At temperatures above the boiling point, A behaves like an ideal gas. In a hypothetical situation, J — Physical Chemistry Chemistry Question
Gas and Liquid
At temperatures above the boiling point, A behaves like an ideal gas. In a hypothetical situation, Jacques Charles performed an experiment about the volume-temperature relationship and obtained the following result (which is not necessarily drawn to scale):
What is the volume of A at 100 oC?
Model Answer
According to the graph, the volume becomes zero at T = –100 oC. This means that the absolute temperature should be calculated as T(K) = oC + 100 (not T (K) = oC + 273.15). If V1 = 15 cm3, T1 = – 50 oC + 100 = 50 K, and T2 = 100 oC + 100 = 200 K, then we have (15 cm3 / 50 K) = (V2 / 200 K). Therefore, V2 = (15 cm3 × 200 K) / (50 K) = 60 cm3.
At equilibrium, the vapor pressures above liquids B and C are 100.1 kPa and 60.4 kPa, respectively. The two liquids B and C are mixed thoroughly at 298 K. What is the vapor pressure above a mixture containing 3 mol B and 4 mol C?
Model Answer
From Dalton’s law, the total vapor pressure is the sum of the individual vapor pressures: ptotal = pB + pC (1)
Using the Raoult’s law, the total pressure may be obtained by substituting each p term with pi° × xi, where pi° is vapor pressure above pure liquid i and xi is mole fraction of liquid i: ptotal = (pB° xB) + (pC° xC) (2)
We know from the question that there are 7 mol of liquid. We obtain the respective mole fractions x: the mole fraction of B is 3/7 and the mole fraction of C is 4/7. Substituting values of xi and pi° into equation (2) yields the total pressure Ptotal as follows:
ptotal = (100.1 kPa × 3/7) + (60.4 kPa × 4/7) = 42.9 kPa + 34.5 kPa = 77.4 kPa
What are the mole fractions of B and C above the mixture explained in question 7.2?
Model Answer
From the definition of mole fraction X, we say
The number of moles ni are directly proportional to the partial pressure pi if we assume that each vapor behaves as an ideal gas (we assume here that T and V are constant). Accordingly, we say
Substituting numbers from question 7.2:
xvapor B = 42.9 kPa / 77.4 kPa = 0.554
The mole fraction of B in the vapor is 0.554, so it contains 55.4% B. The remainder of the vapor must be C, so the vapor also contains (100 - 55.4) % = 44.6 % of C.
Note that the liquid phase comprises 43 % B and 57 % C, but the vapor contains proportionately more of the volatile B. We should expect the vapor to be richer in the more volatile component.