Question

A flask contains 5 moles of an ideal gas at 25C and 2 atm pressure. The temperature of the gas is raised to 50C. What fraction of the gas must be removed to keep the pressure in the flask constant.

Answer 1

Initial Initial molee of ideal gas, na = 5 moles temperature, To 25°C = 25+273 = 298K Final temperature, T2 = 50°c = 50+273 = 32BK flask also Pressure = constant = 2 atm A Flask remain same, so volume of remain same Volume of flask = constant =V. So, according to ideal gas equation, PV=ART Where P and v are constant So ART = constant le HRT, = n₂ R Tz 5 mol x R x 298K = N₂ X RX 323K So, n₂ = 5 mol x &x2984 XX 323 K = 4.613 mol Moles & gas removed = 5-4.613 = 0.387 mol. Mole fraction of gas must removed = 0.387 mol = 0.0774