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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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Answer» `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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