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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)` |
Answer» 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` |
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