Question

Calculate the work done by an ideal gas while going from state A to state C in Fig. 15–28 for each of the following processes:

(a) ADC,

(b) ABC, and

(c) AC directly.

FIGURE 15–28

Problem 68

Short answer

1. The work done in the process ADC is\({P_{\rm{A}}}\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\).
2. The work done in the process ABC is\({P_{\rm{C}}}\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\).
3. The work done in the process AC is \(\frac{1}{2}\left( {{P_{\rm{C}}} + {P_{\rm{A}}}} \right)\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\).

Step-by-step solution

Understanding the work-done evaluation by the PV diagram

The total work done in the PV diagram is the total area of the PV curve. The total work done in any process is the sum of the work done during each process.

In this problem, evaluate the entire area in between the curve to the volume axis. The work done in any cyclic thermodynamic process will be equal to the heat added.

The representation of the PV diagram

The PV diagram can be shown as

Here, P is the pressure on the vertical axis, and V is the volume on the horizontal axis.

(a) Determination of the work done in process ADC

The work done in a process is

\(W = P\Delta V\). … (i)

Here, P is the pressure, and \(\Delta V\) is the change in the volume of the ideal gas.

Now, write the equations of the work done in the processes with the help of equation (i).

In area ADC, the work done is equal to the sum of the work done in process AD and the work done in process CD. Process CD is the constant volume process.

So, the change in the volume of process CD is zero. The volume at state point C is equal to the volume at D state point.

The work done in process ADC can be expressed as

\(\begin{array}{c}{W_{{\rm{ADC}}}} = {P_{\rm{A}}}\Delta {V_{{\rm{AD}}}} + {P_{\rm{D}}}\Delta {V_{{\rm{CD}}}}\\ = {P_{\rm{A}}}\left( {{V_{\rm{D}}} - {V_{\rm{A}}}} \right) + {P_{\rm{A}}} \times 0\\ = {P_{\rm{A}}}{V_{\rm{C}}} - {P_{\rm{A}}}{V_{\rm{A}}} + 0\\ = {P_{\rm{A}}}\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\end{array}\).

Thus, the work done in the process ADC is \({P_{\rm{A}}}\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\).

(b) Determination of the work done in process ABC

In area ABC, the work done is equal to the sum of the work done in process AB and the work done in process BC. The process AB is the constant volume process. So, the change in the volume of process AB is zero.

The volume at state point B is equal to the volume at A state point.

The work done in process ABC can be expressed as

\(\begin{array}{c}{W_{{\rm{ABC}}}} = {P_{\rm{A}}}\Delta {V_{{\rm{AB}}}} + \Delta {P_{\rm{C}}}\Delta {V_{{\rm{BC}}}}\\ = {P_{\rm{A}}} \times 0 + {P_{\rm{C}}}\left( {{V_{\rm{C}}} - {V_{\rm{B}}}} \right)\\ = 0 + {P_{\rm{C}}}{V_{\rm{C}}} - {P_{\rm{C}}}{V_{\rm{A}}}\\ = {P_{\rm{C}}}\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\end{array}\).

Thus, the work done in process ABC is \({P_{\rm{C}}}\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\).

(c) Determination of the work done in process AC

This can be obtained by the area below the curve AC.

In the curve, the closed area (aACb) is a trapezoid. In this trapezoid, (aA) and (bC) are the parallel sides, (ab) is the height, which is the distance between the two parallel sides.

The volume at state point C is equal to the volume at D state point. The pressure at state point D is equal to the pressure at A state point.

The work done in process AC can be expressed as

\(\begin{array}{c}{W_{{\rm{AC}}}} = \frac{1}{2} \times {\rm{sum of parallel sides}} \times {\rm{distance between the parallel sides}}\\ = \frac{1}{2} \times \left( {{P_{\rm{C}}} + {P_{\rm{D}}}} \right) \times \left( {{V_{\rm{D}}} - {V_{\rm{A}}}} \right)\\ = \frac{1}{2}\left( {{P_{\rm{C}}} + {P_{\rm{A}}}} \right)\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\end{array}\).

Thus, the work done in the process AC is \(\frac{1}{2}\left( {{P_{\rm{C}}} + {P_{\rm{A}}}} \right)\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\).