What is the partial pressure (in mm Hg) of neon in a 4.00 L vessel that contains 0.442 mol of methane, 0.414 mol of ethane, and 0.859 mol of neon at a total pressure of 943 mm Hg?
Use mole fraction.
mole fraction Ne = moles Ne/total moles.
partial pressure Ne = mole fraction Ne x total P.
To find the partial pressure of neon (Ne) in the vessel, we need to use the concept of mole fraction.
Step 1: Calculate the total number of moles of gas in the vessel.
Total moles of gas = moles of methane + moles of ethane + moles of neon
Total moles of gas = 0.442 mol + 0.414 mol + 0.859 mol
Total moles of gas = 1.715 mol
Step 2: Calculate the mole fraction of neon.
Mole fraction of neon (X_neon) = moles of neon / total moles of gas
X_neon = 0.859 mol / 1.715 mol
X_neon = 0.501
Step 3: Calculate the partial pressure of neon.
Partial pressure of neon = mole fraction of neon × total pressure
Partial pressure of neon = 0.501 × 943 mm Hg
Partial pressure of neon ≈ 472.743 mm Hg
Therefore, the partial pressure of neon in the 4.00 L vessel is approximately 472.743 mm Hg.
To determine the partial pressure of neon in the vessel, we need to use the concept of mole fraction.
Step 1: Calculate the total number of moles of gas in the vessel.
To find the total number of moles, add the moles of each gas:
Total moles = moles of methane + moles of ethane + moles of neon
Total moles = 0.442 mol + 0.414 mol + 0.859 mol
Step 2: Calculate the mole fraction of neon.
The mole fraction of a gas is the ratio of the moles of that gas to the total moles of all the gases present.
Mole fraction of neon = moles of neon / Total moles
Step 3: Calculate the partial pressure of neon.
The partial pressure of a gas in a mixture is its mole fraction multiplied by the total pressure of the mixture.
Partial pressure of neon = Mole fraction of neon × Total pressure
Given that the total pressure is 943 mm Hg and the moles of neon is 0.859 mol, we can substitute these values into the equation:
Partial pressure of neon = (0.859 mol / (0.442 mol + 0.414 mol + 0.859 mol)) × 943 mm Hg
Now, let's calculate the partial pressure of neon:
Partial pressure of neon = (0.859 mol / (0.442 mol + 0.414 mol + 0.859 mol)) × 943 mm Hg
Partial pressure of neon = (0.859 mol / 1.715 mol) × 943 mm Hg
Dividing 0.859 mol by 1.715 mol gives us a mole fraction of neon of approximately 0.501. Multiplying this by the total pressure of 943 mm Hg gives us the partial pressure of neon.
Partial pressure of neon = 0.501 × 943 mm Hg
Therefore, the partial pressure of neon in the 4.00 L vessel is approximately 472 mm Hg.