Calculate the partial pressure (in atm) of propane in a mixture that contains equal weights of propane (C3H8) and butane (C4H10) at 18°C and 746 mmHg
Ppropane = Xpropane*Ptotal
Ptotal = 746/760 = ?atm (the problem asks for atm).
So all we need to do is to determine Xpropane and plug into the above.
molar mass C3H8 = about 44
molar mass C4H10 = about 58. YOu can do all of this more accurately.
Xpropane = (g/44)/[(g/44)+(g/58)]
Solve for X and substitute into the above. I get something like 420 mm (but the problem asks for atm).
To calculate the partial pressure of propane (C3H8) in the mixture, we need to use Dalton's law of partial pressures. According to Dalton's law, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each individual gas.
First, let's convert the given pressure from mmHg to atm:
1 atm = 760 mmHg
Pressure in atm = 746 mmHg / 760 mmHg/atm
Pressure in atm = 0.98 atm
Next, we need to calculate the mole fraction of propane in the mixture. The mole fraction (X) is the ratio of the number of moles of a component to the total number of moles in the mixture.
Since the mixture contains equal weights of propane and butane, we can assume that the number of moles of propane and butane is equal.
Let's assume we have 100 grams of the mixture, so we have 50 grams of propane and 50 grams of butane.
To calculate the number of moles, we divide the mass of each component by its molar mass:
Molar mass of propane (C3H8) = (3 * atomic mass of carbon) + (8 * atomic mass of hydrogen)
= (3 * 12.01 g/mol) + (8 * 1.01 g/mol)
= 44.1 g/mol
Number of moles of propane = 50 grams / 44.1 g/mol
= 1.134 moles
Since the number of moles of butane is also 1.134 moles, the total number of moles in the mixture is 2.268 moles.
Now, we can calculate the mole fraction of propane:
Mole fraction of propane = moles of propane / total moles
= 1.134 moles / 2.268 moles
= 0.5
Finally, we can calculate the partial pressure of propane using Dalton's law:
Partial pressure of propane = mole fraction of propane * total pressure
= 0.5 * 0.98 atm
= 0.49 atm
Therefore, the partial pressure of propane in the mixture is 0.49 atm.
To calculate the partial pressure of propane (C3H8) in the mixture, we need to use Dalton's law of partial pressures. According to Dalton's law, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each individual gas.
Step 1: Convert the given pressure from mmHg to atm.
1 atm = 760 mmHg
746 mmHg × (1 atm / 760 mmHg) = 0.980 atm
Step 2: Calculate the moles of propane and butane in the mixture.
Since the mixture contains equal weights of propane and butane, we can assume that the mole ratio is also 1:1.
1 mole of propane (C3H8) = 3 molar masses of carbon = 3 × 12.01 g/mol = 36.03 g/mol
1 mole of butane (C4H10) = 4 molar masses of carbon = 4 × 12.01 g/mol = 48.04 g/mol
Step 3: Calculate the moles of propane and butane using their respective molar masses and equal weights.
Moles of propane = Mass of propane (g) / Molar mass of propane (g/mol)
Moles of butane = Mass of butane (g) / Molar mass of butane (g/mol)
Since the weights are equal, let's assume each is 1 gram for simplicity.
Moles of propane = 1 g / 36.03 g/mol = 0.0278 mol
Moles of butane = 1 g / 48.04 g/mol ≈ 0.0208 mol
Step 4: Calculate the partial pressure of propane using Dalton's law of partial pressures.
Partial pressure of propane = Total pressure × (Moles of propane / Total moles)
In this case, the total moles are the sum of the moles of propane and butane.
Total moles = Moles of propane + Moles of butane
Total moles = 0.0278 mol + 0.0208 mol = 0.0486 mol
Partial pressure of propane = 0.980 atm × (0.0278 mol / 0.0486 mol) ≈ 0.560 atm
Therefore, the partial pressure of propane in the mixture is approximately 0.560 atm.