1.

Method 2 of Q In a given process work done on a gas is 40 J and increases in its internal energy is 10J. Find heat given or taken to/from the gas in this process.

Answer» Correct Answer - C
Given, `DeltaU=+10J`
Work done on the gas is 40 J. Therefore, work done by the gas used in the equation, `Q=W+DeltaU` will be `-40J`. Now, putting the values in the equation,
`Q=W+DeltaU`
We have,
`Q=-40+10`
`=-30J`
Here, negative sign indicates that heat is taken out from the gas.


Discussion

No Comment Found

Related InterviewSolutions