A gas expands from I to F along the three paths indicated in the figure below. Calculate the work done on the gas along each of the following paths.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 (3,6), extends vertically down to point B (3,2), then extends horizontally to point F (9,2).The blue path starts at point I (3,6), and extends down and to the right to end at point F (9,2).The orange path starts at point I (3,6), extends horizontally to the right to point A (9,6), then extends vertically down to end at point F (9,2).(a) IAF J(b) IF J(c) IBF J

A gas expands from I to F in the figure below. The energy added to the gas by heat is 405 J when the gas goes from I to F along the diagonal path.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 (2,4), extends vertically down to point B (2,1), then extends horizontally to point F (4,1).The blue path starts at point I (2,4), and extends down and to the right to end at point F (4,1).The orange path starts at point I (2,4), extends horizontally to the right to point A (4,4), then extends vertically down to end at point F (4,1).(a) What is the change in internal energy of the gas?(b) How much energy must be added to the gas by heat for the indirect path IAF to give the same change in internal energy?Step 1(a) The area under the diagonal path IF is a triangle, of height 3.00 atm, sitting atop a rectangle with a height of 1.00 atm. The base of the triangle is the same as the base of the rectangle, 2.00 L. The work done on the gas during this process isW  =  −area under curve =  −area of triangle BIF + area of rectangle (2,0)BF(4,0) =  −121 3 atm2.00 L + 1.00 atm2 2 L =  −4 5 atm · L.Converting to joules, we haveW = −-5.00 5 atm · L1.013 105 N/m21 atm10−3 m31 L = −-507.5 507 J.Step 2Let Q be the energy added to the gas by heat and ΔU be the change in internal energy of the process from the final energy UF to the initial energy UI. We are given the value of the energy added to the gas. Noting the sign on the value calculated for W above and the sign on the change in internal energy ΔU, for the change in internal energy, we haveΔU  =  UF − UI =  Q + W = 1500 405 J − 506.5 J = 993.5 -102 J.Step 3(b) The area under the indirect path IAF is the area of a rectangle of height atm and width L. The work done in joules on the gas during this process isW  =  −area under curve =  − atm · L1.013 105 N/m21 atm10−3 m31 L =  − J.

A gas follows the PV diagram in the figure below. Find the work done on the gas along the paths AB, BC, CD, DA, and ABCDA. (Enter your answers in J.)HINTThe area under the graph in a PV diagram is equal in magnitude to the work done on the gas. The work is positive if the gas is compressed and negative if it expands.Click the hint button again to remove this hint.A rectangular path is plotted on a PV diagram which has a horizontal axis labeled V (m3), and a vertical axis labeled P (105 Pa). The area inside the rectangle is shaded. The path is clockwise and runs through corner points A–D in the following order:A (1.00, 4.00), B (7.00, 4.00), C (7.00, 1.00), D (1.00, 1.00).(a)AB J(b)BC J(c)CD J(d)DA J(e)ABCDA J

Sketch a PV diagram and find the work done by the gas during the following stages.(a) A gas is expanded from a volume of 1.0 L to 7.5 L at a constant pressure of 5.0 atm. J(b) The gas is then cooled at constant volume until the pressure falls to 1.5 atm. J(c) The gas is then compressed at a constant pressure of 1.5 atm from a volume of 7.5 L to 1.0 L. (Note: Be careful of signs.) J(d) The gas is heated until its pressure increases from 1.5 atm to 5.0 atm at a constant volume. J(e) Find the net work done during the complete cycle. J

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

The area under the diagonal path IF is a triangle, of height 3.00 atm, sitting atop a rectangle with a height of 1.00 atm. The base of the triangle is the same as the base of the rectangle, 2.00 L. The work done on the gas during this process isW  =  −area under curve =  −area of triangle BIF + area of rectangle (2,0)BF(4,0) =  −12 atm2.00 L + 1.00 atm L =  − atm · L.Converting to joules, we haveW = − atm · L1.013 105 N/m21 atm10−3 m31 L = − J.