The air within a piston equipped with a cylinder absorbs 570J of heat and expands from an initial volume of 0.12L to a final volume of 0.88L against an external pressure of 1.0 atm. What is the change in internal energy of the air within the piston?

Answer 1

delta E = w + q
q=570 J
V1 = 0.12 L
V2 = 0.88 L
delta V = 0.88 L - 0.12 L = 0.76 L
w = -P x delta V
w = -1.0 atm x 0.76 L = -0.76 L atm
You need to convert L atm to J
-0.76 L atm / 1 L atm x 101.3 J = -76.99 J
delta E = -76.99 J + 570 J = 493 J

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