If 1.93 g of Ar are added to 4.00 atm of He in a 2.00 L cylinder at 27.0 °C, what is the total pressure of the resulting gaseous mixture?

Method 1 :

Consider the Ar in the cylinder :
Pressure, P = Partial pressure Ar, P(Ar) = ? atm
Volume, V = 2.00 L
Absolute temperature, T = (273.2 + 27.0) K = 300.2 K
Mass, m = 1.93 g
Molar mass, M = 39.95 g/mol
Gas constant, R = 0.08206 L atm / (mol K)

PV = nRT and n = m/M
Then, PV = (m/M)RT
P = mRT/(MV)

Partial pressure Ar, P(Ar) = 1.93 × 0.08206 × 300.2 / (39.95 × 2.00) atm = 0.60 atm

Total pressure of the resulting gaseous:
P(Total) = P(Ar) + P(He) = (0.60 + 4.00) atm = 4.60 atm

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Method 2 :

Consider the He in cylinder :
Pressure, P = Partial pressure He , P(He) = 4.00 atm
Volume, V = 2.00 L
Absolute temperature, T = (273.2 + 27.0) K = 300.2 K
Gas constant, R = 0.08206 L atm / (mol K)
No. of moles of He, n = ? mol

PV = nRT
Then, n = PV/(RT) = 4.00 × 2.00 / (0.08206 × 300.2) mol = 0.3247 mol

Molar mass of Ar = 39.95 g/mol
No. of moles of Ar = (1.93 g) / (39.95 g/mol) = 0.0483 mol

Mole fraction of Ar, Xe(He) = 0.3247/(0.3247 + 0.0483) = 0.3247/0.3730

P(He) = P(Total) X(He)
Then, P(Total) = P(He) / X(He)
Total pressure, P(Total) = (4.00 atm) / (0.3247/0.3730) = 4.60 atm