A balloon is inflated to a volume of 7.0 L 7.0 L on a day when the atmospheric pressure is 765 mmHg. The next day, as a storm front arrives, the atmospheric pressure drops to 737 mmHg 737 mmHg . Assuming the temperature remains constant, what is the new volume of the balloon, in liters?

The new volume is 7.27 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 7.0 L

The pressure = 765 mmHg = 765/760 atm = 1.00658 atm

The pressure drops to 737 mmHg = 737/ 760 atm = 0.969737

The temperature remains constant

Step 2: Calculate the new volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure in the balloon = 765 mmHg

⇒with V1 = the initial volume = 7.0L

⇒with P2 = the decreased pressure = 737 mmHg

⇒with V2 = the new volume = TO BE DETERMINED

765 mmHg * 7.0 L = 737 mmHG * V2

V2 = (765 mmHG * 7.0 L) / 737 mmHG

V2 = 7.27 L

The new volume is 7.27 L