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250 mL of nitrogen maintained at 720 mm pressure and 380 mL of oxygen maintained at 650 mm pressure of the misture ?

Answer» <html><body><p></p>Solution :Step <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>. To <a href="https://interviewquestions.tuteehub.com/tag/calculate-412157" style="font-weight:bold;" target="_blank" title="Click to know more about CALCULATE">CALCULATE</a> the partial pressure of nitrogen <br/> `{:("Given Conditions","Final Conditions"),("<a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> "V_(1)=250 mL,V_(2)=1000 mL),("Pressure "P_(1)=720 mm,P_(2)=? mm):}` <br/> Applying Boyle's Law (since the temperature remains constant), <br/> `P_(2)V_(2)=P_(1)V_(1),i.e., "" 1000xxP_(2)=720xx250"or" P_(2)=(720xx250)/(1000)=180 mm` <br/> Thus, the partial pressure due to nitrogen `(p_(n_(2)))=180 mm`. <br/> Step 2. To calculate the partial pressure of oxygen <br/> `{:("Given Conditions","Final Conditions"),(V_(1)=380 mL,V_(2)=1000 mL),(P_(1)=<a href="https://interviewquestions.tuteehub.com/tag/650-331163" style="font-weight:bold;" target="_blank" title="Click to know more about 650">650</a> mm,P_(2)=? mm):}` <br/> Applying Boyle's Law (since the temperature remains constant), <br/> `P_(2)V_(2)=P_(1)V_(1),i.e.,1000xxP_(2)=380xx650 "or" P_(2)=(380xx650)/(1000)=247 mm` <br/> Thus, the partical pressure due to oxygen `(p_(o)_(2))`=247 mm. <br/> Step 3. To calculate the final pressure of the gaseous mixture. <br/> If P is the final pressure of the gaseous mixture, then according to Dalton's Law of Partial Pressures, <br/> `P=P_(n_(2))+P_(o_(2))=180+247=427 mm`.</body></html>


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