   Chapter 12.1, Problem 12.1QQ

Chapter
Section
Textbook Problem

By visual inspection, order the PV diagrams shown in Figure 12.5 from the most negative work done on the system to the most positive work done on the system, (a) a, b, c, d (b) a, c, b, d (c) d, b, c, a (d) d, a, c, b

To determine
The work done on the systems.

Explanation

Section1:

To determine: The area A of diagram (a).

Answer: The Area A is 4×105J .

Explanation:

Given Info:

The width of rectangle is 2.00m3 .

The height of rectangle is 1×105Pa .

The base of triangle is 2 m3 .

The height of the triangle is 2×105Pa .

Formula to calculate the area is,

A=(W1×H1)+12(B1×H2)

• W1 is the width of rectangle
• H1 is the height of rectangle
• B1 is the base of triangle
• H2 is the height of triangle

Substitute 2.00m3 for width, 1×105Pa for height for rectangle and 2×105Pa for height and 2 m3 for base for triangle to find A.

A=(2m3)(1×105Pa)+12(2m3)(2Pa)=4×105J

The finial volume is larger than the initial volume, the work done is negative. Therefore, A=4×105J

Section2:

To determine: The area A of diagram (b).

Answer: The Area A is 3×105J .

Explanation:

Given Info:

The width of rectangle A1 is 1.00m3 .

The height of rectangle A1 is 1×105Pa .

The width of rectangle A2 is 1.00m3 .

The height of rectangle A2 is 2×105Pa .

Formula to calculate the area is,

A=(W1×H1)+(W2×H2)

• W1 is the width of rectangle A1
• W1 is the height of rectangle A1
• W1 is the width of rectangle A2
• W1 is the height of rectangle A2

Substitute 1.00m3 for width, 1×105Pa for height for rectangle A1 and 2×105Pa for height and 1.00m3 for width for triangle to find A.

A=(1m3)(1×105Pa)+(1m3)(2×105Pa)=3×105J

The finial volume is smaller than the initial volume, the work done is positive. Therefore, A=3×105J

Section3:

To determine: The area A of diagram (c).

Answer:  The Area A is 3×105J .

Explanation:

Given Info:

The width of rectangle is 2.00m3

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