You’re at the zoo and have a big red 1.80 L helium balloon. The barometric pressure today is 785 mmHg. Then you hear the roar of a lion. Startled, you accidentally release the balloon. It flies away. By the time it reaches the clouds, the atmospheric pressure that high is only 3.00 atmospheres. What would the volume of the balloon up there? (Temperature is constant)

It is important to express the initial and the final pressure in the same unit. Either express both in atm or in mmHg. What ever the case is, we'll still arrive at same answer. Here, we shall be converting from mmHg to atm. This is illustrated below:

760mmHg = 1atm

Therefore, 785 mmHg = 785/760 = 1.03 atm

Step 3:

Determination of the final volume. This is illustrated below.

We shall be applying the Boyle's law equation since the temperature is constant.

P1V1 = P2V2

Initial Volume (V1) = 1.80 L

Initial pressure (P1) = 1.03 atm

Final pressure (P2) = 3.00 atm

Final volume (V2) =?

P1V1 = P2V2

1.03 x 1.8 = 3 x V2

Divide both side by 3

V2 = (1.03 x 1.8) /3

V2 = 0.62 L

Therefore, the new volume of the balloon is 0.62 L

lcoley8 lcoley8 · Answer 2 · 2020-05-05T04:51:38+02:00

Answer:

The volume of the balloon up there will be 0.618 L

Explanation:

Given:

V₁ = 1.8 L

P₁ = 785 mmHg = 1.03 atm

P₂ = 3 atm

Question: What would the volume of the balloon up there, V₂ = ?

According the ideal gas law:

PV = nRT

However, in this case, the number of moles of helium balloon is constant. The temperature will also be assumed to be constant. Therefore, the expression of the ideal gases is as follows:

P₁V₁ = P₂V₂

Solving for V₂

[tex]V_{2} =\frac{P_{1}V_{1} }{P_{2} } =\frac{1.03*1.8}{3} =0.618L[/tex]