Two moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has
ID: 1502947 • Letter: T
Question
Two 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.
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
(a)
WIAF = WIA + WAF = 0 + PA*dV = 1.5*1.013*10^5*(0.8-0.3)*10^-3 = 75.975 J
WIBF = WAB + WBF = PB*(dV) + 0 = 2*1.013*10^5*(0.8-0.3)*10^-3 = 101.3 J
WIF = area aunder IF = (1/2)*AF*AI + AF*PA
WIF = ((1/2)*0.5*10^-3*0.5*10^5)+(0.5*10^-3*1.5*1.013*10^5) = 88.475 J
(b)
QIAF = dU + W = (182-91) + 75.975 = 166.975 J
QIBF = dU + W = (182-91) + 101.3 = 192.3 J
QIF = (182-91) + 88.475 = 179.475 J
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