Question

The pressure of a mixture of 64 g of oxygen, 28 g of nitrogen and 132 g of carbon dioxide gases in a closed vessel is P. Under isothermal conditions if entire oxygen is removed from the vessel, the pressure of the mixture of remaining two gases is

• A. P
• B. \(\frac{3P}{2}\)
• C. \(\frac{P}{3}\)
• D. \(\frac{2P}{3}\)

Hint:

In problems involving gas mixtures where conditions (like removing a component) change, focus on the total number of moles. The ratio of final pressure to initial pressure will be the same as the ratio of the final total moles to the initial total moles, provided volume and temperature are constant.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
According to the ideal gas law (\(PV=nRT\)) and Dalton's law of partial pressures, the total pressure of a gas mixture in a fixed volume at a constant temperature is directly proportional to the total number of moles of gas present. If some moles of gas are removed, the pressure will decrease proportionally.
Step 2: Key Formula or Approach:
Since Volume (V) and Temperature (T) are constant, \(P \propto n\), where \(n\) is the total number of moles.
This means \(\frac{P_{final}}{P_{initial}} = \frac{n_{final}}{n_{initial}}\).
We need to calculate the number of moles (\(n = \text{mass} / \text{molar mass}\)) for each gas initially and finally.
Step 3: Detailed Explanation:
First, we need the molar masses of the gases:
- Molar mass of Oxygen (O₂): \(2 \times 16 = 32\) g/mol.
- Molar mass of Nitrogen (N₂): \(2 \times 14 = 28\) g/mol.
- Molar mass of Carbon Dioxide (CO₂): \(12 + 2 \times 16 = 44\) g/mol.
Calculate the initial number of moles (\(n_{initial}\)):
- Moles of O₂: \(n_{O_2} = \frac{64 \text{ g}}{32 \text{ g/mol}} = 2\) mol.
- Moles of N₂: \(n_{N_2} = \frac{28 \text{ g}}{28 \text{ g/mol}} = 1\) mol.
- Moles of CO₂: \(n_{CO_2} = \frac{132 \text{ g}}{44 \text{ g/mol}} = 3\) mol.
- Initial total moles: \(n_{initial} = n_{O_2} + n_{N_2} + n_{CO_2} = 2 + 1 + 3 = 6\) mol.
Calculate the final number of moles (\(n_{final}\)):
The entire oxygen is removed. The remaining gases are Nitrogen and Carbon Dioxide.
- Final total moles: \(n_{final} = n_{N_2} + n_{CO_2} = 1 + 3 = 4\) mol.
Calculate the final pressure (\(P_{final}\)):
Using the proportionality \(P \propto n\):
\[ \frac{P_{final}}{P_{initial}} = \frac{n_{final}}{n_{initial}} \] \[ \frac{P_{final}}{P} = \frac{4 \text{ mol}}{6 \text{ mol}} = \frac{2}{3} \] \[ P_{final} = \frac{2}{3}P \] Step 4: Final Answer:
The pressure of the remaining mixture is \(\frac{2}{3}P\). Therefore, option (D) is correct.