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
Correct Answer: D
Solution :
| [d] \[{{P}_{T}}\](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) |
| \[\frac{{{P}_{A}}^{o}}{4}+\frac{3{{P}_{B}}^{o}}{8}>1atm\Rightarrow {{P}_{A}}^{o}+3{{P}_{B}}^{o}>4atm\] |
| \[P'{{'}_{T}}\](1 mole of A, 3 moles of B and 4 moles of C) |
| And \[\frac{{{P}_{A}}^{o}}{8}+\frac{3{{P}_{B}}^{o}}{8}+\frac{4{{P}_{C}}^{o}}{8}=1atm\] |
| \[\Rightarrow {{P}_{A}}^{o}+3{{P}_{B}}^{o}+4{{P}_{C}}^{o}=8atm\] |
| So \[\Rightarrow {{P}_{A}}^{o}+3{{P}_{B}}^{o}=(8-4\times 0.8)atm\] |
| \[=4.8atm\,\,({{P}_{C}}^{o}=0.8atm)\] |
| Hence \[{{P}_{B}}^{o}=1.4atm\] |
| \[{{P}_{A}}^{0}=0.6atm\] |
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