A substance undergoes the cyclic process shown in the figure below. Work output occurs along path AB while work input is required along path BC, and no work is involved in the constant volume process CA. Energy transfers by heat occur during each process involved in the cycle.A pressure-volume graph consists of points and line segments plotted on a coordinate plane, where the horizontal axis is V (liters) and the vertical axis is P (atm). Three points are plotted:point A at (10.0, 5.00)point B at (50.0, 1.00)point C at (10.0, 1.00)Line segments connect the three points to form a triangle. Arrows along the line segments point from A to B, from B to C, and from C to A.(a) What is the work output during process AB? J(b) How much work input is required during process BC? J(c) What is the net energy input Q during this cycle? J
Question
A substance undergoes the cyclic process shown in the figure below. Work output occurs along path AB while work input is required along path BC, and no work is involved in the constant volume process CA. Energy transfers by heat occur during each process involved in the cycle.A pressure-volume graph consists of points and line segments plotted on a coordinate plane, where the horizontal axis is V (liters) and the vertical axis is P (atm). Three points are plotted:point A at (10.0, 5.00)point B at (50.0, 1.00)point C at (10.0, 1.00)Line segments connect the three points to form a triangle. Arrows along the line segments point from A to B, from B to C, and from C to A.(a) What is the work output during process AB? J(b) How much work input is required during process BC? J(c) What is the net energy input Q during this cycle? J
Solution
(a) The work done on a gas (or by a gas) is given by the area under the pressure-volume curve. For process AB, this is a trapezoid. The area of a trapezoid is given by the formula:
Area = 1/2 * (Base1 + Base2) * Height
Here, Base1 is the initial volume (V1 = 10.0 L), Base2 is the final volume (V2 = 50.0 L), and the Height is the pressure difference (P = P1 - P2 = 5.00 atm - 1.00 atm = 4.00 atm).
So, the work done (W) during process AB is:
W_AB = 1/2 * (10.0 L + 50.0 L) * 4.00 atm = 120.0 L*atm
To convert this to Joules, we use the conversion factor 1 L*atm = 101.3 J:
W_AB = 120.0 Latm * 101.3 J/Latm = 12156 J
(b) For process BC, the work done is the area of a rectangle, given by the formula:
Area = Base * Height
Here, the Base is the volume difference (V = V2 - V1 = 50.0 L - 10.0 L = 40.0 L), and the Height is the pressure (P = 1.00 atm).
So, the work done (W) during process BC is:
W_BC = 40.0 L * 1.00 atm = 40.0 L*atm
Again, converting to Joules:
W_BC = 40.0 Latm * 101.3 J/Latm = 4052 J
(c) The net energy input Q during the cycle is the difference between the heat added to the system and the work done by the system. Since the system returns to its initial state at the end of the cycle, the total change in internal energy is zero. Therefore, the heat added to the system (Q) is equal to the total work done by the system (W).
The total work done by the system is the work done during process AB minus the work done during process BC:
Q = W_AB - W_BC = 12156 J - 4052 J = 8104 J
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