Question

A balloon is inflated to a volume of 8.0 L on a day when the atmospheric pressure is 1.013 bar . The next day, a storm front arrives, and the atmospheric pressure drops to 0.968 bar . Assuming the temperature remains constant, what is the new volume of the balloon, in liters

Answer 1

`V_2=8.4L`

Step-by-step explanation:

Hello there!

In this case, according to the definition of the Boyle's law, which describes de pressure-volume behavior as an inversely proportional relationship, it is possible for us to write:

`P_1V_1=P_2V_2`

Thus, since we are given the initial pressure and temperature, and the final pressure, we are able to calculate the final volume as shown below:

`baV_2=(P_1V_1)/(P_2)\\\\V_2=(8.0L*1.013bar)/( 0.968bar)\\\\V_2=8.4L`

Regards!