Question

3. Saturate vapor pressure of pure C&H=C1(1) and pure C6H5Br(1) are 125238 Pa and 66104 Pa, respectively. Assuming that the ideal solution of the two liquids boils at 413.2K and 1 atm, estimate the composition of the ideal solution and the composition of the vapor above the boiling solution when there is only one drop of liquid phase vaporized

Answer 1

Solution:​​​​

Let subscript 'A' stand for C₆H₅Cl

Let subscript 'B' stands for C₆H₅Br

Given Presure, P = 1 atm = 101325 Pa

Given bubble point temperature, T= 413.2 K

Vapor Pressure of C₆H₅Cl, P^{_{​​​​​​}​​​​​​s}_{​​​​​A}= 125238 Pa

Vapor Pressure of C₆H₅Br,  P^{​​​​​​s}​​​​​_{​​​​​B} = 66104 Pa

The composition of the first vapour produced from a liquid of boiling is calculated as follows.

We know Total pressure is the sum of partial pressures (Dalton's Law)

P = p_{​​​​​​A} + p_{​​​​​​B}

where   p_{​​​​A} ​​​​​​= partial pressure of C₆H_{​​​​​​​​}​​​​₆Cl

p_{​​​​​​B} = partial pressure of C₆H₅Br

We know  p​​​​​​A = X​​​​​​A * P​​​​​​s​​A ------ ( Raoult's law )

Similarly p_{​​​​​​B} = X_{​​​​​B}  * P​​​​​^​s_B

Where X_{​​​​​​A}is the mole fraction of C₆H₅Cl in liquid mixture and X_{​​​​​​B} is the mole fraction of C₆H_{​​​​5}​​​Br in liquid mixture.

P = X_{​​​​​A} * P^{​​​​​​s}_A + X​​​​_​​​B * P^s_B

P= X_{​​​​​​A} * P^{​​​​​​s}_{​​​​​A} + (1-X_A) * P^{​​​​​​s}_B

Therefore X_{​​​​​​A} = (P - P^{​​​​​​s}_{​​​​​B}) / ( P^​s_{​​​​A} - P^​s_B)

X_{​​​​​​A} = (101325 - 66104) / (125238 - 66104) = 0.5956

X_{​​​​​B} = 1 - X_{​​​​​​A} = 1 - 0.5956 = 0.4044

By Applying Dalton's Law, the partial pressure of component gas is in an ideal gas mixture is the product of its mole fraction and the total pressure.

_{​​​​​​}y_{​​​​​​A} = p_{​​​​​​A} / P

y_{​​​​​​B} = p_{​​​​​​B} / P

Where y_{​​​​​​A}and y_{​​​​​​B}are the mole fraction of C₆H₅Cl and C₆H₅Br respectively.

y_{​​​​​​A} =( X_A * P^{​​​​​​s}_​A ) / P

y_{​​​​​​A} = (0.5956 * 125238) / 101325 = 0.7362

y_{​​​​​​B} = 1 - y_{​​​​​​A} = 1 - 0.7362 = 0.2638

VAPOR COMPOSITION

C₆H₅Cl = 73.62%

C₆H₆Br = 26.38%

LIQUID COMPOSITION

C₆H₅Cl = 59.56%

C₆H₅Br = 40.44%