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PhysicsCollege Physics11th Edition

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Chapter 12, Problem 76AP
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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Solutions

Chapter
Section
Problem 1CQ
Problem 2CQ
Problem 3CQ
Problem 4CQ
Problem 5CQ
Problem 6CQ
Problem 7CQ
Problem 8CQ
Problem 9CQ
Problem 10CQ
Problem 11CQ
Problem 12CQ
Problem 13CQ
Problem 14CQ
Problem 15CQ
Problem 16CQ
Problem 17CQ
Problem 1P
Problem 2P
Problem 3P
Problem 4P
Problem 5P
Problem 6P
Problem 7P
Problem 8P
Problem 9P
Problem 10P
Problem 11P
Problem 12P
Problem 13P
Problem 14P
Problem 15P
Problem 16P
Problem 17P
Problem 18P
Problem 19P
Problem 20P
Problem 21P
Problem 22P
Problem 23P
Problem 24P
Problem 25P
Problem 26P
Problem 27P
Problem 28P
Problem 29P
Problem 30P
Problem 31P
Problem 32P
Problem 33P
Problem 34P
Problem 35P
Problem 36P
Problem 37P
Problem 38P
Problem 39P
Problem 40P
Problem 41P
Problem 42P
Problem 43P
Problem 44P
Problem 45P
Problem 46P
Problem 47P
Problem 48P
Problem 49P
Problem 50P
Problem 51P
Problem 52P
Problem 53P
Problem 54P
Problem 55P
Problem 56P
Problem 57P
Problem 58P
Problem 59P
Problem 60P
Problem 61P
Problem 62AP
Problem 63AP
Problem 64AP
Problem 65AP
Problem 66AP
Problem 67AP
Problem 68AP
Problem 69AP
Problem 70AP
Problem 71AP
Problem 72AP
Problem 73AP
Problem 74AP
Problem 75AP
Problem 76AP
BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A diatomic ideal gas expands from a volume of VA = 1.00 m3 to VB = 3.00 m3 along the path shown in Figure P12.76. If the initial pressure is PA = 2.00 × 105 Pa and there are 87.5 mol of gas, calculate (a) the work done on the gas during this process, (b) the change in temperature of the gas, and (c) line change in internal energy of the gas. (d) How much thermal energy is transferred to the system?

images

Figure P12.76

(a)

To determine
The work done on the gas during the process.

Explanation

Given Info:

The width of rectangle is 2 m3 .

The height of rectangle is 2×105 Pa

The base of triangle is 2 m3 .

The height of the triangle is 2×105 Pa

The area under the graph in a PV diagram is equal in magnitude to the work done on the substance. If the finial volume is greater than the initial volume, then the work done on the substance is negative.

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 m3<

(b)

To determine
The change in temperature of the gas.

(c)

To determine
The change in internal energy of the gas.

(d)

To determine
How much energy is transferred to the system.

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Chapter 12,
Problem 75AP
Chapter 13.1,
Problem 13.1QQ
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Chapter 12 Solutions

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Sect-12.1 P-12.1QQSect-12.3 P-12.2QQSect-12.4 P-12.3QQSect-12.5 P-12.4QQSect-12.5 P-12.5QQCh-12 P-1CQCh-12 P-2CQCh-12 P-3CQCh-12 P-4CQCh-12 P-5CQ
Ch-12 P-6CQCh-12 P-7CQCh-12 P-8CQCh-12 P-9CQCh-12 P-10CQCh-12 P-11CQCh-12 P-12CQCh-12 P-13CQCh-12 P-14CQCh-12 P-15CQCh-12 P-16CQCh-12 P-17CQCh-12 P-1PCh-12 P-2PCh-12 P-3PCh-12 P-4PCh-12 P-5PCh-12 P-6PCh-12 P-7PCh-12 P-8PCh-12 P-9PCh-12 P-10PCh-12 P-11PCh-12 P-12PCh-12 P-13PCh-12 P-14PCh-12 P-15PCh-12 P-16PCh-12 P-17PCh-12 P-18PCh-12 P-19PCh-12 P-20PCh-12 P-21PCh-12 P-22PCh-12 P-23PCh-12 P-24PCh-12 P-25PCh-12 P-26PCh-12 P-27PCh-12 P-28PCh-12 P-29PCh-12 P-30PCh-12 P-31PCh-12 P-32PCh-12 P-33PCh-12 P-34PCh-12 P-35PCh-12 P-36PCh-12 P-37PCh-12 P-38PCh-12 P-39PCh-12 P-40PCh-12 P-41PCh-12 P-42PCh-12 P-43PCh-12 P-44PCh-12 P-45PCh-12 P-46PCh-12 P-47PCh-12 P-48PCh-12 P-49PCh-12 P-50PCh-12 P-51PCh-12 P-52PCh-12 P-53PCh-12 P-54PCh-12 P-55PCh-12 P-56PCh-12 P-57PCh-12 P-58PCh-12 P-59PCh-12 P-60PCh-12 P-61PCh-12 P-62APCh-12 P-63APCh-12 P-64APCh-12 P-65APCh-12 P-66APCh-12 P-67APCh-12 P-68APCh-12 P-69APCh-12 P-70APCh-12 P-71APCh-12 P-72APCh-12 P-73APCh-12 P-74APCh-12 P-75APCh-12 P-76AP

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