Thermodynamics: Calculating Pressure Increase From Work

In summary: I'll double check my calculations and see where the error is. Thank you for your help!In summary, the conversation revolved around estimating the pressure increase required to impart 1 J of mechanical work in compressing 1 mol of silver and alumina at room temperature. The molar volumes and coefficients of compressibility for both substances were given, and the student attempted to solve for the required pressure rise. However, there was some discrepancy in the results, and the student later realized that they had used the wrong value of beta in their calculation for silver. After correcting this error, they obtained the correct answer of 440 atm for silver and 979 atm for alumina.
  • #1
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!
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  • #2
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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  • #3
Sorry, thought I attached the image to the original post

unnamed.jpg
 

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  • #4
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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  • #5
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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  • #6
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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  • #7
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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  • #8
Chestermiller said:
It looks like you used the wrong value of beta in your calculation. Otherwise, nicely done.
Okay, thank you!
 

Related to Thermodynamics: Calculating Pressure Increase From Work

1. What is thermodynamics?

Thermodynamics is the branch of physics that deals with the relationship between heat, energy, and work. It studies the behavior of systems that involve temperature, pressure, and volume.

2. How is pressure increase calculated from work?

The formula for calculating pressure increase from work is P = W/V, where P is the pressure, W is the work done, and V is the change in volume. This formula is based on the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

3. What are the units of measurement for pressure and work?

Pressure is typically measured in units of Pascal (Pa) or Newtons per square meter (N/m^2). Work is measured in units of Joules (J). However, other units such as atmosphere (atm) or pounds per square inch (psi) may also be used.

4. What factors can affect the pressure increase in a system?

The pressure increase in a system can be affected by several factors, including the amount of work done on the system, the temperature, and the volume of the system. Additionally, the type of gas or substance in the system and any external forces or constraints can also affect the pressure increase.

5. How is thermodynamics used in real-world applications?

Thermodynamics has many practical applications, such as in the design and operation of engines, refrigerators, and air conditioners. It is also used in industrial processes, such as power generation and chemical reactions. Additionally, the principles of thermodynamics are essential in understanding weather patterns and climate change.

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