Question

a balloon filled with 0.500 l of air at sea level is submerged in the water to a depth that produces a pressure of 3.25 atm. what is the volume of the balloon at this depth?
options:
6.50 l
0.154 l
0.615 l
1.63 l
none of the above

Explanation:

Step1: Identify the gas law

This problem involves Boyle's Law, which states that for a given amount of gas at constant temperature, the product of pressure and volume is constant, i.e., \( P_1V_1 = P_2V_2 \). At sea level, the pressure \( P_1 \) is approximately 1 atm, and the initial volume \( V_1 = 0.500 \, \text{L} \). The final pressure \( P_2 = 3.25 \, \text{atm} \), and we need to find the final volume \( V_2 \).

Step2: Rearrange Boyle's Law formula

From \( P_1V_1 = P_2V_2 \), we can solve for \( V_2 \):
\( V_2 = \frac{P_1V_1}{P_2} \)

Step3: Substitute the values

Substitute \( P_1 = 1 \, \text{atm} \), \( V_1 = 0.500 \, \text{L} \), and \( P_2 = 3.25 \, \text{atm} \) into the formula:
\( V_2 = \frac{1 \, \text{atm} \times 0.500 \, \text{L}}{3.25 \, \text{atm}} \)

Step4: Calculate the result

Simplify the expression:
\( V_2 = \frac{0.500}{3.25} \, \text{L} \approx 0.1538 \, \text{L} \approx 0.154 \, \text{L} \)

Answer:

0.154 L (corresponding to the option "0.154L")