2 liquids a and b have vapour pressure of 500 mm hg and 200 mm hg .calculate mole fraction of a at which 2 liquids have equal partial pressure
I'm not sure I understand the problem. I interpret this problem as unmixed. If that is true, then you want
Pa = Xa*Poa = Xa*500
Pb = Xb*Pob = Xb*200 and since Pa = Pb, then
Xa*500 = Xb*200
Xa + Xb = 1 and solve for Xa and Xb. I get something like 143 mm Hg but that is approximate. Let me know if this is not the correct interpretation.
I also think so
To calculate the mole fraction of liquid A at which two liquids have equal partial pressure, we need to use Raoult's law.
Raoult's law states that the partial pressure of a component in a liquid mixture is equal to the product of its mole fraction and the vapor pressure of the pure component.
Let's denote the mole fraction of liquid A as xA.
According to Raoult's law:
Partial pressure of A (PA) = xA * vapor pressure of A (P°A)
Partial pressure of B (PB) = (1 - xA) * vapor pressure of B (P°B)
Since the partial pressures of A and B are equal, we can set up the equation:
PA = PB
xA * P°A = (1 - xA) * P°B
Now let's substitute the given values:
P°A = 500 mmHg
P°B = 200 mmHg
Solving the equation:
xA * 500 mmHg = (1 - xA) * 200 mmHg
500xA = 200 - 200xA
700xA = 200
xA = 200 / 700
xA ≈ 0.286
Therefore, the mole fraction of liquid A at which the two liquids have equal partial pressure is approximately 0.286.