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# Engineering

The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture.  The mole fraction of liquids B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :
 

  • Option 1)

    450\; mmHg,0.5,0.5

  • Option 2)

    450\; mmHg,0.4,0.6

  • Option 3)

    500\; mmHg,0.4,0.6

     

  • Option 4)

    500\; mmHg,0.5,0.5

Answers (1)
S solutionqc

P_{A}^{o}=400\; mmHg\; \; \; \; \; \; \; P_{B}^{o}=600\; mmHg

X_{B}=0.5\; \; \; \; \; X_{A}=0.5

\rightarrow P_{s}=X_{A}P_{A}^{o}+X_{B}P_{B}^{o}=0.5\times400+0.5\times600=500\; mmHg

P_{A}=x_{A}P_{A}^{o}=Y_{A}P_{s}^{o}(for \; vapour\; phase)

                                    \downarrow

                                  mole fraction of A in vapour phase

Y_{A}=\frac{P_{A}}{P_{S}}=\frac{x_{A}P_{A}^{o}}{P_{s}}=\frac{0.5\times400}{500}=0.4

Y_{B}=1-0.4=0.6

P_{s}=500\; mmHg\; \; \; Y_{A}=0.4\; \; \; Y_{B}=0.6


Option 1)

450\; mmHg,0.5,0.5

Option 2)

450\; mmHg,0.4,0.6

Option 3)

500\; mmHg,0.4,0.6

 

Option 4)

500\; mmHg,0.5,0.5

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