   Chapter 12, Problem 18P

Chapter
Section
Textbook Problem

A quantity of a monatomic ideal gas undergoes a process in which both its pressure and volume are doubled as shown in Figure P12.18. What is the energy absorbed by heat into the gas during this process? Hint: The internal energy of a monatomic ideal gas at pressure P and occupying volume V is given by U  =  3 2 P V . Figure P12.18

To determine
The energy absorbed by heat in to the gas during the process.

Explanation

Section1:

To determine: the work done on the gas

Answer: the work done on the gas is W=32P0V0

Explanation:

Given Info:

Initial volume of the gas is V0

Final volume of the gas is 2V0

Initial pressure of the gas is P0

Finial pressure of the gas is 2P0

The PV diagram of the gas is,

The area under the path on a PV diagram is equal to the magnitude of the work done on the gas.

Formula to calculate the area under PV diagram is,

A=(W1×H1)+12(B1×H2)

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

Substitute V0 for W1 , P0 for H1 , V0 for B1 and P0 for H2 to find the area of the PV diagram,

A=P0V0+12P0V0=32P0V0       (1)

Since, the finial volume is higher than the initial volume; the work done on the gas is negative.

Thus, the work done on the gas is,   W=32P0V0 .

Section 2:

To determine: To determine the change in internal energy of the ideal gas.

Answer: The change in internal energy of the ideal gas is 92P0V0 .

Explanation:

Given Info:

The internal energy of a monatomic ideal gas at pressure P and volume Vis given by U=32PV

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