- #1

jeff.berhow

- 17

- 0

## Homework Statement

Air that initially occupies 0.140m

^{3}(V

_{1}) at a gauge pressure of 103.0 kPa (p

_{1}) is expanded isothermally to a pressure of 101.3 kPa (p

_{2}) and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air.

## Homework Equations

W

_{isothermal}= nRTln(V

_{2}/V

_{1})

W

_{isobaric}= p

_{2}ΔV = p

_{2}(V

_{1}- V

_{2})

W

_{isothermal}+ W

_{isobaric}= W

_{total}

## The Attempt at a Solution

We can find V

_{2}easily enough by multiplying p

_{1}and V

_{1}and dividing by p

_{2}because both p

_{1}V

_{1}and p

_{2}V

_{2}equal nRT. I can now fill in most of the formula:

nRTln(V

_{2}/V

_{1}) + p

_{2}(V

_{1}- V

_{2}) = W

_{total}

Here, I thought I could then just back substitute nRT for either p

_{1}V

_{1}or p

_{2}V

_{2}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:

W

_{total}= (103kPa)(0.140m

^{3})ln(0.140m

^{3}/.141m

^{3}) + (101.3 kPa)(.140m

^{3}- .141m

^{3})

Unfortunately, if I use the actual pressure, I still don't get the desired answer:

W

_{total}= (204kPa)(0.140m

^{3})ln(0.140m

^{3}/.141m

^{3}) + (202.3 kPa)(.140m

^{3}- .141m

^{3})

What am I fundamentally missing here? Thanks in advance.

*** ANSWER ***

Doc Al said:They 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!)

W

_{total}= (204kPa)(0.140m

^{3})ln(0.140m

^{3}/.141m

^{3}) + (101.3 kPa)(.140m

^{3}- .141m

^{3}) = 5.6 kPa

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