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One moles of an ideal gas which `C_(V) = 3//2 R` is heated at a constant pressure of `1 atm` from `25^(@)C` to `100^(@)C`. Calculate `DeltaU, DeltaH` and the entropy change during the process. |
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Answer» As `(C_(P) - C_(V)) = R rArr C_(P) = C_(V) +R` `rArr C_(P) = (3)/(2)R+R =(5)/(2)R` Heat given at constnat pressure `(DeltaH) = nC_(P) DeltaT =1 xx (5)/(2)R xx (373-298)` `rArr (DeltaH) = 1xx (5)/(2) xx 1.987 xx 75 = 372.56 cal` Work done in the process `=- P DeltaV = - P (V_(2)-V_(1))` `=- P ((nRT_(2))/(P)-(nRT_(1))/(P))` as `(PV = nRT)` `=nRT (T_(2)-T_(1)) =- 1 xx 1.987 xx (373-298) =- 149.05 cal` From the first law of thermodynamics, `DeltaU = q +w = 372.56 - 149.05 = 223.51 cal` |
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