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?

A. 3.55 atm
B. 2.89 atm
C. 13.0 atm
D. 5.42 atm

Answer 1

D. 5.42 atm.

Step-by-step explanation:

The total pressure of the gas mixture = P of He + P of Ar after cooling.

P of He = 1.87 atm.

• Firstly, we need to find the P of Ar after cooling:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

• If n is constant, and have different values of P, V and T:

(P₁V₁T₂) = (P₂V₂T₁)

• Knowing that:

P₁ = 3.9 atm, V₁ = 5.0 m³, T₁ = 425.0 K,

P₂ = ??? atm, V₂ = 3.1 m³, T₂ = 240.0 K,

• Applying in the above equation

(P₁V₁T₂) = (P₂V₂T₁)

∴ P₂ = (P₁V₁T₂)/(V₂T₁) = (3.9 atm)(5.0 m³)(240 K)/(3.1 m³)(425.0 K) = 3.552 atm.

∴ The total pressure of the gas mixture = P of He + P of Ar after cooling.

P of He = 1.87 atm & P of Ar after cooling = 3.552 atm.

∴ The total pressure of the gas mixture = 1.87 atm + 3.552 atm = 5.422 atm ≅ 5.42 atm.

So, the right choice is: D. 5.42 atm.