Question

If 17.06 mol of helium gas is at 10.1 ∘C and a gauge pressure of 0.311 atm .

Part A

Calculate the volume of the helium gas under these conditions.

Part B

Calculate the temperature if the gas is compressed to precisely half the volume at a gauge pressure of 1.26 atm .

Answer 1

Universal gas constant = R = 8.314 J/(mol.K)

Atmospheric pressure = P_atm = 1 atm = 101325 Pa

Number of moles of the helium gas = n = 17.06 mol

Initial temperature of the helium gas = T₁ = 10.1 ^oC = 10.1 + 273 K = 283.1 K

Initial gauge pressure of the helium gas = P_1g = 0.311 atm = 0.311 x 101325 Pa = 31512.1 Pa

Initial absolute pressure of the helium gas = P₁

P₁ = P_1g + P_atm

P₁ = 31512.1 + 101325

P₁ = 132837.1 Pa

Initial volume of the helium gas = V₁

P₁V₁ = nRT₁

(132837.1)V₁ = (17.06)(8.314)(283.1)

V₁ = 0.3023 m³

Now the gas is compressed to half the volume at a gauge pressure of 1.26 atm

New volume of the helium gas = V₂

V₂ = V₁/2 = 0.3023/2 = 0.15115 m³

New gauge pressure of the helium gas = P_2g = 1.26 atm = 1.26 x 101325 Pa = 127669.5 Pa

New absolute pressure of the helium gas = P₂

P₂ = P_2g + P_atm

P₂ = 127669.5 + 101325

P₂ = 228994.5 Pa

New temperature of the helium gas = T₂

P₂V₂ = nRT₂

(228994.5)(0.151) = (17.06)(8.314)T₂

T₂ = 244.03 K

T₂ = 244.03 - 273 ^oC

T₂ = -28.97 ^oC

A) Volume of the helium gas under the given conditions = 0.3023 m³

B) Temperature of the gas if it is compressed to half the volume at a gauge pressure of 1.26 atm = -28.97 ^oC