The vapor pressure of water at 80 degrees c is 355mmHg. A 100cm3 vessel contained wateraturated oxygen at 80 degrees c., the total gas pressure being 760 mmHg. The contents of the vessel were transferred by pumping to a 50cm3 vessel at the same temperature. What were the partial pressures of oxygen and water vapor and what was the total pressure in the final equilibrated state. Please explain reasoning, Thanks!
To solve this problem, we need to use the ideal gas law and Dalton's law of partial pressures. Let's break down the problem step by step.
Step 1: Determine the initial partial pressure of water vapor and oxygen.
Given that the water vapor is saturated at 80 degrees Celsius, we can find the vapor pressure of water at this temperature, which is 355 mmHg.
Since the total gas pressure in the 100 cm3 vessel is 760 mmHg, we can subtract the vapor pressure of water from the total pressure to find the partial pressure of oxygen.
Partial pressure of oxygen = Total pressure - Vapor pressure of water
= 760 mmHg - 355 mmHg
= 405 mmHg
So, the initial partial pressure of oxygen is 405 mmHg.
Step 2: Determine the final partial pressures of water vapor and oxygen, as well as the total pressure, after transferring the contents to the 50 cm3 vessel.
As the contents are transferred to the new vessel, the total number of gas molecules remains the same. Therefore, the total pressure will not change.
Since the temperature remains the same, the mole fractions of oxygen and water vapor will also remain constant. Therefore, the partial pressure of oxygen will also remain the same.
Partial pressure of oxygen = 405 mmHg
Now, let's calculate the partial pressure of water vapor in the final vessel. To do this, we will use the idea that the mole fraction of water vapor remains constant.
Mole fraction of water vapor = Initial moles of water vapor / Initial total moles
Now, we can determine the initial moles of water vapor and total moles using the given volumes and known molar volume of ideal gases. The molar volume is the volume occupied by one mole of gas at a given temperature and pressure, which is approximately 22.4 L/mol at standard temperature and pressure (STP).
Initial moles of water vapor = Initial volume of vessel (in liters) / Molar volume of water vapor at 80 degrees Celsius
Initial volume of vessel = 100 cm3 = 100/1000 L = 0.1 L
Molar volume of water vapor at 80 degrees Celsius = Molar volume at STP × (Initial temperature / STP temperature)
= 22.4 L/mol × (80 + 273) K / 273 K
= 22.4 L/mol × 353/273
= 28.97 L/mol
Initial moles of water vapor = 0.1 L / 28.97 L/mol
≈ 0.00345 mol
Now, we need to calculate the initial total moles of gas in the vessel. The ideal gas law can be used here:
PV = nRT
Where:
P = pressure
V = volume
n = moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
n = (PV) / (RT)
n = (760 mmHg × 0.1 L) / (0.0821 L·atm/(mol·K) × (80 + 273) K)
n ≈ 0.0355 mol
So, the initial total moles of gas in the vessel ≈ 0.0355 mol.
Now, let's determine the final partial pressure of water vapor in the new vessel.
Partial pressure of water vapor = Mole fraction of water vapor × Total pressure
The mole fraction of water vapor remains constant:
Mole fraction of water vapor = Initial moles of water vapor / Initial total moles
= 0.00345 mol / 0.0355 mol
≈ 0.0972
The total pressure remains the same:
Total pressure = 760 mmHg
Partial pressure of water vapor = 0.0972 × 760 mmHg
≈ 73.75 mmHg
So, the final partial pressure of water vapor is approximately 73.75 mmHg.
Finally, the total pressure in the final equilibrated state is the same as before:
Total pressure = 760 mmHg
To summarize:
- The partial pressure of oxygen remains the same at 405 mmHg.
- The partial pressure of water vapor in the final vessel is approximately 73.75 mmHg.
- The total pressure in the final equilibrated state remains the same at 760 mmHg.