#1. Boyle's Law. A gas with a volume of 4.0 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L.
What is the pressure in the container if the temperature remains constant? (Answer is 68.3 kPa).

Respuesta :

Explanation:

According to Boyles law

[tex]pressure \: \alpha \frac{1}{volume} [/tex]

P1V1 = P2V2

205 (4) = P2 (12)

[tex]P2 = \frac{205 \times 4}{12} [/tex]

P2= 820/12

P2= 68.3 KPa

The pressure in the container after the expanssion of gas, if the temperature remains constant is 68.3 kPa.

What is Boyle's Law?

According to the Boyle's Law, at constant temperature pressure of any gas is inversely proportional to the volume of the gas, i.e.

P ∝ 1/V

Required equation for the calculation of the pressure is:

P₁V₁ = P₂V₂, where

P₁ = initial pressure = 205 kPa

V₁ = initial volume = 4 L

P₂ = final pressure = ?

V₂ = final volume = 12 L

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

P₂ = (205)(4) / (12) = 68.3 kPa

Hence resultant pressure is 68.3 kPa.

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

https://brainly.com/question/469270