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Method 2 of `DeltaU`
Work done by a gas in a given process is `-20J`. Heat given to the gas is 60J. Find change in internal energy of the gas.

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`DeltaU=Q-W`
Substituting the values we have,
`DeltaU=60-(-20)`
`=80J`
`DeltaU` is positive. Hence, internal erergy of the gas is increasing.

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