A sample of oxygen gas is collected by displacement of water at 25°C and 1.11 atm total pressure. If the vapor pressure of water is 23.756 mm Hg at 25°C, what is the partial pressure of the oxygen gas in the sample?
Ptotal = pO2 + pH2O
You know Ptotal and pH2O, solve for pO2.
zaAZA
To find the partial pressure of the oxygen gas in the sample, we need to subtract the vapor pressure of water from the total pressure.
Given:
Total pressure (Ptotal) = 1.11 atm
Vapor pressure of water (Pwater) = 23.756 mm Hg
First, let's convert the vapor pressure of water to atm:
Pwater = 23.756 mm Hg * (1 atm / 760 mm Hg) = 0.0312487 atm
Now, subtract the vapor pressure of water from the total pressure to find the partial pressure of oxygen gas:
Partial pressure of oxygen gas (Poxygen) = Ptotal - Pwater
Poxygen = 1.11 atm - 0.0312487 atm
Poxygen ≈ 1.07976 atm
Therefore, the partial pressure of the oxygen gas in the sample is approximately 1.07976 atm.
To find the partial pressure of the oxygen gas in the sample, we need to subtract the vapor pressure of water from the total pressure.
Given:
Total pressure (Ptotal) = 1.11 atm
Vapor pressure of water (Pwater) = 23.756 mm Hg
We need to convert the vapor pressure of water from mm Hg to atm, as the total pressure is given in atm.
1 mm Hg is equal to 0.00131579 atm. So, we can convert the vapor pressure of water as follows:
Pwater (converted to atm) = 23.756 mm Hg × 0.00131579 atm/mm Hg
Pwater (converted to atm) ≈ 0.03129 atm
Now that we have the vapor pressure of water in atm, we can find the partial pressure of the oxygen gas by subtracting the vapor pressure from the total pressure:
Partial pressure of oxygen gas (PO2) = Ptotal - Pwater
PO2 = 1.11 atm - 0.03129 atm
PO2 ≈ 1.0797 atm
Therefore, the partial pressure of the oxygen gas in the sample is approximately 1.0797 atm.