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.
Question
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.
Solution
The question seems to be incomplete, especially in Step 3(b) where the values for height and width of the rectangle are missing. However, I can guide you on how to approach it once you have the complete information.
Step 3(b) is about calculating the work done on the gas during the indirect path IAF. The work done (W) is equal to the negative of the area under the curve on a PV diagram. In this case, the area under the curve is the area of a rectangle. The area of a rectangle is calculated as height multiplied by width.
In the context of a PV diagram, the height would be the pressure difference (in atm) and the width would be the volume difference (in L). Once you have these values, you can calculate the area of the rectangle.
The work done is then converted to joules using the conversion factors provided in the question.
Finally, to find out how much energy must be added to the gas by heat for the indirect path IAF to give the same change in internal energy, you would use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system plus the work done on it. In this case, you already know the change in internal energy from part (a), and you've calculated the work done, so you can solve for the heat added.
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