Question

A 19.0 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

Answer 1

Using the ideal gas law, the volume of the balloon when it stops inflating and the pressures equalize would be 494.0 L, as calculated from the initial pressure and volume of the tank and the atmospheric pressure.

Step-by-step explanation:

The question concerns the behavior of gases under different conditions and relates to the ideal gas law, which is a fundamental concept in chemistry. When a balloon is filled with helium from a tank with a pressure of 26.0 atm and a volume of 19.0 L, the balloon will inflate until the pressure inside the balloon equals the outside atmospheric pressure. At this point, the volume of the balloon could be derived using the ideal gas law, which states that for a fixed amount of gas at constant temperature, the product of the pressure and volume (P1V1) will be equal to the product of the final pressure and volume (P2V2). In this scenario, assuming the temperature remains constant and the atmospheric pressure is 1 atm, we apply the equation P1V1 = P2V2.

Given that P1 is 26.0 atm and V1 is 19.0 L, and P2 is 1.0 atm (atmospheric pressure), we can solve for V2 as follows: V2 = (P1V1/P2) = (26.0 atm * 19.0 L) / 1.0 atm = 494.0 L. The volume of the balloon when it stops inflating and the pressures equalize would be 494.0 L.

Answer 2

The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.

The balloon expands by am additional 475 L.

Step-by-step explanation:

Assuming Helium behaves like an ideal gas and temperature is constant.

According to Boyle's law for ideal gases, at constant temperature,

P₁V₁ = P₂V₂

P₁ = 26 atm

V₁ = 19.0 L

P₂ = 1 atm (the balloon is said to expand till the pressure matches the pressure of the atmpsphere; and the pressure of the atmosphere is 1 atm)

V₂ = ?

P₁V₁ = P₂V₂

(26 × 19) = 1 × V₂

V₂ = 494 L (it is assumed the balloon never bursts)

The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.

The balloon expands by am additional 475 L.

Hope this Helps!!!