A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it rises to a location where the pressure is 0.720 atm and the temperature is -11.7°C. What is the volume of the balloon at that new location?

The new volume is 63583 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 5.37 * 10^4 L

The initial pressure = 0.995 atm

The initial temperature = 32.0 °C = 305.15 K

The pressure decreased to 0.720 atm

The temperature decreased to -11.7 °C = 261.45 K

Step 2: Calculate the new volume

P1*V1 / T1 = P2*V2/T2

⇒with P1 = the initial pressure = 0.995 atm

⇒with V1 = the initial volume = 5.37 *10^4 L

⇒with T1 = the initial temperature = 305.15 K

⇒with P2 = the decreased pressure = 0.720 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

(0.995 * 5.37*10^4)/305.15 = (0.720 * V2) / 261.45

V2 = 63583 L

The new volume is 63583 L

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