What is the entropy change of gas during an internal combustion engine test?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the change in entropy of nitrogen gas during a test of an internal combustion engine, where the gas undergoes a sudden and irreversible compression. The problem involves concepts from thermodynamics, particularly the behavior of gases under changing volume and temperature conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need for the number of moles of gas and whether both temperature and volume changes must be considered in the entropy calculation. There are attempts to clarify the correct application of formulas and the importance of using absolute temperatures.

Discussion Status

There are multiple interpretations of how to approach the problem, with some participants suggesting different methods and equations. Guidance has been offered regarding the use of specific heat and the treatment of temperature in Kelvin. The discussion is ongoing, with participants sharing their calculations and questioning each other's assumptions.

Contextual Notes

Some participants express uncertainty about the application of the formulas provided and the necessity of converting temperatures to Kelvin. There is also mention of a reference sheet for further clarification on entropy calculations.

sparkle123
Messages
172
Reaction score
0

Homework Statement



During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 C was compressed suddenly (and irreversibly) to 0.500L by driving in a piston. In the process the temperature of the gas increased to 28.1 C. Assume ideal behavior. What is the change in entropy of the gas?

Homework Equations



dS=nRln(V2/V1)
dS=Cln(T2/T1)
C=5R/2

The Attempt at a Solution


I think that you need the number of moles of gas to be able to solve this...
The answer provided is -14.6 J/K but how do you do this? Please help!
 
Last edited:
Physics news on Phys.org
I would use your second equation, noting that the specific heat with constant pressure = 5R/2. Since R is a constant and the two temperatures are given, you can solve that without knowing n. Don't forget to convert to Kelvin, and do note that ds in your equations is not a differential change in entropy, but rather a (delta-s) change in entropy.
 
So I don't need to take into account the change in volume?
Thanks!
 
sparkle123 said:
So I don't need to take into account the change in volume?
Thanks!

I don't think so, but then, I'm rusty with some of this thermo stuff.
 
You do need to take account of both. Since entropy is a state function, you can imagine any process from state A to state B, and the entropy change will be the same. So you can imagine first an isothermal compression from 3.0L to 0.5L, followed by a change in temperature from 18.5 to 21.8C at constant volume. When do those formulas you gave apply?
 
I plugged the numbers in as follows:

[tex] c_v = 20.81 \frac{J}{K\ mol}[/tex]

[tex] R = 8.314472 \frac{J}{mol\ K}[/tex]

[tex] dS = c_v ln(\frac{T2}{T1}) + R\ ln(\frac{2}{V1}) = -14.22 \frac{J}{mol\ K}[/tex]

The result is close to, but not the same as, the supposed solution value. The solution also seems to have mucked up the units.

A reasonable reference sheet for this stuff is http://www.grc.nasa.gov/WWW/K-12/airplane/entropy.html"
 
Last edited by a moderator:
I plugged in the same values and got a different answer. In your formula, you have an "ln(2/V1)". I assume you mean ln(V2/V1) which is ln(0.5/2), right?
 
I think I found your error. You forgot to convert celsius to Kelvin when you did T2/T1, didn't you? You can only use an absolute temperature any time you're dividing temperatures in thermodynamics.
 
LeonhardEuler said:
I plugged in the same values and got a different answer. In your formula, you have an "ln(2/V1)". I assume you mean ln(V2/V1) which is ln(0.5/2), right?

Right.
 

Similar threads

Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 14 ·
Replies
14
Views
8K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 60 ·
3
Replies
60
Views
11K
  • · Replies 10 ·
Replies
10
Views
2K