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Calculate the maximum work done (in multiple of 10^(3)) in expanding 16g of oxygen at 300K and occupying a volume of 5 dm^(3) isothermally, until the voume becomes 25 dm^(3) (Give you anwer as nearest integer)

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Solution :Maximum work (or) slow process terms indicates that the process must be reversible. `w= - 2.303n TR "long" (v_(2))/(v_(1))`
`= -2.303 xx (16)/(32) (300) (2) "log" (25)/(5) = -2 xx 10^(-3)J`


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