posted by savannah on .
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?
delta E = w + q
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
Mastering Chemistry accepts this answer as Correct