Work Done by Air "pressure difference" 1. The problem statement, all variables and given/known data Air that initially occupies .14 m^3 at a gauge pressure of 103.0kPa is expanded isothermally to a pressure to 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.) 2. Relevant equations W = V2 [tex]\int[/tex] V1 (pdv) pV=nRT 3. The attempt at a solution The professor did some simplification on the board and it came down to nRT(ln)(V_2 / V_1) then I solved for V2 and V1 V=(nRT)/p "the nRT cancels" and then I have this equation piV1(ln)(pi / pf) = .239 kJ =Work The answer in the back of the book is 5.6 kJ, so if someone could help me out where I'm going wrong this would be highly appreciated.