Respuesta :
Explanation:
The given data is as follows.
Initial volume ([tex]V_{1}[/tex]) = 100 [tex]in^{3}[/tex]
Final volume ([tex]V_{2}[/tex]) = 10 [tex]in^{3}[/tex]
Initial pressure ([tex]P_{1}[/tex]) = 50 psia = 50 [tex]\frac{pounds}{in^{2}}[/tex]
Temperature = [tex]100^{o}F[/tex]
So, we assume that vapors are also ideal gas.
Hence, the work done for ideal gas will be calculated as follows.
W = [tex]P \int dV[/tex]
Since, it is given that the temperature is constant so, it is an isothermal process.
Therefore, work done for isothermal process is as follows.
W = [tex]P_{1}V_{1} ln (\frac{V_{2}}{V_{1}})[/tex]
Putting the values into the above formula as follows.
W = [tex]P_{1}V_{1} ln (\frac{V_{2}}{V_{1}})[/tex]
= [tex]50 pounds/in^{2} \times 100 in^{3} \times ln (\frac{10}{100})[/tex]
= -11512.925 lbf-in
As there are 12 inch present in 1 ft. So, converting lbf-in into lbf-ft as follows.
W = [tex]\frac{-11512.925}{12} lbf-ft[/tex]
= -959.41 lbf-ft
The negative sign means work is supplied.
Thus, we can conclude that the work done is -959.41 lbf-ft.
