A given mass of gas occupies 2 dm3 at 300k. At what temperature will its volume be doubled keeping the pressure constant?

Answer Details

This is a problem in the field of thermodynamics, specifically dealing with the ideal gas law. The ideal gas law is a formula that relates the pressure, volume, and temperature of an ideal gas. It is written as follows: PV = nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas in Kelvin. In this problem, we are told that the mass of the gas is constant, so the number of moles of gas is also constant. Therefore, we can write the ideal gas law as: P1V1 = P2V2 where P1 and V1 are the initial pressure and volume of the gas, and P2 and V2 are the final pressure and volume of the gas. We are told that the pressure is constant, so we can simplify the equation to: V1/T1 = V2/T2 where T1 is the initial temperature of the gas and T2 is the final temperature of the gas. We are given that the initial volume of the gas is 2 dm^3, and we want to find the temperature at which its volume will be doubled. Therefore, we can set V2 = 4 dm^3. Plugging in the known values, we get: 2/300 = 4/T2 Simplifying this equation, we get: T2 = (300 x 4) / 2 = 600 K Therefore, the answer is option D: 600 K. To summarize, we used the ideal gas law and the fact that the number of moles of gas is constant to write an equation relating the initial and final temperature and volume of the gas. We then solved for the final temperature when the volume is doubled, given the initial volume and temperature of the gas.