Question

3. A toy balloon at 25.0 °C has an internal pressure of 1.05 atm and a volume of 5.0 mL. The balloon is released and floats to an altitude of 35,000 ft where the pressure is 0.45 atm and a temperature of -15.0 °C. What is the balloon's new volume? i’m

Answer 1

We can use the combined gas law to solve this problem:

(P1 × V1) / (T1) = (P2 × V2) / (T2)

Where:

P1 = 1.05 atm (pressure at sea level)

V1 = 5.0 mL (volume at sea level)

T1 = 25.0 °C + 273.15 = 298.15 K (temperature at sea level, converted to Kelvin)

P2 = 0.45 atm (pressure at 35,000 ft)

V2 = ? (new volume at 35,000 ft, what we are solving for)

T2 = -15.0 °C + 273.15 = 258.15 K (temperature at 35,000 ft, converted to Kelvin)

Plugging in the values, we get:

(1.05 atm × 5.0 mL) / (298.15 K) = (0.45 atm × V2) / (258.15 K)

Simplifying and solving for V2, we get:

V2 = (1.05 atm × 5.0 mL × 258.15 K) / (0.45 atm × 298.15 K)

V2 = 5.44 mL

Therefore, the balloon's new volume at 35,000 ft is 5.44 mL.