A 1L flask is filled with 1.25 g of argon at 25 degrees celcium. A sample of ethane vapor is added to the same flask until the total pressure is 1.45 atm.
What is the partial pressure of argon in the flask?
mols Ar = 1.25/39.95 = 0.0313
Use PV = nRT to calculate pressure Ar.
Remember that the partial pressure of a gas doesn't depend upon what other gas is present. In effect, we ignore the ethane vapor although the partial pressure of the ethane COULD be calculated, also.
To find the partial pressure of argon in the flask, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = moles of gas
R = ideal gas constant (0.0821 L * atm / mol * K)
T = temperature in Kelvin
First, let's calculate the moles of argon in the flask:
n = mass / molar mass
n = 1.25 g / 39.948 g/mol
n = 0.0313 mol (rounded to four decimal places)
Now, let's convert the temperature from Celsius to Kelvin:
T(K) = T(C) + 273.15
T(K) = 25°C + 273.15
T(K) = 298.15 K
Next, rearrange the ideal gas law equation to solve for partial pressure:
P = (n * R * T) / V
Since the volume (V) is constant at 1 liter and the total pressure (P) in the flask is given as 1.45 atm, we can substitute the known values into the equation to solve for the partial pressure of argon:
P(argon) = (n(argon) * R * T) / V
P(argon) = (0.0313 mol * 0.0821 L * atm / mol * K * 298.15 K) / 1 L
P(argon) ≈ 0.794 atm (rounded to three decimal places)
Therefore, the partial pressure of argon in the flask is approximately 0.794 atm.