**ONLY NEED HELP WITH PART B PLEASE!! Five moles of gas initially at a pressure

Question

**ONLY NEED HELP WITH PART B PLEASE!! Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 80.0 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 180 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

The work done on the gas = Integral of PdV. For the path IAF, there is no change in volume during IA. Hence work done in this part is zero. For AF, the pressure is constant. So the work done = P*(change in volume) = P2*(V2-V1) For the path IBF, work is done only during IB. So the work done = P1*(V2-V1) For the path IF, we have to define P in terms of V.The change in internal energy is independent of the path. P = 2.0-(2.0-1.5)*(V-0.3)/(0.8-0.3) = 2.3 - V.You will have to convert this to proper units Hence work done = int PdV within the limits 0.3 to 0.8 litres b) The nett energy transferred to the gas by heat = change in internal energy + work done on the gas as calculated above for each path. The change in internal energy is independent of the path.

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