How to find the thermal compressibility of a gas

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Homework Help Overview

The discussion revolves around determining the thermal compressibility of a gas described by a specific equation of state and undergoing a reversible process. The original poster seeks to understand whether the thermal compressibility tends to a constant value at very high temperatures.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods to derive the thermal compressibility, including substituting the equation of state into the given process equation and differentiating with respect to pressure. There are questions about the presence of terms that prevent simplification to a constant value at high temperatures.

Discussion Status

Several participants are actively engaging with the problem, sharing their calculations and questioning each other's approaches. There is a focus on verifying results and identifying potential errors in reasoning or calculations, but no consensus has been reached on the correct interpretation or outcome.

Contextual Notes

Participants are working under the constraints of the provided equations and definitions, and there is an emphasis on the behavior of the thermal compressibility as temperature increases. The constants involved in the equations are noted but not fully defined within the discussion.

PKM
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Homework Statement

:
[/B]
A gas obeying the equation of state PV=RT undergoes a hypothetical reversible process <br /> PV^\frac{5}{3} e^\frac{-PV}{E_0} = c_1 Can we prove that the thermal compressibility of the gas undergoing this process tends to a constant value at very high temperature? Here, E_0 and c_1 are constants with dimensions.

Homework Equations


The thermal compressibility of a gas is given as \kappa = \frac{-1}{V} \frac{\delta V}{\delta P}

The Attempt at a Solution


First I tried to find the thermal compressibility using the above differential equation, considering the reversible process given. I made use of the equation of state PV = RT, to substitute RT for PV. My result contains a term P, which cannot be cancelled, or substituted to yield a constant value, at very high temperatures.
Any solution, or comment?
 
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What did you get for ##\kappa##? I'll tell you if I confirm.
 
Chestermiller said:
What did you get for ##\kappa##? I'll tell you if I confirm.
I proceeded in two ways:
First, in the equation <br /> PV^\frac{5}{3} e^\frac{-PV}{E_0} = c_1, I put PV=RT. Then partially differentiated it w.r.t. P. It yields \frac{1}{\sqrt[3] {V}}\frac{dV}{dP} = const
Secondly, I differentiated the equation
<br /> PV^\frac{5}{3} e^\frac{-PV}{E_0} = c_1 w.r.t. P. The result appears to be -\frac{1}{V}\frac{\delta V}{\delta P} = \kappa = \frac{1-E_0}{P}+\frac{5}{3V}
Maybe somewhere I've gone wrong (it would be much kind of you if you point it out); but if my calculations are correct, how may I proceed?
 
PKM said:
I proceeded in two ways:
First, in the equation <br /> PV^\frac{5}{3} e^\frac{-PV}{E_0} = c_1, I put PV=RT. Then partially differentiated it w.r.t. P. It yields \frac{1}{\sqrt[3] {V}}\frac{dV}{dP} = const
Secondly, I differentiated the equation
<br /> PV^\frac{5}{3} e^\frac{-PV}{E_0} = c_1 w.r.t. P. The result appears to be -\frac{1}{V}\frac{\delta V}{\delta P} = \kappa = \frac{1-E_0}{P}+\frac{5}{3V}
Maybe somewhere I've gone wrong (it would be much kind of you if you point it out); but if my calculations are correct, how may I proceed?
I get $$\kappa=\frac{1}{P}\frac{\left(\frac{PV}{E_0}-1\right)}{\left(\frac{PV}{E_0}-\frac{5}{3}\right)}$$At high temperatures, this approaches ##\kappa=1/P##.
 

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