- #1
Saladsamurai
- 3,020
- 7
`Air that initially occupies [itex].140 m^3[/itex] 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 [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
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