- #1

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Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)

So I am going with [tex]\sum W=W_{12}+W_{23}[/tex]

1-->2 is isothermal so [tex]W_{12}=nRT\ln\frac{V_2}{V_1}[/tex](1)

to find [itex]V_2[/itex] I used [itex]V_2=\frac{p_1}{p_2}*V_1[/itex]. I think I need to use

*absolute pressure*though.

So [tex]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[/tex]

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

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

So [itex]\sum W=-242+243= 1 J[/itex]

The answer in the text is 5.6 kJ !!

Where is my error(s)?!

Thanks,

Casey