Question

A sample of argon fills a volume of 5 m3 at 425 K with a pressure of 3.9 atm. The argon is cooled to 240 K and shrinks to a volume of 3.1 m3. After cooling it is added to a tank of helium with a pressure of 1.87 atm. What is the total pressure of the gas mixture?

Answer 1

Step-by-step explanation:

Given parameters:

Inital volume of argon, V = 5m³

Initial temperature of argon, T = 425K

Initial pressure of argon, P = 3.9atm

Final temperature when cooled T₂ = 240K

Final volume when cooled V₂ = 3.1m³

Pressure of the helium gas = 1.87atm

Uknown parameters:

Total mixture of the gas = Ptotal

Solution

To find the total pressure of the gas mixture, we must determined the final pressure of the cooled Ar gas. Then the total pressure can be obtained using the expression below:

Ptotal = P₁ + P₂

Where P₁ and P₂ are the partial pressure exterted by Ar and He in the mixture.

Using the general gas law, we find the final pressure exterted when the argon gas cooled:

`(P_(1) V_(1) )/(T_(1) )` =
`(P_(2) V_(2) )/(T_(2) )`

The unkown here is P₂ and simply make it the subject of the expression above:

P₂ =
`(P_(1) V_(1) T_(2) )/(V_(2) T_(1) )`

P₂ =
`(3.9 x 240 x 5)/(3.1 x 425)` = 3.55atm

Partial pressure exerted by Ar is 3.55atm

Ptotal = 3.55atm + 1.87atm = 5.42atm