Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L ha
ID: 2067067 • Letter: F
Question
Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 92.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 178 J.
(a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas.
WIAF = J
WIBF = J
WIF = J
(b) For the paths IAF, IBF, and IF in the figure above, calculate the net energy transferred to the gas by heat in the process.
QIAF = J
QIBF = J
QIF = J
Explanation / Answer
Part A)
Work = PV (which is the area under the curve)
P must be in Pascals and V in cubic meters
WIAF = (1.5)(1.013 X 105)(.5)(1 X 10-3) = 76.0 J
WIBF = (2)(1.013 X 105)(.5)(1 X 10-3) = 101.3 J
WIF = (1.5)(1.013 X 105)(.5)(1 X 10-3) + (.5)(.5)(1.013 X 105)(.5)(1 X 10-3) = 88.6 J
Part B)
U = Q - W
Q = U - W
QIAF = (178 - 92) - (-76) = 162 J
QIBF = (178 - 92) - (-101.3) = 187.3 J
QIF = (178 - 92) - (-88.6) = 174.6 J
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