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Chapter 15: Q68GP (page 412)

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

(a) ADC,

(b) ABC, and

(c) AC directly.

FIGURE 15鈥�28

Problem 68

Short Answer

Expert verified
  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

01

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.

02

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.

03

(a) Determination of the work done in process ADC

The work done in a process is

\(W = P\Delta V\). 鈥� (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 \({P_{\rm{A}}}\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\).

04

(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 \({P_{\rm{C}}}\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\).

05

(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 \(\frac{1}{2}\left( {{P_{\rm{C}}} + {P_{\rm{A}}}} \right)\left( {{V_{\rm{C}}} - {V_{\rm{A}}}} \right)\).

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Most popular questions from this chapter

Question: (II) Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.4 atm. Then the gas expands at constant pressure, from a volume of 5.9 L to 9.3 L, where the temperature reaches its original value. See Fig.15鈥�22. Calculate (a) the total work done by the gas in the process, (b) the change in internal energy of the gas in the process, and (c) the total heat flow into or out of the gas.

Question: (a) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 750掳C and 440掳C, and of the second 415掳C and 270掳C. If the heat of combustion of coal is \({\bf{2}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{7}}}\;{{\bf{J}} \mathord{\left/{\vphantom {{\bf{J}} {{\bf{kg}}}}} \right.} {{\bf{kg}}}}\) at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the ideal (Carnot) efficiency. (b) Water is used to cool the power plant. If the water temperature is allowed to increase by no more than 4.5 C掳, estimate how much water must pass through the plant per hour.

(II) Sketch a PV diagram of the following process: 2.5 L of ideal gas at atmospheric pressure is cooled at constant pressure to a volume of 1.0 L, and then expanded isothermally back to 2.5 L, whereupon the pressure is increased at constant volume until the original pressure is reached.

A particular car does work at the rate of about\({\bf{7}}{\bf{.0}}\;{\bf{kJ/s}}\)when traveling at a steady\({\bf{21}}{\bf{.8}}\;{\bf{m/s}}\)along a level road. This is the work done against friction. The car can travel 17 km on 1.0 L of gasoline at this speed (about 40 mi/gal). What is the minimum value for\({{\bf{T}}_{\bf{H}}}\)if\({{\bf{T}}_{\bf{L}}}\)is 25掳C? The energy available from 1.0 L of gas is\({\bf{3}}{\bf{.2 \times 1}}{{\bf{0}}{\bf{7}}}\;{\bf{J}}\).

(II) When\({\bf{5}}{\bf{.80 \times 1}}{{\bf{0}}{\bf{5}}}\;{\bf{J}}\)of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from\({\bf{1}}{\bf{.9}}\;{{\bf{m}}{\bf{3}}}\)to\({\bf{4}}{\bf{.1}}\;{{\bf{m}}{\bf{3}}}\). Calculate

(a) the work done by the gas, and

(b) the change in internal energy of the gas.

(c) Graph this process on a PV diagram.
See all solutions

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