science
posted by mahatta on .
15. Two 5 dm3 flasks are connected by by a narrow tube of negligible volume. Initially the two flasks are both at 27 oC and contain a total of 2 moles of an ideal gas. One flask is heated to a uniform temperature of 127 oC while the other is kept at 27 oC. Assuming their volume does not alter, calculate the number of moles of gas in each flask of the gas and the final pressure.

Two 5 dm3 flasks are connected by a narrow tube of negligible volume. Initially the two flasks are both at 27oC and contain a total of 2 moles of an ideal gas. One flask is heated to a uniform temperature of 127oC while the other is kept at 27oC. Assuming their volume does not alter, calculate the number of moles of gas in each flask of the gas and the final pressure.
Solution: As their volumes are kept constant, P then varies directly with T. A number of 1 mole (= 2/1) of gas is comprehensively distributed within each flask.
In the initial condition on ideal gas, PV = nRT or P x (2x5) = 2 x 0.082 x (273+27) => P equals 4.92 atm as an initial one.
Anyway, as the flask of our consideration is heated to the temperature of 127 degree Celcius, a number of total moles are kept constant at 2 moles, but leaking from the flask of our consideration to the other one, for example X mol, leaving the flask 1  X mole, and the other one 1 + X mol.
Heated flask: P’V = nRT
=> P’x V = (1  X) x R x (273+127)
Likewise, temptcontrolling flask: P’V = nRT
=> P’ x V = (1 + X) x R x (273+27)
Then, X = 0.143 mol. That is, heated flask will contain gas of 1  0.143 = 0.857 mol and the other one 1.143 mol.
Considering the heated flask, PV = nRT => P x 5 = 0.857 x 0.082 x (273+127) = 5.62 atm as a final pressure gauged. We can check for the correctness by considering the other flask; PV = nRT => P x 5 = 1.143 x 0.082 x (273+27) = 5.62 atm.