Compute Work Done by Air: 103kPa to 101.3kPa

[Saladsamurai]

`Air that initially occupies $.140 m^3$ at a gauge pressure of 103.0 kPa is expanded isothermally to a pressure of 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.)

So I am going with $$\sum W=W_{12}+W_{23}$$

1-->2 is isothermal so $$W_{12}=nRT\ln\frac{V_2}{V_1}$$(1)
to find $V_2$ I used $V_2=\frac{p_1}{p_2}*V_1$. I think I need to use absolute pressure though.

So $$V_2=\frac{103*10^3+1.0*10^5}{101.3*10^3+1.0*10^5}*0.14 m^3=0.1412 m^3$$

So from (1) and using the fact that $p_1V_1=nRT$ I get $W_{12}=p_1V_1\ln\frac{.1412}{.14}=(103*10^3+1.0*10^5)(.14)\ln\frac{.1412}{.14}=243 J$

From2-->3 since pressure is constant I used $W_{23}=p*\Delta V=(101.3*10^3+1.0*10^3)(.14-.1412)=-242 J$

So $\sum W=-242+243= 1 J$

The answer in the text is 5.6 kJ !

Where is my error(s)?!

Thanks,
Casey

 

[Saladsamurai]

I know it is a lot to go through, but I would appreciate a hand. I have my final tomorrow and I thought I understood this perfectly fine.

Any help is appreciated!
Casey

 

[Saladsamurai]

So I have gone over this again today and I still do not know where my mistake is? Nobody else can see it either? This thread has been up for 3 days...I thought it was pretty well-organized?

Casey