What is the Efficiency of a Cyclic Process?

Click For Summary

Homework Help Overview

The discussion revolves around the efficiency of a cyclic process in thermodynamics, specifically focusing on the relationship between work done, heat input, and the specific heats of gases.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the efficiency formula using work and heat input definitions, while questioning the absence of specific heat values. Other participants explore the relationship between specific heats and molecular degrees of freedom, particularly for monoatomic gases.

Discussion Status

Participants are actively engaging with the problem, discussing the implications of varying degrees of freedom on specific heats and questioning the completeness of the problem statement regarding the type of gas. Some guidance has been offered regarding the calculations and assumptions made.

Contextual Notes

There is uncertainty regarding the specific heat values and the type of gas involved, which affects the calculations and interpretations of pressure. The problem does not specify the degrees of freedom, leading to multiple interpretations.

Saitama
Messages
4,244
Reaction score
93

Homework Statement


(see attachment, ignore the arrows made with the pen)


Homework Equations





The Attempt at a Solution


Efficiency of a cycle is defined as ##\eta=\frac{W}{Q}## where W is work done and Q is heat input.

W can be easily calculated by finding the area enclosed within the loop shown in the graph which is equal to ##(P-P_o)V_o##.

Heat input occurs only in the processes, B->A (##Q_1##) and B->C (##Q_2##).
##Q_1=nC_v\Delta T_1## where ##\Delta T_1=\frac{V_o}{nR}(P_o-P)##
##\Rightarrow Q_1=\frac{C_vV_o}{R}(P_o-P)##

##Q_2=nC_p\Delta T_2## and ##\Delta T_2=\frac{P_oV_o}{nR}##
##\Rightarrow Q_2=\frac{C_pP_oV_o}{R}##

\eta=\frac{W}{Q_1+Q_2}=\cfrac{(P_o-P)V_o}{\cfrac{C_vV_o}{R}(P_o-P)+\cfrac{C_pP_oV_o}{R}}
I don't have ##C_p## and ##C_v##, how am I supposed to solve this? :confused:

Any help is appreciated. Thanks!
 

Attachments

  • efficiency.jpg
    efficiency.jpg
    14.2 KB · Views: 1,375
Physics news on Phys.org
What is the relation between Cp and Cv, and how does Cv depend on the degrees of freedom of the molecules in the gas? Assume monoatomic gas first.

ehild
 
ehild said:
What is the relation between Cp and Cv, and how does Cv depend on the degrees of freedom of the molecules in the gas? Assume monoatomic gas first.

ehild

The molar specific heat at constant volume is fR/2 and at constant pressure is (f/2+1)R where f is the degrees of freedom. For a monoatomic gas, f=3, hence Cv=3R/2 and Cp=5/2. Substituting the values give ##P=2P_o/7## which is not correct. :confused:
 
f is not given, you get different pressures for different values of f. f=5 for diatomic gas and f=6 for three or more atomic molecules. For monoatomic gas, your solution should be correct.

ehild
 
ehild said:
f is not given, you get different pressures for different values of f. f=5 for diatomic gas and f=6 for three or more atomic molecules. For monoatomic gas, your solution should be correct.

ehild

I thought that it would be a bit difficult to put different values of f and find P. The answer in the key is A i.e ##0.3P_o##. I substituted it in the expression I ended up with and replaced ##C_v## and ##C_p## with fR/2 and (f/2+1)R respectively. Solving I got f=50/17. Is the question wrong? :confused:
 
2/7 is 0.3 with one significant digit. As you had to choose among different data, you should have figured out which was possible.

ehild
 
ehild said:
2/7 is 0.3 with one significant digit. The question is wrong in the sense that it did not indicate the type of the gas. ((mono-atomic).

ehild

Thank you ehild! :smile:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
Replies
35
Views
2K
Replies
1
Views
1K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 18 ·
Replies
18
Views
4K