A 13.0-L helium tank is pressurized to 26.0 atm. When connected to this tank, a

Question

A 13.0-L helium tank is pressurized to 26.0 atm. When connected to this tank, a balloon will inflate because the pressure inside the tank is greater than the atmospheric pressure pushing on the outside of the balloon. Assuming the balloon could expand indefinitely and never burst, the pressure would eventually equalize causing the balloon to stop inflating. What would the volume of the balloon be when this happens? Assume atmospheric pressure is 1.00 atm. Also assume ideal behavior and constant temperature. i got 338L for he whole thing but that is the volume of the entire sample of helium. But you need to consider that 13.0 liters of that is still in the 13.0-L tank. A helium tank is able to inflate balloons if the inside pressure is greater than the atmospheric pressure. can you explain how to do this

Explanation / Answer

(13.0 L) x (26.0 / 1.00) = 338 L total at 1 atm

But there are still 13 L in the tank, so the volume of the balloon is 338- 13 = 325 L

V1P1=V2P2 by ideal gas eq. the initial state is when the volume is 13 (v1) and the pressure is 26 (p1), the final state is P2= 1 atm, this is because the pressure of the outside the balloon is 1 atm, the balloon can expand indefinitey without burst, then the balloon will stop growing until P inside equals the P from outside which is 1 atm.

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