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Figure shows the pV-diagram for an isothermal expansion of

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 38P Chapter 19

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 38P

Figure shows the ?pV?-diagram for an isothermal expansion of 1.50 mol of an ideal gas, at a temperature of 15.0°C. (a) What is the change in internal energy of the gas? Explain. (b) Calculate the work done by (or on) the gas and the heat absorbed (or released) by the gas during the expansion. Figure:

Step-by-Step Solution:

Solution 38P Step 1: (a).Internal energy is due to motion of particles in a system. As internal energy depends on temperature. As we know temperature in isothermal process is constant so the internal energy will also be constant thus the change in internal energy will be zero. U = 0 Step 2: (b).The ideal gas equation PV = nRT V2 Work done by the gas is given by W = nRT/V dV V 1 W = nRT ln(V /V2) d1 W = 1.50 × (8.314)(288 K)ln(0.02/0.01) W = 1.50 × (8.314)(288 K) × 0.693 W = 2489.01 J

Step 3 of 3

Chapter 19, Problem 38P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Figure shows the pV-diagram for an isothermal expansion of

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