

Let suppose that gas inside the balloon behaves ideally. From the Equation of State for Ideal Gases, we know that pressure ($P$), in atmospheres, is inversely proportional to volume ($V$), in liters, and directly proportional to temperature ($T$), in Kelvin. Based on this fact, we construct the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressures, in atmospheres.

$V_{1}, V_{2}$ - Initial and final volumes, in liters.

$T_{1}, T_{2}$ - Initial and final temperatures, in Kelvin.

If we know that $P_{1} = 1\,atm$, $V_{1} = 1.80\,L$, $T_{1} = 293.15\,K$, $P_{2} = 0.667\,atm$, $T_{2} = 263.15\,K$, then the final pressure is:

$V_{2} = V_{1} \cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot \left(\frac{T_{2}}{T_{1}} \right)$

$V_{2} = 2.422\,L$

The new volume of the balloon is 2.422 liters.