Question

A 6.1 L balloon is filled with water at 1.57 atm. If the balloon is squeezed into a 4.28 L beaker and does not burst, what is the pressure of the water in the balloon?

a) 2.24 atm
b) 1.10 atm
c) 41.0 atm
d) 0.0345 atm

Answer 1

The pressure of the water in the balloon when it is compressed into a 4.28 L volume is calculated using Boyle's Law and found to be 2.24 atm, corresponding to option (a).

Step-by-step explanation:

The student's question asks about the pressure change of water in a balloon when the volume decreases. This scenario can be analyzed using Boyle's Law, which states that for a given mass of gas at constant temperature, the pressure and volume are inversely proportional. In equation form, Boyle's Law is expressed as P1V1 = P2V2. To find the new pressure P2 when the volume changes from 6.1 L to 4.28 L:

1. Using the initial conditions (P1 = 1.57 atm, V1 = 6.1 L), apply Boyle's Law: P1V1 = P2V2
2. Substitute the known values to find P2: 1.57 atm * 6.1 L = P2 * 4.28 L
3. Solve for P2: P2 = (1.57 atm * 6.1 L) / 4.28 L = 2.24 atm

Thus, the pressure of the water in the balloon when squeezed into a 4.28 L beaker is 2.24 atm, which corresponds to option (a).