A gas expands from `3 dm^(3)` to `5 dm^(3)` against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mol of water at a temperature of 290 K. Calculate final temperature of water. Specific heat of water `=4.184 J g^(-1)K^(-1)`

Explanation

Work done `=PxxdV=3.0xx(5.0-3.0)`
`=6.0litre-atm=6.0xx101.3J`
`=607.8J` ltBrgt let `DeltaT` be the change in temperature
Heat absorbed `=mxxsxxDeltaT`
`=10.0xx18xx4.184xxDeltaT`
Given, `PxxdV=mxxsxxDeltaT`
or `DeltaT=(PxxdV)/(mxxs)=(607.8)/(10.0xx18.0xx4.184)=0.807`
Final temperature `=290+0.807=290.807K`