NEET 2025 Physics Ideal Gas Law MCQ Question

To solve the problem, we need to determine the equilibrium pressure after removing the partition between two chambers containing ideal gases. We can use the ideal gas law and the principle of conservation of moles to find the final pressure.

Given Data

• Volume of chamber 1, $V_1 = 2 \, \text{litres}$
• Volume of chamber 2, $V_2 = 3 \, \text{litres}$
• Moles of gas in chamber 1, $n_1 = 5 \, \text{moles}$
• Moles of gas in chamber 2, $n_2 = 4 \, \text{moles}$
• Pressure in chamber 1, $p_1 = 1 \, \text{atm}$
• Pressure in chamber 2, $p_2 = 2 \, \text{atm}$

Step 1: Calculate the initial total number of moles

The total number of moles before the partition is removed is:

$$n_{\text{total}} = n_1 + n_2 = 5 + 4 = 9 \, \text{moles}$$

Step 2: Calculate the initial total pressure

Using the ideal gas law, we can express the pressures in each chamber. The ideal gas law is given by:

$$PV = nRT$$

Assuming $R$ and $T$ are constant (since the system is insulated thermally), we can express:

For chamber 1:

$$p_1V_1 = n_1RT \implies p_1 = \frac{n_1RT}{V_1}$$

For chamber 2:

$$p_2V_2 = n_2RT \implies p_2 = \frac{n_2RT}{V_2}$$

Step 3: Calculate the equilibrium pressure after removing the partition

After removing the partition, the total volume available for the gases becomes:

$$V_{\text{total}} = V_1 + V_2 = 2 + 3 = 5 \, \text{litres}$$

Using the ideal gas law, the final pressure $p_f$ can be calculated using the total number of moles and the total volume:

$$p_f = \frac{n_{\text{total}}RT}{V_{\text{total}}}$$

To isolate $p_f$, we can express it in terms of the initial conditions:

$$p_f = \frac{(n_1 + n_2)RT}{V_1 + V_2}$$

Step 4: Calculate the effective pressure using the individual contributions

The final pressure can also be found by considering the individual contributions from each chamber. The pressure contributed by each gas can be calculated using the ideal gas law in each chamber:

Using the volumes of the gases:

1. For gas in chamber 1 after mixing: $$p_{1f} = \frac{n_1RT}{V_{\text{total}}} = \frac{5RT}{5} = RT$$
2. For gas in chamber 2 after mixing: $$p_{2f} = \frac{n_2RT}{V_{\text{total}}} = \frac{4RT}{5}$$

To find the final pressure:

$$p_f = \frac{5RT + 4RT}{5} = \frac{9RT}{5}$$

Now, we also know that $p_f$ should correspond to the weighted average of pressures based on their initial contributions:

From the pressures:

$$p_f = \frac{p_1V_1 + p_2V_2}{V_{\text{total}}} = \frac{(1 \, \text{atm} \times 2) + (2 \, \text{atm} \times 3)}{5} = \frac{2 + 6}{5} = \frac{8}{5} = 1.6 \, \text{atm}$$

Conclusion

Thus, the equilibrium pressure after the partition is removed is:

$$\boxed{1.6 \, \text{atm}}$$

Why Other Options are Incorrect

• 1.3 atm: This pressure would imply a greater total volume contribution from the gases than what is calculated based on the moles and pressures.
• 1.4 atm: Similar reasoning, the calculated pressure does not align with the pressures and volumes provided.
• 1.8 atm: This pressure is higher than the calculated average, suggesting an imbalance in