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 .
a) Partial pressure of argon is 0.673 atm.
b) Partial pressure of ethane is 0.427 atm.
According to the ideal gas equation:'
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 =
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.
Thus partial pressure of ethane is 0.427 atm.
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
Partial Pressure = Total Pressure * Mol Fraction
Partial Pressure = (1.2 atm)(0.562) = 0.674 atm