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Topic: 1st Law Thermodynamics  (Read 4578 times)

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  • Guest
1st Law Thermodynamics
« on: September 02, 2004, 05:22:05 PM »
I have been working on this problem for hours now, and have not gotten very far.  This question is a 4 part question.  Part 1 we had to determine the pressure  in the ocean at a depth of 2500 meters.  This was found to be 2.48x10^5 atm.  However part 2 is where i get stuck.  I was given an isothermal compressibility of 49.5x10^-6 atm^-1 and the equation Beta=(-deltaV)/Vi x deltaP.  I am to determine the percentage difference in the density at sea level compared to this depth at 2500 meters.    I also know that the density of sea water at sea level is 1.025kgL^-1 and the pressure at the surface is 1 atm.  I believe i can solve parts 3 and 4 if i figure this step out.  PLEASE *delete me*  thanks so much.


  • Guest
Re:1st Law Thermodynamics
« Reply #1 on: September 03, 2004, 03:16:37 PM »
I think your pressure, 2.48 EE 5 is off. It probably should be 2.48 EE 2 atm, considering 1 atm per 10 meter a general guide.
So, delta P is 248 - 1 = 247 atmospheres.
Assuming Beta value is correct:
49.5x10^-6 =(-deltaV)/Vi x deltaP =(-deltaV)/Vi x 247

(deltaV)/Vi =-.012
deltaV = -.012Vi
indicates a 1.2% contraction in volume
final density/ini density =Vi/Vf = Vi/.988Vi = 1.012
final dens = 1.012(1.025kg/L)

« Last Edit: September 05, 2004, 03:30:32 PM by Demotivator »

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