Nov 20, 2000

PHY 113 -- Solution to Extra PHY 113 - Solution to Extra Practice Problem for Chapters 19-22

Note: Please send email if you have any questions about this solution. (Extra credit will be awarded for any errors you find in the solution.)

Since this process involves an ideal gas, we know that

PV = nRT
and
Eint = 1
g-1
 nRT = 1
g-1
 PV
throughout the process.

  1. Using the First Law of Thermodynamics, we can complete the table according to:

    DEint DQ DW
    A ® B [1/( g-1)](P2-P1)V1 [1/( g-1)](P2-P1)V1 0
    B ® C [1/( g-1)]P2(V2-V1) [(g)/( g-1)]P2(V2-V1) P2(V2-V1)
    C ® D -[1/( g-1)](P2-P1)V2 -[1/( g-1)](P2-P1)V20
    D ® A -[1/( g-1)]P1(V2-V1) -[(g)/( g-1)]P1(V2-V1) -P1(V2-V1)

  2. From the ideal has law, we see that
    T = PV
    nR
    		 ,
    so that the highest temperature corresponds to the largest PV product which occurs at step C, and the lowest temperature corresponds to the lowest PV product which occurs at step A.
  3. The efficiency of this process can be determined from
    e = Wnet
    Qinput
    		 = WAB+WBC+WCD+WDA
    QAB+QBC
    		 .
    After a little algebra, this result simplifies to the following:
    e = g-1
    V1
    V2-V1
    		 + g P2
    P2-P1
    		.

File translated from TEX by TTH, version 2.20.
On 20 Nov 2000, 14:53.