1.

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.

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


Discussion

No Comment Found

Related InterviewSolutions