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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?

Answer»

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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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