A weather balloon was initially at a pressure of 0.950 atm, and its volume was 35.0 L. The pressure decreased to 0.750 atm, without loss of gas or change in temperature. What was the change in the volume?
A)Increased by 44.3 L
B)Increased by 9.3 L
C)Increased by 7.4 L
D)Decreased by 27.6 L
E)Decreased by 7.4 L

Question

Quizplus · Accepted Answer

The volume change can be calculated using Boyle's Law, which states that P1V1 = P2V2 for a given amount of gas at constant temperature. Rearranging for V2 gives V2 = (P1V1)/P2. Substituting the given values: V2 = (0.950 atm * 35.0 L) / 0.750 atm = 44.3 L. The change in volume is the final volume minus the initial volume: 44.3 L - 35.0 L = 9.3 L.

A Weather Balloon Was Initially at a Pressure of 0