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

250 mL of nitrogen maintained at 720 mm pressure and 380 mL of oxygen

1.	250 mL of nitrogen maintained at 720 mm pressure and 380 mL of oxygen maintained at 650 mm pressure of the misture ?
Answer» Solution :Step 1. To CALCULATE the partial pressure of nitrogen `{:("Given Conditions","Final Conditions"),("VOLUME "V_(1)=250 mL,V_(2)=1000 mL),("Pressure "P_(1)=720 mm,P_(2)=? mm):}` Applying Boyle's Law (since the temperature remains constant), `P_(2)V_(2)=P_(1)V_(1),i.e., "" 1000xxP_(2)=720xx250"or" P_(2)=(720xx250)/(1000)=180 mm` Thus, the partial pressure due to nitrogen `(p_(n_(2)))=180 mm`. Step 2. To calculate the partial pressure of oxygen `{:("Given Conditions","Final Conditions"),(V_(1)=380 mL,V_(2)=1000 mL),(P_(1)=650 mm,P_(2)=? mm):}` Applying Boyle's Law (since the temperature remains constant), `P_(2)V_(2)=P_(1)V_(1),i.e.,1000xxP_(2)=380xx650 "or" P_(2)=(380xx650)/(1000)=247 mm` Thus, the partical pressure due to oxygen `(p_(o)_(2))`=247 mm. Step 3. To calculate the final pressure of the gaseous mixture. If P is the final pressure of the gaseous mixture, then according to Dalton's Law of Partial Pressures, `P=P_(n_(2))+P_(o_(2))=180+247=427 mm`.