Question

A metallic container of fixed volume of 2.5×10−3m3 immersed in a large tank of temperature 27°C contains two compartments separated by a freely movable wall. Initially, the wall is kept in place by a stopper so that there are 0.02 mol of the nitrogen gas on one side and 0.03 mol of the oxygen gas on the other side, each occupying half the volume. When the stopper is removed, the wall moves and comes to a final position. The movement of the wall is controlled so that the wall moves in infinitesimal quasi-static steps.

(a) The final volumes of the two sides, assuming ideal gas behavior for the nitrogen and oxygen gases, are:
a. 0.00125 m³ and 0.00125 m³
b. 0.001 m³ and 0.0015 m³
c. 0.0015 m³ and 0.001 m³
d. 0.001 m³ and 0.001 m³

Answer 1

The final volumes of the two compartments would be 0.00125 m³ each.

Step-by-step explanation:

Initially, the metallic container is divided into two compartments with equal volumes. One side contains 0.02 mol of nitrogen gas and the other side contains 0.03 mol of oxygen gas. When the stopper is removed, the wall moves and the gases mix. Since the gases are ideal, they follow the ideal gas law equation PV=nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Using the ideal gas law equation, we can calculate the final volumes of the two sides. Since the total volume of the container is fixed at 2.5x10⁻³ m³, and the nitrogen gas occupies half the initial volume, its final volume would be 0.00125 m³. Similarly, the final volume of the oxygen gas would also be 0.00125 m³.

Therefore, the correct answer is (a) 0.00125 m³ and 0.00125 m³.