A balloon has an internal pressure of 1.95 atm and a volume of 5.0 L. If the temperature where the balloon is released is 20 °C, what will happen to the volume when the balloon rises to an altitude where the pressure is
0.65 atm and the temperature is -15 °C?

13.1 L

Explanation:

Using the combined gas law:

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

Based on the information provided in this question;

P1 = 1.95 atm

P2 = 0.65 atm

V1 = 5.0 L

V2 = ?

T1 = 20 °C = 20 + 273 = 293K

T2 = -15 °C = -15 + 273 = 258K

Using P1V1/T1 = P2V2/T2

1.95 × 5/293 = 0.65 × V2/258

9.75/293 = 0.65V2/258

0.0333 = 0.00252V2

V2 = 0.0333 ÷ 0.00252

V2 = 13.1 L