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A gas expands from I to F in the figure below. The energy added to the gas by heat is 424 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? J(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? J

Question

A gas expands from I to F in the figure below. The energy added to the gas by heat is 424 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? J(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? J

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Solution

The question seems to be missing the diagram, but I'll try to answer it based on the description provided.

(a) The change in internal energy of the gas (ΔU) can be calculated using the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system (Q) minus the work done by the system (W). In this case, we know that Q = 424 J, but we don't have information about the work done by the system.

If we assume that the process is adiabatic (no heat is exchanged with the surroundings), then the work done by the system is equal to the heat added, so W = Q = 424 J. Therefore, the change in internal energy would be ΔU = Q - W = 424 J - 424 J = 0 J.

(b) For the indirect path IAF, the work done by the system is different. The work done on a gas in a quasi-static process is given by the area under the PV curve. In this case, the work done along path IA is zero (since the volume doesn't change), and the work done along path AF is the area of the rectangle with sides (4-2) L and (4-1) atm, which is W = 2 L * 3 atm.

To convert this to joules, we can use the conversion factor 1 L*atm = 101.3 J, so W = 2 * 3 * 101.3 J = 606.6 J.

Therefore, to achieve the same change in internal energy (ΔU = 0 J), the heat added to the system must be equal to the work done by the system, so Q = W = 606.6 J.

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