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

Answer:

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.

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!!!

Answer 2

The volume of the balloon when this happens will be 494L

In order to get the volume that would eventually equalize causing the balloon to stop inflating, we will use Boyle's law expressed as;

$P_1V_1=P_2V_2$

Given the following parameters

P₁ = 26.0atm

V₁ = 19.0L

P₂ = 1.0atm

Required

Final volume V₂

Substitute the given values into the formula;

$V_2=\frac{P_1V_1}{P_2}\\V_2=\frac{26 \times 19}{1} \\V_2 =494L$

Hence the volume of the balloon when this happens will be 494L

Learn more here: https://brainly.com/question/22625142