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When 1 mole of gas is heated at constant volume. Temperature is raised from `298` to `308K` . Heat supplied to the gas is `500J` . Then which stamenet is correct?A. `q=-W=500J,DeltaU=0`B. `q=DeltaU=500J,W=0`C. `q=-W=500J,DeltaU=0`D. `DeltaU=0,q=W=-500J` |
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Answer» Correct Answer - C We Know that, `DeltaH=DeltaE+PDeltaV` When, `DeltaV=0` , `:. DeltaH=DeltaE` From the first law of thermodynamics, `DeltaE=q-W` In the given problem,`DeltaH=500J` `-W=- PDeltaV,DeltaV=0` So, `DeltaE=q=500J` |
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