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

Question

Chemistry

Zander Sherman

27 September, 13:31

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

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Answers (1)

Know the Answer?

Arely Whitehead · Answer 1

The volume of the ballon is 325L.

Explanation:

Boyle's law express that the pressure of a gas is inversely proportional to its volume. That means if the pressure increases, the volume decreases. The formula is:

P₁V₁ = P₂V₂

Where P represents pressure and V volume of 1, initial state and 2, final state of the gas.

In the problem, the volume of the tank is 13.0L and the final pressure of the ballon is 1atm - The atmospheric pressure-. As 1atm of gas is in the ballon, the pressure of the tank is 26.0atm - 1.0atm = 25.0atm.

Replacing in Boyle's law expression:

25.0atm*13.0L = 1atmV₂

325L = V₂

The volume of the ballon is 325L.