What is the partial pressure of argon, PAr, in the flask?

What is the partial pressure of argon, PAr, in the flask? A 1.00 L flask is filled with 1.10 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.350 atm .

Answers

a) Partial pressure of argon is 0.673 atm.

b) Partial pressure of ethane is 0.427 atm.

Explanation:

According to the ideal gas equation:'

PV=nRT

P = Pressure of the argon gas = ?

V= Volume of the gas = 1.00 L

T= Temperature of the gas = 25°C = 298 K       (0°C = 273 K)

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas = frac{text {given mass}}{text {Molar mass}}=frac{1.10g}{40g/mol}=0.0275moles

P=frac{nRT}{V}=frac{0.0275times 0.0821times 298}{1.00}=0.673atm

Thus the partial pressure of argon is 0.673 atm.

b) According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_1+p_2

p_{total}=p_{Ar}+p_{ethane}

1.100=0.673+p_{ethane}

1.100-0.673=p_{ethane}

p_{ethane}=0.427atm

Thus partial pressure of ethane is 0.427 atm.

Assuming ideal gas, we can solve the total number of moles in the system as:

PV=nRT
Solving for n,
n = PV/RT = (1.2 atm)(1 L)/(0.0821 L·atm/mol·K)(25 + 273 K)
n = 0.049 

Now, compute for he moles of Argon knowing that its molar mass is 39.95 g/mol.

Mol Ar: 1.10 g/39.95 g/nol = 0.0275 mol
Mol fraction of Ar = 0.0275/0.049 = 0.562

Thus,
Partial Pressure = Total Pressure * Mol Fraction
Partial Pressure = (1.2 atm)(0.562) = 0.674 atm

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