The initial volume and pressure of a given mass of gas is V and 3.P. What is its pressure if its volume is increased to 2V at constant temperature?

The pressure of the gas will decrease to \(frac{3}{2}\) P when its volume is increased to 2V at constant temperature. This can be explained using Boyle's law, which states that the pressure of a gas is inversely proportional to its volume, when temperature is held constant. In other words, as the volume of a gas increases, its pressure will decrease proportionally and vice versa. Given that the initial volume and pressure of the gas are V and 3.P, respectively, we can write: P1V1 = P2V2 where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume. Substituting the given values, we get: (3.P) V = P2 (2V) Simplifying this equation, we get: P2 = (3.P) V / (2V) P2 = \(frac{3}{2}\) P Therefore, the pressure of the gas will decrease to \(frac{3}{2}\) P when its volume is increased to 2V at constant temperature.