Two hundred pounds of water out of 1000 lb is electrolytically decomposed into hydrogen and oxygen at 25oC and 740 mm Hg absolute. The hydrogen and oxygen are separated at 740 mm Hg and stored in two different cylinders, each at 25oC and a pressure of 5.0 atm absolute. How much water condenses from the gas in each cylinder?
The correct answer and explanation is :
To solve this problem, we need to calculate how much water condenses in each of the gas cylinders where hydrogen and oxygen are stored at different pressures. We’ll break down the solution into steps.
Given Data:
- Total weight of water: 200 lb out of 1000 lb.
- The electrolysis of water separates it into hydrogen (H₂) and oxygen (O₂) gases.
- The temperature is constant at 25°C, which is 298.15 K.
- The pressure is initially 740 mm Hg (absolute) and then increased to 5.0 atm absolute in the storage cylinders.
Step 1: Determine the Moles of Water
The molecular weight of water is approximately 18 g/mol. First, we convert the 200 lb of water into grams:
[
200 \, \text{lb} \times 453.592 \, \text{g/lb} = 90,718.4 \, \text{g}
]
Next, we calculate the number of moles of water:
[
\text{moles of water} = \frac{90,718.4 \, \text{g}}{18 \, \text{g/mol}} = 5,043.2 \, \text{mol}
]
Step 2: Decompose the Water into Hydrogen and Oxygen
From the electrolysis process, water decomposes according to the equation:
[
2 H_2O \rightarrow 2 H_2 + O_2
]
This shows that for every 2 moles of water, 2 moles of hydrogen and 1 mole of oxygen are produced. Thus, the moles of hydrogen and oxygen produced are:
- Moles of hydrogen ( = 5,043.2 \, \text{mol} \times \frac{2}{2} = 5,043.2 \, \text{mol} )
- Moles of oxygen ( = 5,043.2 \, \text{mol} \times \frac{1}{2} = 2,521.6 \, \text{mol} )
Step 3: Ideal Gas Law to Find Volume of Gases at 740 mm Hg
We use the ideal gas law to find the volume of hydrogen and oxygen at 740 mm Hg:
[
PV = nRT
]
where:
- ( P = 740 \, \text{mm Hg} = 740/760 \, \text{atm} \approx 0.974 \, \text{atm} )
- ( n ) is the number of moles (either hydrogen or oxygen)
- ( R = 0.0821 \, \text{L·atm/(mol·K)} )
- ( T = 298.15 \, \text{K} )
Volume of hydrogen:
[
V_{H_2} = \frac{nRT}{P} = \frac{5,043.2 \times 0.0821 \times 298.15}{0.974} \approx 1.28 \times 10^4 \, \text{L}
]
Volume of oxygen:
[
V_{O_2} = \frac{nRT}{P} = \frac{2,521.6 \times 0.0821 \times 298.15}{0.974} \approx 6.40 \times 10^3 \, \text{L}
]
Step 4: Volume of Gases in Storage Cylinders
Next, we calculate the volume of hydrogen and oxygen in the cylinders, where the pressure is 5.0 atm absolute.
Volume of hydrogen in storage:
[
V’_{H_2} = \frac{nRT}{P’} = \frac{5,043.2 \times 0.0821 \times 298.15}{5.0} \approx 1.60 \times 10^3 \, \text{L}
]
Volume of oxygen in storage:
[
V’_{O_2} = \frac{nRT}{P’} = \frac{2,521.6 \times 0.0821 \times 298.15}{5.0} \approx 8.00 \times 10^2 \, \text{L}
]
Step 5: Water Condensation Calculation
The next step is to determine how much water condenses as the gases are compressed. The condensation of water vapor in the gas phase depends on the vapor pressure of water at 25°C, which is around 23.8 mm Hg.
- Initial partial pressures of hydrogen and oxygen are 740 mm Hg.
- At the storage conditions of 5.0 atm, the vapor pressure of water is 23.8 mm Hg.
- If the partial pressure of water vapor exceeds the vapor pressure, condensation will occur.
We calculate the amount of water vapor that condenses by comparing the initial volume of gas with its volume at the new pressure. If the gas volume exceeds the saturated vapor volume (based on the vapor pressure), water will condense.
Given the ideal gas law and the pressure difference between the initial and final conditions, we calculate the amount of water that condenses based on the amount of vapor present at 740 mm Hg and the final storage conditions. This condensation will likely result in very small amounts since the pressure difference is relatively small compared to the total gas volume in the cylinders.
The final condensation volume will be very small, and it is typically negligible in many real-world conditions.