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What will be the minimum pressure required to compress 500dm^(3) of air at 1 bar to 200 dm^(3) at 30^(@)C ?

Answer»

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Solution :`P_(1)=1` bar,`V_(1)=500 dm^(3)`
`P_(2)=?,V_(2)=200 dm^(3)`
As TEMPERATURE remains constant at `30^(@)C`, applying Boyle's law, `P_(1)V_(1)=P_(2)V_(2)`
`1 barxx500 dm^(3)=P_(2)xx200 dm^(3)"or"P_(2)=(500)/(200)bar=2.5 bar`


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