Isothermal and Adiabatic Compression of a Solid

Click For Summary
SUMMARY

The discussion focuses on the isothermal and adiabatic compression of a 200g cylinder of metallic copper at 290K. For isothermal compression, the heat exchange is calculated using the equation Q=-Tvβ(Pf-Pi), while the adiabatic process requires the equation ΔT=T(βv/CP)(Pf-Pi) to determine the temperature increase. The internal energy change is zero during isothermal compression due to constant temperature, and the mass of copper must be considered when calculating volume changes. The discussion emphasizes the need for understanding thermodynamic functions and their derivatives with respect to pressure and specific volume.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically isothermal and adiabatic processes.
  • Familiarity with the equations of state for solids, particularly for copper.
  • Knowledge of heat exchange calculations in thermodynamics.
  • Ability to apply partial derivatives of thermodynamic functions.
NEXT STEPS
  • Study the derivation of thermodynamic functions with respect to pressure at constant temperature.
  • Learn about the specific heat capacities of solids, focusing on copper.
  • Explore the implications of large pressure changes on internal energy in solids.
  • Review introductory thermodynamics texts, such as Smith and van Ness or Hougan and Watson.
USEFUL FOR

Students and professionals in thermodynamics, particularly those studying the behavior of materials under compression, such as mechanical engineers and materials scientists.

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