Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L ha
ID: 1369514 • 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 91.0 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 182 J.
(a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas. (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.
WIAF = WIBF = WIF =P(atm) 2.00 1.50- A V (liters) 0.300 0.800
Explanation / Answer
A)
Work done W = p*dV
p is the pressure and dV is the change in volume
WIAF = 0+(1.5*1.013*10^5*(0.8-0.3)/1000) = 75.975 J
WIBF = (2*1.013*10^5*(0.8-0.3)/1000) = 101.3 J
WIF = 0.5*2*1.013*10^5*0.5/1000 = 50.65 J
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B) QIAF = (Uf-Ui) + WIAF = (182-91) + 75.975 = 166.975 J
QIBF = (Uf-Ui)+WIBF = (182-91) + 101.3 = 192.3 J
QIF = (Uf-Ui)+WIF = (182-91)+50.65 = 141.65 J
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