Work Done by Air pressure difference

[just.karl]

Work Done by Air "pressure difference"

Homework Statement

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.)

Homework Equations

W = V2 \int V1 (pdv)

pV=nRT

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.

 

[just.karl]

help? please?

 

[Redbelly98]

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.

239 J is the work done during isothermal expansion. I would think we need to account for the constant pressure compression as well.

(Even so, 5.6 kJ doesn't look right to me.)

 

[RTW69]

For a polytropic process of an ideal gas with isothermal expansion (n=1) you can also use the expression for work=P1*V1*ln(V2/V1) which gives a similar result of 284.2 J However that is the total area under the curve from pt 1 to pt 2. There is also the area (work) under the curve when the air is cooled at constant pressure to the initial volume that must be subtracted from the above work to get the area of the complete process. If that is the case it get a very small total work of about 1.4 J. In any case 5.6 KJ seems way to large.