Liquids A and B form an ideal solution. A certain solution contains 25 mole % of A, whereas the vapor in equilibrium with the solution at 25 degrees C contains 50. mole % of A. The heat of vaporization of A is 5 kcal/mol; that of B is 7 kcal/mol.
a) Calculate the ratio of the vapor pressure of pure A to that of pure B at 25 degrees C.
b) Calc. value for the same ratio at 100.degrees C.
My first thought was this:
P1(a)=X1(a)P1*(a)
P2(a)=X2(a)P2*(a)
P1(b)=X1(b)P1*(b)
P2(b)=X2(b)P2*(b)
where P1(a)=P2(a)=P1(b)=P2(b)
but if I did
.50P2*a=.25P1*a,
P1/2*a would = 0, as would P2/1*b if I did the same for b.
So how do I really approach this?
To solve this problem, you can use Raoult's law, which states that the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the liquid phase.
a) Let's start by calculating the mole fraction of component A in the liquid phase at 25 degrees Celsius. The solution contains 25 mole % of A, which corresponds to a mole fraction of 0.25.
Now, we can set up the equation using Raoult's law for component A:
P1(a) = X1(a) * P*(a)
where P1(a) is the vapor pressure of A in the equilibrium solution, X1(a) is the mole fraction of A in the liquid phase, and P*(a) is the vapor pressure of pure A at the given temperature.
Similarly, since the vapor in equilibrium with the solution at 25 degrees Celsius contains 50 mole % of A, the mole fraction of A in the vapor phase, X2(a), is 0.50.
So, we have:
P2(a) = X2(a) * P*(a)
Since in both equations, P*(a) is the same, we can equate the two equations and solve for the ratio of vapor pressures:
P1(a) / P2(a) = X1(a) / X2(a) = 0.25 / 0.50 = 0.5
Therefore, the ratio of the vapor pressure of pure A to that of pure B at 25 degrees Celsius is 0.5.
b) To calculate the ratio of vapor pressures at 100 degrees Celsius, we can use the same approach. However, we need to consider the effect of temperature on the vapor pressures.
Using Raoult's law, the equations become:
P1(a) = X1(a) * P*(a)
P2(a) = X2(a) * P*(a)
Now, we need to determine the new mole fractions of A in the liquid and vapor phases at 100 degrees Celsius.
To do this, we can use the formula:
X2(a) = X1(a) * P1*(a) / P2*(a)
where P1*(a) and P2*(a) are the vapor pressures of pure A at 25 degrees Celsius and 100 degrees Celsius, respectively.
Given that the heat of vaporization of A is 5 kcal/mol, we can use the Clausius-Clapeyron equation to relate the vapor pressures at different temperatures:
ln(P2*(a) / P1*(a)) = (ΔHvap(a) / R) * (1 / T1 - 1 / T2)
where ΔHvap(a) is the heat of vaporization of A, R is the ideal gas constant, T1 is the initial temperature (25 degrees Celsius), and T2 is the final temperature (100 degrees Celsius).
Substituting the values, we can calculate the ratio of vapor pressures at 100 degrees Celsius.