1.

When one mole of a monoatomic ideal gas at initialtemperature T K expandsadiabatically from 1 litre to 2litres , the final temperature in Kelvin would be

Answer»

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`(T)/( 2^(2//3))`
`T - (2)/( 3xx 0.0821)`
`T + (2)/( 3 XX 0.0821)`

Solution :`RV^(gamma-1)=` constant, i.e.,`T_(1)V_(1)^(gamma-1)= T_(2)V_(2)^(gamma-1)`
or `(T_(2))/(T_(1))= ((V_(1))/(V_(2)))^(gamma-1)` or`(T_(2))/(T) = ((1)/(2))^((5)/(3)-1)= ( 1)/( 2^(2//3))`
( for monoatomic ideal gas , `gamma=(C_(p))/(C_(v))=((5//2)R)/((3//2)R)=(5)/(3))`
`:. T_(2) =(T)/( 2^(2//3))`


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