# Total internal energy change of a gas

1. Mar 6, 2012

### wumple

1. The problem statement, all variables and given/known data
If ΔWi denotes the amount of work done by a gas in being compressed isothermally to a given volume, ΔQ the amount of heat absorbed by the gas in this process, and ΔWa the amount of work done by the gas in being expanded adiabatically back to its original volume, then the change ΔE in the internal energy of the gas as a result of the complete process is given by (multiple choice)

2. Relevant equations
I know the first law of thermodynamics tells you dE = dQ - DW ...

3. The attempt at a solution
...so I suppose it would be ΔQ - ΔWa - ΔWi (which my prof gave as the right answer). But I don't understand why the two works have the same sign. If the volume decreases in the first case, then shouldn't 'dV' in the integral to find the work be negative for that case, but positive in the case of increasing volume? So then the two works would have opposite sign?

Thanks for any help!!