Question

What will be the partial pressure of $He$ and $ O_{2} $ respectively, if $200\,mL$ of $He$ at $0.66 \,atm$ and $400 \,mL$ of $ O_{2} $ at $0.52 \,atm$ pressure are mixed in $400 \,mL$ vessel at $ 20^{\circ}C $ ?

• A. 0.33 and 0.56
• B. 0.33 and 0.52
• C. 0.38 and 0.52
• D. 0.25 and 0.45

Answer 1

The Correct Option is B

Total pressure, $p =\frac{\text { milliequiv. of } He +\text { milliequiv. } \text { of } O _{2}}{\text { total volume }}$ $=\frac{200 \times 0.66+400 \times 0.52}{400}$ $=\frac{132+208}{400} $ $=\frac{340}{400}=0.85$ Partial pressure of $ He =\frac{\text { milliequiv of } He }{\text { total milliequiv }} \times p$ $=\frac{132}{340} \times 0.85 $ $=0.33 \,atm $ Similarly, partial pressure of $ O _{2} =\frac{208}{340} \times 0.85 $ $=0.52 \,atm $