At constant temperature, a sample of helium at 760. torr in a closed container was compressed from 5.00 l to 3.00 l, with no change in moles or temperature. what was the new pressure exerted by the helium on its container?

To solve this problem, we will use Boyle's law which states that:
At constant temperature, P1V1 = P2V2 where:
P1 is the initial pressure = 760 torr
V1 is the initial volume = 5 liters
P2 is the final pressure that we want to calculate
V2 is the final volume = 3 liters

Substitute in the above equation to get P2 as follows:
(760)(5) = P2 (3)
P2 = (760*5) / (3) = 1266.667 torr

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nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-05-03T06:32:18+02:00

At constant temperature, a sample of helium at 760. torr in a closed container was compressed from 5L to 3L, then the new pressure is 1,266.6 torr.

What is Boyle's Law?

Boyle's law explains that pressure of any gas is inversely proportional to the volume of gas, and it will be represented as P₁V₁ = P₂V₂.

P₁ = initial pressure = 760 torr

V₁ = initial volume = 5 L

P₂ = final pressure = ?

V₂ = final volume = 3 L

On putting values on the above equation, we get

P₂ = (760)(5) / (3) = 1,266.6 torr

Hence required pressure of helium is

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

https://brainly.com/question/469270

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