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 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.Three paths are plotted on a PV diagram, which has a horizontal axis labeled V (liters), and a vertical axis labeled P (atm).The green path starts at point I (0.300,2.00), extends vertically down to point A (0.300,1.50), then extends horizontally to point F (0.800,1.50).The blue path starts at point I (0.300,2.00), and extends down and to the right to end at point F (0.800,1.50).The orange path starts at point I (0.300,2.00), extends horizontally to the right to point B (0.800,2.00), then extends vertically down to end at point F (0.800,1.50).(a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas.WIAF = JWIBF = JWIF = 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 = JQIBF = JQIF = J
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 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.Three paths are plotted on a PV diagram, which has a horizontal axis labeled V (liters), and a vertical axis labeled P (atm).The green path starts at point I (0.300,2.00), extends vertically down to point A (0.300,1.50), then extends horizontally to point F (0.800,1.50).The blue path starts at point I (0.300,2.00), and extends down and to the right to end at point F (0.800,1.50).The orange path starts at point I (0.300,2.00), extends horizontally to the right to point B (0.800,2.00), then extends vertically down to end at point F (0.800,1.50).(a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas.WIAF = JWIBF = JWIF = 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 = JQIBF = JQIF = J
Solution
(a) The work done on the gas can be calculated using the formula W = -PΔV, where P is the pressure and ΔV is the change in volume.
For path IAF: The pressure is constant at 2.00 atm = 2.00 * 101.3 J/L = 202.6 J/L. The change in volume is 0.800 L - 0.300 L = 0.500 L. So, WIAF = -PΔV = -202.6 J/L * 0.500 L = -101.3 J.
For path IBF: The work done is the sum of the work done from I to B and from B to F. From I to B, the volume changes but the pressure is constant at 2.00 atm = 202.6 J/L. The change in volume is 0.800 L - 0.300 L = 0.500 L. So, the work done from I to B is -202.6 J/L * 0.500 L = -101.3 J. From B to F, the volume is constant so no work is done. Therefore, WIBF = -101.3 J + 0 = -101.3 J.
For path IF: The work done is the area under the curve on the PV diagram. Since the path is a straight line from I to F, the work done is the area of the trapezoid formed by the points I, F, and the points where the path intersects the axes. This area is given by the formula A = 1/2 * (base1 + base2) * height = 1/2 * (2.00 atm + 1.50 atm) * (0.800 L - 0.300 L) = 1.75 atm * 0.500 L = 0.875 atmL. Converting to joules, WIF = -0.875 atmL * 101.3 J/(atm*L) = -88.6 J.
(b) The net energy transferred to the gas by heat can be calculated using the first law of thermodynamics, ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat transferred to the system, and W is the work done on the system.
For path IAF: ΔU = 182 J - 91.0 J = 91.0 J. So, QIAF = ΔU + W = 91.0 J + (-101.3 J) = -10.3 J.
For path IBF: ΔU = 182 J - 91.0 J = 91.0 J. So, QIBF = ΔU + W = 91.0 J + (-101.3 J) = -10.3 J.
For path IF: ΔU = 182 J - 91.0 J = 91.0 J. So, QIF = ΔU + W = 91.0 J + (-88.6 J) = 2.4 J.
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