Isothermal and Adiabatic Compression of a Solid

Click For Summary

Homework Help Overview

The problem involves the isothermal and adiabatic compression of a metallic copper cylinder, focusing on changes in internal energy, heat exchange, and temperature increase under different conditions. The context is thermodynamics, specifically relating to solids and their behavior under pressure changes.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the relationship between internal energy and temperature for solids, questioning whether the internal energy change can be considered zero during isothermal compression. They also raise concerns about the role of mass and volume in calculating heat exchange and temperature changes during adiabatic processes.

Discussion Status

Participants are exploring the appropriate equations and concepts related to internal energy changes under constant temperature and pressure. Some guidance has been offered regarding the nature of internal energy for solids and the need to consider pressure changes, but no consensus has been reached on specific equations or methods to apply.

Contextual Notes

There is a mention of the complexity introduced by large pressure changes and the potential need for specific thermodynamic equations that may not be covered in introductory texts. The original poster's level of study in thermodynamics is also uncertain, which may affect the discussion.

Spiral1183
Messages
5
Reaction score
0

Homework Statement


A 200g cylinder of metallic copper is compressed isothermally and quasi-statically at 290K in a high-pressure cell.
A) Find the change in internal energy of the copper when the pressure is increased from 0 to 12kbar.
B) How much heat is exchanged with the surrounding fluid?
C) If the process is instead carried out adiabatically, find the temperature increase of the copper.
For copper,
CP=16J(mol*K)-1, β=32x10-6K-1, κ=0.73x10-6atm-1, and v=7cm3mol-1

Homework Equations


For the isothermal compression, we found Q=-Tvβ(Pf-Pi)

For the adiabatic change in temperature, ΔT=T(βv/CP)(Pf-Pi)

The Attempt at a Solution


For part A, it is asking me to find the internal energy, which is the heat minus the work done correct? If there is no change in temperature, wouldn't this be zero?

For part B in the isothermal process, is it as simple as plugging in the known variables to find the heat exhange? I'm thinking that I should be taking into account the mass of the copper which should change the volume, correct?

Part C I am looking at the same way, but not sure if I need to factor in the mass of the copper.
 
Last edited:
Physics news on Phys.org
The internal energy is a function of temperature only for an ideal gas. It is also a good approximation for solids, but in this case, the pressure change is very large, and you are being asked to determine the change in internal energy as a function of pressure at constant temperature.
 
So in that case what equation would I use to determine the change in internal temperature? Everything I am finding seems to deal with ideal gas or changes in temperature with constant pressure, not the other way around.
 
Spiral1183 said:
So in that case what equation would I use to determine the change in internal temperature? Everything I am finding seems to deal with ideal gas or changes in temperature with constant pressure, not the other way around.

I don't know what level of thermo you are taking. But even many introductory texts derive expressions for the partial derivatives of the thermodynamic functions with respect to pressure at constant temperature (or the partial derivatives with respect to specific volume at constant temperature). Hopefully your textbook covers this. If not, see Smith and van Ness, or Hougan and Watson.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
4K
Replies
3
Views
10K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
8K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K