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What will be the minimum pressure required to compress 500 dm^(3) of air at 1 bar to 200 dm^(3) at 30^(@)C? |
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Answer» <P> Solution :THUS, `P_(1)= 1 "bar" P_(2) = ?``v_(1)=500 dm^(3) "" v_(2)=200 dm^(3)` TEMPERATURE remain constant According to Boyle.s law `P_(1)v_(1)=P_(2)v_(2) or P_(2) =(P_(1)v_(1))/(v_(2))` `p_(2)=(1 "bar" xx500 dm^(3))/(200 dm^(3))=2*5 "bar"` |
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