   Chapter 12, Problem 15P

Chapter
Section
Textbook Problem

A gas expands from I to F in Figure P12.5. The energy added to the gas by heat is 418 J when the gas goes from I to F along the diagonal path. (a) What is the change in internal energy of the gas? (b) How much energy must be added to the gas by heat for the indirect path IAF to give the same change in internal energy?

(a)

To determine
The change in internal energy of the gas.

Explanation

Section1:

To determine: To determine the work done on the path IF.

Answer: The work done on the gas in the path IF is 507J .

Explanation:

Given Info:

The width of rectangle is 2L .

The height of rectangle is 1atm .

The base of triangle is 2L .

The height of the triangle is 3atm .

The energy added to the gas by heat is 418J .

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

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 2L for width, 1atm for height for rectangle and 3atm for height and 2L for base for triangle to find A,

A=(2L)[1atm(1

(b)

To determine
how much energy must be added to the gas by heat.

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