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.

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

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

Consider an ideal gas that occupies 2 L at 1 bar. The gas is compressed isothermallyto 0.5 L at a constant pressure of 4 bar. This is followed by another isothermalcompression to 0.25 L at a constant pressure of 8 bar. Calculate the work involved inthis two-step compression. Calculate and compare the work if the same gas wascompressed 2 L to 0.25 L in a reversible, isothermal manner.

1 mole of an ideal gas is made to expandfrom an initial volume of 2.24 3dm to11.2 3dm at 300 K. Calculate the maximumamount of work that can be produced inthe surroundings by this process

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.