Thermodynamics: Calculating Pressure Increase From Work

[Matt James]

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!

 

[Chestermiller]

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!

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

 

[Matt James]

Sorry, thought I attached the image to the original post

 

[Matt James]

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

 

[Chestermiller]

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.

 

[Matt James]

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

 

[Chestermiller]

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.

 

[Matt James]

It looks like you used the wrong value of beta in your calculation. Otherwise, nicely done.

Okay, thank you!