A gas expands from `3 dm^(3)` to `5dm^(3)` anainst a constant pressure of `3atm`. The work done during the expansion if used to heat `10mol` of water at temperature `290K`. Find the final temperature of water, if the specific heat of water `= 4.18g^(-1)K^(-1)`.

Answer

290.81 K

Explanation

Work is done against constant P
`Delta V = 5 - 3= 2dm^(3) = 2` litre, `P = 3` atm.
`:. W = - P.Delta V = - 3 xx 2` litre atm.
`= - (6 xx 4.184 xx 1.987)/(0.0821) J = - 607.57 J`
Now this work is used up in heating water
`:. W = n xx C xx Delta T`
`607.57 = 10 xx 4.184 xx 18 xx Delta T, Delta T = 0.81`
Final temperature ` = T_(1) + Delta T = 290 + 0.81 = 290.81 K`