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.

Explanation

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`