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InterviewSolution
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Work done in expansion of an idela gas from `4L` to `6L` against a constant external pressure of `2.5 atm` was used to heat up `1mol` of water at `293K`. If specific heat of water is `4.184J g^(-1)K^(-1)`, what is the final temperature of water. |
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Answer» Since work is done against constant pressure and is done in one step, therefore, it is irreversible `:. W =- P_(ex) DeltaV` `DeltaV = (6 -4) = 2L` `W =- 2.5 xx 2=- 5L atm` `(1L atm = 101.262J) =- 5 xx 101.262J =- 506.31J` or `= (5 xx 1.987)/(0.0821) cal = (5 xx 1.987 xx 4.184)/(0.0821) = - 506.31J` Now this work is done in heating `1mol` of `H_(2)O`. `w = n xx` Specific heat `xx DeltaT` `506.31 = 1 xx 4.184 xx 18 xx DeltaT` Specific heat `= 4.184J g^(-1) = 4.184 xx 18J mol^(-1)]` `DeltaT = 6.723` Final temperature =` T_(1) + DeltaT = 293 + 6.723 = 299.723K` |
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