Solutions Question 80
Question: An ideal mixture of liquids A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to the second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of C, $ P_c^{0} $ =0.8 atm. Calculate the vapour pressures of pure A and pure B
Options:
A) $ P_A^{o} $ =1.4 atm, $ P_B^{o} $ =0.7 atm
B) $ P_A^{o} $ =1.2atm, $ P_B^{o} $ =0.6atm
C) $ P_A^{o} $ =1.4atm, $ P_B^{o} $ =0.6atm
D) $ P_A^{o} $ = 0.6 atm, $ P_B^{o} $ = 1.4 atm
Show Answer
Answer:
Correct Answer: D
Solution:
$ P_{T} $ (2 moles of A and 2 moles of B) is $ \frac{P_A^{o}}{2}+\frac{P_B^{o}}{2}=1atm\Rightarrow P_A^{o}+P_B^{o}=2atm $ $ P{’ _{T}} $ (1 mole of A and 3 moles of B) $ \frac{P_A^{o}}{4}+\frac{3P_B^{o}}{8}>1atm\Rightarrow P_A^{o}+3P_B^{o}>4atm $ $ P’{’ _{T}} $ (1 mole of A, 3 moles of B and 4 moles of C) And $ \frac{P_A^{o}}{8}+\frac{3P_B^{o}}{8}+\frac{4P_C^{o}}{8}=1atm $
$ \Rightarrow P_A^{o}+3P_B^{o}+4P_C^{o}=8atm $
So
$ \Rightarrow P_A^{o}+3P_B^{o}=(8-4\times 0.8)atm $
$ =4.8atm(P_C^{o}=0.8atm) $
Hence $ P_B^{o}=1.4atm $
$ P_A^{0}=0.6atm $
BETA
