Four moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L ha

Question

Four moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 89.0 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 158 . P (atm) 2.00- 1.50 V (liters) 0.300 0.800 (a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas. WIAF 75 99 The response you submitted has the wrong sign. WIRE 101.325 The response you submitted has the wrong sign. ] 88.159 The response you submitted has the wrong sign. (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-145 Q1BF-170325 ] QIF 157 159

Explanation / Answer

(a) First through IAF process
WIAF = WIA + WAF
Since IA is constant volume process therefore WIA = 0
WIAF = 0 + WAF = WAF = -(1.5*101.325*103*(0.8-0.5)*10-3) = -75.99 J
-ve sign indicate that work is done on the gas
Similarly for IBF
WIBF = area below curve IB
= - (2*101.325*103)*(0.8-0.5)*10-3 = -101.325 J
Now process IF
WIF = area under the curve IF
= Area of triangle + WAF
= - [{(1/2)*(0.5*101.325*103)*(0.8-0.3)*10-3} + 75.99]
= -88.66 J

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