A) 0.33 and 0.52
B) 0.33 and 0.56
C) 0.25 and 0.45
D) 0.38 and 0.52
Correct Answer: A
Solution :
\[PV=nRT\] \[{{n}_{He}}=\frac{0.66\times 200}{RT}\] \[{{n}_{{{O}_{2}}}}=\frac{0.52\times 400}{RT}\] \[PV=nRT\] \[P=\frac{({{n}_{He}}+{{n}_{{{O}_{2}}}})RT}{400}\] \[{{P}_{T}}=\frac{0.66\times 200+400\times 0.52}{400}\] \[{{X}_{He}}=\frac{{{n}_{He}}}{{{n}_{{{O}_{2}}}}+{{n}_{He}}}\] \[{{X}_{He}}=\frac{0.66\times 200}{0.66\times 200+400\times 0.52}\] \[=\frac{132}{132+208}\] \[{{P}_{He}}={{X}_{He}}{{P}_{T}}=\frac{132}{340}\times \frac{340}{400}\] \[{{P}_{He}}=0.33\] \[{{X}_{{{O}_{2}}}}=\frac{208}{400}\] So, \[{{P}_{{{O}_{2}}}}=0.52\] Hence, the correct option is (a).
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