A) 8 and 5 atom
B) 9.6 and 4 atm
C) 4.8 and 2 atm
D) 6.4 and 4 atm
Correct Answer: C
Solution :
[c] Moles of A, \[({{n}_{A}})=\frac{{{P}_{A}}{{V}_{A}}}{RT}=\frac{8\times 12}{RT}=\frac{96}{RT}\] Moles of B, \[({{n}_{B}})=\frac{{{P}_{B}}{{V}_{B}}}{RT}=\frac{8\times 5}{RT}=\frac{40}{RT}\] Total pressure \[\times \] total volume \[=({{n}_{A}}+{{n}_{B}})\times RT\] \[p\times (12+8)=\frac{1}{RT}(96+40)RT\] \[P=6.8\] Partial pressure of A = P \[\times \] mole fraction of A \[=6.8\left( \frac{96}{RT}/\frac{96+40}{RT} \right)\] =4.8 aim Partial pressure of \[B=6.8-4.8=2\text{ }atm.\]
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