Question 6.4.3: Bubble- and Dew-Point Calculations 1. Calculate the temperat......

Let A = benzene and B = toluene.
1. Equation 6.4-4 may be written in the form

P= x_{A} p^{*}_{A}(T_{bp} ) + x_{B}p_{B}^{*}(T_{bp} ) +…             (6.4-4)

f(T_{bp} ) = 0.400 p^{*}_{A}(T_{bp} ) + 0.600 p^{*}_{B}(T_{bp} )  –  760  mm  Hg = 0

The solution procedure is to substitute for p^{*}_{A} and p^{*}_{B} either the Antoine equation (Table B.4) or the APEx function AntoineP, and then use any equation solving program or (on a spreadsheet) Excel’s Solver or the Goal Seek tool to determine the boiling point. The solution is \boxed{T_{bp} = 95.1°C} . At this temperature, Equation 6.4-1 yields

Raoult’s Law:             p_{A}\equiv \boxed{y_{A} P = x_{A}p^{*}_{A}(T)}           (6.4-1)

p_{A} = 0.400 (1181  mm  Hg)= 472.5  mm  Hg \\ p_{B}= 0.600 (479  mm  Hg)=287.5  mm  Hg \\\left. \Large{\Downarrow} \right. \\ P =(472.5 + 287.5)  mm  Hg = 760  mm  Hg

Furthermore, from Equation 6.4-5,

y_{i}= \frac{p_{i}}{P_{bp}}=\frac{x_{i} p^{*}_{i}(T)}{P_{bp}}              (6.4-5)

y_{A}= \frac{472.5}{760.0} = \boxed{0.622  mol  benzene/mol} \\ y_{B} = 1  –  y_{A} = \boxed{0.378  mol  toluene/mol}

Since the composition of the liquid was given, this was a \boxed{bubble-point} calculation.
2. Equation 6.4-7 may be written as

x_{A} + x_{B} +x_{C} + … = 1\\ \left. \Large{\Downarrow} \right. Equation  6.4-6 \\ \frac{y_{A}P}{p^{*}_{A} (T_{dp})}+ \frac{y_{B}P}{p^{*}_{B} (T_{dp})} + … = 1              (6.4-7)

f(T_{dp}) = \frac{(0.100)(760  mm  Hg)}{p^{*}_{A} (T_{dp})} + \frac{(0.100)(760  mm  Hg)}{p^{*}_{B} (T_{dp})}  –  1.00 = 0

A solution procedure similar to that in Part (1) leads to the result \boxed{T_{dp} = 52.4°C} , at which temperature p^{*}_{A} = 297.4 mm Hg and p^{*}_{B} = 102.1 mm Hg. Then, from Equation 6.4-6,

x_{i} = \frac{y_{i}P}{p_{i}^{*} (T_{dp})},      i= A, B, C, …   excluding  G             (6.4-6)

x_{A} = \frac{0.100(760  mm  Hg)}{p^{*}_{A} (52.4°C)} = \boxed{0.256  mol  benzene/mol} \\ x_{B} = 1  –  x_{A} = \boxed{0.744  mol  toluene/mol}

The composition of the vapor was given and that of the liquid was calculated; therefore, this was a \boxed{dew-point} calculation.
3. The vapor pressures of benzene and toluene at 80°C are determined from the Antoine equation to be 757.7 mm Hg and 291.2 mm Hg, respectively. Assuming that nitrogen is insoluble in the condensate, Equation 6.4-8 gives

P_{dp} = \frac{1}{\frac{y_{A}}{p^{*}_{A}(T)}+ \frac{y_{B}}{p^{*}_{B}(T)} + \frac{y_{C}}{p^{*}_{C}(T)} + …}              (6.4-8)

P= \frac{1}{(0.150/757.7  mm  Hg)+(0.100/291.2  mm Hg)} = \boxed{1847  mm  Hg} \\ x_{A} = \frac{y_{A} P}{p^{*}_{A}} = \frac{0.150(1847  mm  Hg)}{757.7  mm  Hg}= \boxed{0.366  mol  benzene/mol} \\ x_{B} = 1  –  x_{A} = \boxed{0.634  mol  toluene/mol}