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: Calculate Moles of Each Component: - Molar Mass of \(O_2\) = 32 g/mol. Moles of \(O_2\) (\(n_1\)) = \( \frac{64}{32} = 2 \) mol. - Molar Mass of \(N_2\) = 28 g/mol. Moles of \(N_2\) (\(n_2\)) = \( \frac{28}{28} = 1 \) mol. - Molar Mass of \(CO_2\) = 44 g/mol. Moles of \(CO_2\) (\(n_3\)) = \( \frac{132}{44} = 3 \) mol. Total initial moles, \( n_{\text{total}} = 2 + 1 + 3 = 6 \) moles.
Step 2: Relation between Pressure and Moles: According to Dalton's Law and the Ideal Gas Equation \( PV = nRT \), at constant volume and temperature, pressure is directly proportional to the number of moles. \[ P \propto n \] Initial Pressure \( P \propto 6 \).
Step 3: Calculate Pressure After Removal: Oxygen (2 moles) is removed. Remaining moles \( n_{\text{final}} = n_2 + n_3 = 1 + 3 = 4 \) moles. Let the new pressure be \( P' \). \[ P' \propto 4 \]
Step 4: Ratio of Pressures: \[ \frac{P'}{P} = \frac{4}{6} = \frac{2}{3} \] \[ P' = \frac{2P}{3} \]
Step 5: Final Answer: The new pressure is \( \frac{2P}{3} \).