Air that initially occupies 0.140m3 (V1) at a gauge pressure of 103.0 kPa (p1) is expanded isothermally to a pressure of 101.3 kPa (p2) and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air.
Wisothermal = nRTln(V2/V1)
Wisobaric = p2ΔV = p2(V1 - V2)
Wisothermal + Wisobaric = Wtotal
The Attempt at a Solution
We can find V2 easily enough by multiplying p1 and V1 and dividing by p2 because both p1V1 and p2V2 equal nRT. I can now fill in most of the formula:
nRTln(V2/V1) + p2(V1 - V2) = Wtotal
Here, I thought I could then just back substitute nRT for either p1V1 or p2V2 and I'd be golden but the answer I'm getting is different from the back of the book (5.6 kJ). I'm getting .001 kJ with plugging these in:
Wtotal = (103kPa)(0.140m3)ln(0.140m3/.141m3) + (101.3 kPa)(.140m3 - .141m3)
Unfortunately, if I use the actual pressure, I still don't get the desired answer:
Wtotal = (204kPa)(0.140m3)ln(0.140m3/.141m3) + (202.3 kPa)(.140m3 - .141m3)
What am I fundamentally missing here? Thanks in advance.
*** ANSWER ***
Wtotal = (204kPa)(0.140m3)ln(0.140m3/.141m3) + (101.3 kPa)(.140m3 - .141m3) = 5.6 kPaThey may be pulling a sneaky one on you. Only the first pressure given is specified as gauge pressure. Treat the second pressure as actual pressure. (That would make this a much more realistic problem!)