1.

Four grams of helium is expanded from 1 atm to one-tenth of its origiinal pressure at 30^(@)C. Change in entropy (assuming ideal gas behaviour) is :

Answer»

`38.3 JK^(-1)`
`76.6 JK^(-1)`
`19.15 JK^(-1)`
`100 JK^(-1)`

Solution :For isothermal EXPANSION of gas:
`DELTA S=nR "in" (P_(1))/(P_(2))`
`n= (4.0)/(4) = 1.0` mol, R= 8.314 `p_(1) =1` atm
`p_(2) = 1//10` atm
`Delta S= (1.0) xx(8.314) "log" (1)/(1//10)`
`=19.15JK^(-1)`


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