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Pressure increase required for 1 Joule of work

  • Thread starter Bergman
  • Start date
  • #1
Bergman

Homework Statement


Estimate the pressure increase required to impart one joule of mechanical work in reversibly compressing 1 mol of silver at room temperature.
Same process for alumina.

Homework Equations


W=-PdV
dV=V(alpha)dT-V(beta)dP
dT=0

The Attempt at a Solution


I used the work equation and the equation for dV to find the following:

delta W=-P[-V(beta)dP]
delta W= V(beta) integral (PdP)
W= VBeta/2 * (P2^2-P1^2)


When I try to use this equation with the values in my text book, I am not getting the correct answer. My book gives the molar volume as 10.27 cc/mol, and the beta value for silver as 9e7.
At this point I'm not entirely sure if my equation is correct and I'm simply making a mistake in my calculations or units, or I'm very far off.


Any advice would be greatly appreciated.
 

Answers and Replies

  • #2
34,381
10,470
When I try to use this equation with the values in my text book, I am not getting the correct answer.
It would help if you could say what you get and what the book gets.
 
  • #3
Bergman
I have continued working on the problem, and found the relationship 1J=9.8699 cc atm. When I use this I get:

2* 9.8699 cc atm / (10.27 cc * 9e-7) = P2^2-P1^2

In which case I get 1461.4 atm

The book gave the compressibility as 9 e7, but I think it should be 9 e-7

The answer in the back of the textbook is 9 e6 atm
 
  • #4
34,381
10,470
Neglecting atmospheric pressure:
This page quotes 0.00993 GPa-1 for silver and confirms the molar volume.

At 1461*105 Pa, we compress it by 0.00145, and I get 1.09 J, that fits.

At 9 million times the atmospheric pressure (???) silver doesn't behave linearly any more.
 
  • #5
Bergman
Thank you for the help, I really appreciate it. I may contact my professor to ask if the textbook may be wrong, considering that I have been able to do the same calculation for alumina, and get the expected value.
 

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