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)`.

Explanation

Since work is done against constant `P` and thus, irreversible
`DeltaV=5-3=2 dm^(3)=2 litre, P=3 atm`
`:. W=-P.DeltaV=-3xx2 litre atm`
`=(6xx4.184xx1.987)/0.0821=-607.57` joule
Now, this work is used up in heating water
`:. W=nxxCxxDeltaT`
`607.57=10xx4.184xx18xxDeltaT`
`:. DeltaT=0.81`
`:.` Final temperature `=T_(1)+DeltaT=290+0.81`
`290.81 K`