Question

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5. -/1 points SerCP11 12.3.P.030 My Notes Ask Your Teacher Four 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 3. 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 P (atm) 2.00 150 V (liters) 0.300 0.800 (a) For the paths IAF, 1BF, end IF in the figure above, calculate the work done on the gas. War Winr Wr (b) For the paths IAF, 1BF, and IF in the figure above, calculate the net energy transferred to the gas by heat in the process. Qr Need Help? Read H

Answer 1

WE=_ a) The work done is the area below the graph: WAF = (1.50 * 1.013 X 105)(0.8 -0.3) x 10-3 = 75.975 = 76.0) Wibe = (2.0 * 1.013 X 10 (0.8 – 0.3) x 10-3 = 101.3) 2.00 + 1.500 2* 1.013 X 105(0.8 -0.3) x 10-3 = 88.63 = 88.6J b) Let's use the first thermodynamics law QIAF = WAF + AU = 76.0 + (182 - 91) = 167 ) QIBE = WIBE + AU = 101.3 + (182-91) = 192.3 Qif = WE + AU = 88.6 + (182-91) = 179.6) 2