Respuesta :
Answer:
The amount of Work done during the process = 0.01026 BTU
Explanation:
Initial Pressure ([tex]P_{1}[/tex]) = 60 [tex]\frac{lbf}{in^{2} }[/tex]
Initial volume ( [tex]V_{1}[/tex] )= 1.78 [tex]ft^{3}[/tex]
Final Pressure ([tex]P_{2}[/tex]) = 20 [tex]\frac{lbf}{in^{2} }[/tex]
Final Volume ( [tex]V_{2}[/tex] ) = We have to calculate.
The relationship between pressure and volume during the process is = P [tex]V^{1.3}[/tex]
⇒ [tex]P_{1}[/tex] × [tex]V_{1}^{1.3}[/tex] = [tex]P_{2}[/tex] × [tex]V_{2}^{1.3}[/tex]
⇒ [tex]\frac{V_{2} }{V_{1} }[/tex] = [tex][\frac{P_{1} }{P_{2}}]^{0.77} }[/tex]
⇒ [tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]3^{0.77}[/tex]
⇒ [tex]\frac{V_{2} }{V_{1} }[/tex] = 2.33
⇒ [tex]V_{2}[/tex] = 2.33 × 1.78
⇒ [tex]V_{2}[/tex] = 4.147 [tex]ft^{3}[/tex] ------------ (1)
The work done during the process is given by W = [tex]\frac{P_{1}V_ {1}- P_{2}V_ {2} }{\gamma - 1}[/tex]
Put all the values in the above formula we get
⇒ W = [tex]\frac{106.8 - 82.94}{1.3 - 1}[/tex]
⇒ W = 79.54 lb f
We know that 1 lb f = 0.00129 BTU
⇒ W = 0.01026 BTU
This is the amount of Work done during the process in BTU
