1.

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`

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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