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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)

Sagot :

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 ([tex]P[/tex]), in atmospheres, is inversely proportional to volume ([tex]V[/tex]), in liters, and directly proportional to temperature ([tex]T[/tex]), in Kelvin. Based on this fact, we construct the following relationship:

[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex] (1)

Where:

[tex]P_{1}, P_{2}[/tex] - Initial and final pressures, in atmospheres.

[tex]V_{1}, V_{2}[/tex] - Initial and final volumes, in liters.

[tex]T_{1}, T_{2}[/tex] - Initial and final temperatures, in Kelvin.

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

[tex]V_{2} = V_{1} \cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot \left(\frac{T_{2}}{T_{1}} \right)[/tex]

[tex]V_{2} = 2.422\,L[/tex]

The new volume of the balloon is 2.422 liters.

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