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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)`. |
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Answer» 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` |
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