# Work done by a gas

1. Dec 10, 2007

`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

2. Dec 10, 2007

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

3. Dec 12, 2007