Thermodynamics: Calculating Pressure Increase From Work

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SUMMARY

The forum discussion focuses on calculating the pressure increase required to impart 1 J of mechanical work in the reversible compression of 1 mol of silver and alumina at room temperature. The correct pressure increase for silver is approximately 9.6 x 106 atm, while for alumina it is 978 atm. Several users identified errors in calculations, particularly in the conversion of mechanical work from Joules to cm3 atm, which significantly affected the results. The importance of using the correct coefficient of compressibility (BETA) values for both materials was emphasized in achieving accurate outcomes.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically mechanical work and pressure-volume relationships.
  • Familiarity with the concept of compressibility and its coefficient (BETA).
  • Basic knowledge of unit conversions, particularly between Joules and cm3 atm.
  • Proficiency in applying the ideal gas law and related equations in thermodynamic calculations.
NEXT STEPS
  • Study the derivation and application of the mechanical work equation: Mechanical Work = -PdV.
  • Learn about the significance of the coefficient of compressibility (BETA) in material science.
  • Research unit conversion techniques between different pressure and volume units.
  • Explore advanced thermodynamic concepts such as isothermal and adiabatic processes in gas compression.
USEFUL FOR

This discussion is beneficial for students and professionals in chemistry and materials science, particularly those involved in thermodynamics, mechanical engineering, and anyone working with pressure-volume relationships in gases and solids.

Matt James
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Homework Statement


Estimate the pressure increase required to impart 1 J of mechanical work in reversibly compressing 1 mol of silver at room temperature. What pressure rise would be required to impart 1 J of work to 1 mol of alumina at room temperature? For alumina take the molar volume to be 25.715 (cc/mol) and (BETA)=8*10^(-7) (atm)^(-1).
For silver, the molar volume is 10.27 (cc/mol) and (BETA)= 9.93*10^(-6)
Beta is the coefficient of compressibility

Homework Equations


Mechanical Work= -PdV
dV=V(ALPHA)dt-V(BETA)dP

The Attempt at a Solution


I assumed that temperature remained constant during this process. I know that the answer should be 9.6*(10)^6 atm for silver and 978 atm for alumina. I have been getting 140041 atm for silver and 311800 atm for alumina. The attempt is attached below. Any ideas as to what I'm doing wrong? Thanks in advance!
0
 
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Matt James said:

Homework Statement


Estimate the pressure increase required to impart 1 J of mechanical work in reversibly compressing 1 mol of silver at room temperature. What pressure rise would be required to impart 1 J of work to 1 mol of alumina at room temperature? For alumina take the molar volume to be 25.715 (cc/mol) and (BETA)=8*10^(-7) (atm)^(-1).
For silver, the molar volume is 10.27 (cc/mol) and (BETA)= 9.93*10^(-6)
Beta is the coefficient of compressibility

Homework Equations


Mechanical Work= -PdV
dV=V(ALPHA)dt-V(BETA)dP

The Attempt at a Solution


I assumed that temperature remained constant during this process. I know that the answer should be 9.6*(10)^6 atm for silver and 978 atm for alumina. I have been getting 140041 atm for silver and 311800 atm for alumina. The attempt is attached below. Any ideas as to what I'm doing wrong? Thanks in advance!
0
I can't open your attachment. Have you uploaded the file?
 
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Sorry, thought I attached the image to the original post

unnamed.jpg
 

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Chestermiller said:
I can't open your attachment. Have you uploaded the file?

I just realized what I was doing wrong. I needed to convert the mechanical work (in Joules) to units of cm^3 atm. This gives me the right answer for alumina, but I'm getting 1461.34 atm for silver. Attached below is the work with the conversions in mind

unnamed-2.jpg
 

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Matt James said:
I just realized what I was doing wrong. I needed to convert the mechanical work (in Joules) to units of cm^3 atm. This gives me the right answer for alumina, but I'm getting 1461.34 atm for silver. Attached below is the work with the conversions in mind

View attachment 214291
This result doesn't seem to agree with your result shown on the paper. Please show your work, including the substitutions.

I confirm the 979 atm for alumina. For silver, from the data given, I get 440 atm.
 
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Chestermiller said:
This result doesn't seem to agree with your result shown on the paper. Please show your work, including the substitutions.

I confirm the 979 atm for alumina. For silver, from the data given, I get 440 atm.
Here's the work for the silver sample, I'm still getting 1461 atm
unnamed-3.jpg
 

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Matt James said:
Here's the work for the silver sample, I'm still getting 1461 atm
View attachment 214295
It looks like you used the wrong value of beta in your calculation. Otherwise, nicely done.
 
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Chestermiller said:
It looks like you used the wrong value of beta in your calculation. Otherwise, nicely done.
Okay, thank you!
 

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