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Air having a pressure of 40 psig and a volume of 8 cu ft expands isothermally to a pressure of 10 psig. Find the external work performed during the expansion.

Answer :

boffeemadrid

Answer:

357.6 lb-ft

Explanation:

V = Volume = 8 ft³

dP = Change in pressure = (40-10) = 30 psig

Work done is given by

[tex]W=VdP\\\Rightarrow W=8\times (40-10)\\\Rightarrow W=240\ psig.ft^3[/tex]

[tex]30\ psig=44.7\ psi\\\Rightarrow 1\ psi=\dfrac{30}{44.7}[/tex]

So, converting to ft-lb

[tex]\dfrac{240}{\dfrac{30}{44.7}}=357.6\ lb-ft[/tex]

The external work performed during the expansion is 357.6 lb-ft

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