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A gas expands from 10 litre to 20 litre at a constant pressure of 2 atmospheres. What is the work done by the gas during the expansion ?

Answer»

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Solution :`P=2" atmospheres"=2xx1.013xx10^(5)= 2.026xx10^(5) NM^(2)`
`V_(1)= 10" litre" = 10xx10^(3) cm^(3) = 10^(-2) m^(3)`,
`V_(2)=20 "litre" = 2xx10^(-2) m^(3)`. WORK DONE = `P(V_( 2)-V_(1)) = 2.026xx10^(5)`
`(2xx10^(-2)-1xx10^(-2)) = 2.026xx10^(3) J.`


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