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A balloon at sea level has 2.40 mL of air at 1,550.50 atm of pressure. After it rises
into the air, the pressure changes to 1.49 atm. What is the new volume? Record your
final answer rounded to two decimal places and provide units where applicable.
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Answer :

GinnyAnswer
To solve the problem, we'll use Boyle's Law, which states that the product of the initial pressure and volume of a gas is equal to the product of the final pressure and volume of the gas, given the temperature remains constant. The formula for Boyle's Law is:

[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]

Where:
- [tex]\( P_1 \)[/tex] is the initial pressure
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( P_2 \)[/tex] is the final pressure
- [tex]\( V_2 \)[/tex] is the final volume

We are given:
- [tex]\( P_1 = 1550.50 \)[/tex] atm
- [tex]\( V_1 = 2.40 \)[/tex] mL
- [tex]\( P_2 = 1.49 \)[/tex] atm

We need to find the final volume [tex]\( V_2 \)[/tex]. Rearranging the formula for [tex]\( V_2 \)[/tex]:

[tex]\[ V_2 = \frac{P_1 \times V_1}{P_2} \][/tex]

Substituting the given values into the formula:

[tex]\[ V_2 = \frac{1550.50 \, \text{atm} \times 2.40 \, \text{mL}}{1.49 \, \text{atm}} \][/tex]

First, calculate the numerator:

[tex]\[ 1550.50 \times 2.40 = 3721.2 \, \text{mL} \cdot \text{atm} \][/tex]

Then, divide by the final pressure:

[tex]\[ V_2 = \frac{3721.2 \, \text{mL} \cdot \text{atm}}{1.49 \, \text{atm}} \][/tex]

[tex]\[ V_2 \approx 2497.45 \, \text{mL} \][/tex]

Hence, the new volume of the balloon, after rising and experiencing the pressure change, is approximately [tex]\( 2497.45 \)[/tex] mL.

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