Ahelium balloon with an internal pressure of 1.00 atm and a volume of 4.50 l at 20.0 °c is released. what volume will the balloon occupy at an altitude where the pressure is 0.600 atm and the temperature is –20.0 °c? *

The balloon will occupy a volume of 6.48 L

Explanation:

Step 1: Data given

internal pressure = 1.00 atm

volume = 4.50 L

Temperature = 20.0 °C

Step 2: Calculate new volume via the ideal gas law

P*V = n*R*T

(P1*V1)/ T1 = (P2*V2)/T2

⇒ with P1 = 1.00 atm

⇒ with V1 = 4.50 L

⇒ with T1 = 20.0 °C = 293 Kelvin

⇒ with P2 = 0.600 atm

⇒ with V2 = TO BE DETERMINED

⇒ with T2 = -20°C = 253 Kelvin

(1.00atm * 4.50 L)/293 Kelvin = (0.600 atm*V2) / 253 Kelvin

0.01536 = 0.00237 V2

V2 = 6.48 L

The balloon will occupy a volume of 6.48 L

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