In one city, a balloon with a volume of 6.0 L is filled with air at 101 kPa
pressure. The balloon is then taken to a second city at a much higher
altitude. At this second city, atmospheric pressure is only 91 kPa. If the
temperature is the same in both places, what will be the new volume of the
balloon?

Respuesta :

Answer: The new volume of the balloon is 6.6 L

Explanation:

To calculate the new tempearture, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.

[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.

We are given:

[tex]P_1=101kPa\\V_1=6.0L\\P_2=91kPa\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]101\times 6.0L=91\times V_2\\\\V_2=6.6L[/tex]

Thus new volume of the balloon is 6.6 L

The new volume of ballon if it is taken to a second city at a much higher

altitude is 6.6 L.

What is Boyle's Law?

Boyle's Law of gas states that at constant temperature, pressure of the gas is inversely proportional to the volume of the gas.

P ∝ 1/V.

For the given question required equation will be:
P₁V₁ = P₂V₂, where

P₁ = pressure of ballon in first city = 101 kPa

V₁ = volume of ballon in first city = 6 L

P₂ = pressure of ballon at another city = 91 kPa

V₂ = volume of ballon at another city = to find?

On putting all these values on the above equation, we get

V₂ = (101) (6) / (91) = 6.6 L

Hence, required volume is 6.6 L.

To know more about Boyle's Law, visit the below link:

https://brainly.com/question/1437490