A 1.00 L flask is filled with 1.30 g of argon at 25 °C. A sample of ethane vapor is added to the same flask at constant volume and temperature until the total pressure is 1.350 atm.
Part A:
What is the partial pressure of argon in the flask?
The partial pressure of Ar will be the pressure of Ar as it is alone in the flask. So we will use the ideal gas equation to find the pressure.
Par * V = n * R * T
Where P is the partial pressure of argon, V is the volume, n is the number of moles, R is a constant and T is the temperature.
Par = ?
V = 1.00 L
R = 0.082 atm*L/(mol*K)
T = 273.15 + 25 = 298.15 K
T = 298.15 K
n is the number of moles of the sample. We know that the mass of the sample is 1.30 g, we can convert that mass into moles using the molar mass of Ar.
molar mass of Ar = 39.95 g/mol
n = mass of Ar/molar mass of Ar
n = 1.30 g/(39.95 g/mol)
n = 0.03254 mol
Finally we can replace these values into the formula and find the answer to our problem.
Par * V = n * R * T
Par = n * R * T/V
Par = 0.03254 mol * 0.082 atm*L/(mol*K) * 298.15 K/(1.00 L)
Par = 0.796 atm
Answer: the partial pressure of argon in the flask is 0.796 atm.
Part B:
What is the partial pressure of ethane in the flask?
According to Dalton's Law the total pressure inside the flask is the sum of the partial pressures.
Total pressure = Partial pressure of argon + Partial pressure of ethane
Pethane = Total pressure - Par
Pethane = 1.350 atm - 0.796 atm
Pethane = 0.554 atm
Answer: the partial pressure of ethane in the flask is 0.554 atm.