How to find the thermal compressibility of a gas

AI Thread Summary
The discussion focuses on finding the thermal compressibility of a gas described by the equation of state PV=RT during a reversible process. The thermal compressibility is defined as κ = -1/V (δV/δP), and attempts to derive it using the given equations yield results that include pressure terms. The calculations suggest that at high temperatures, the thermal compressibility approaches κ = 1/P, indicating it tends to a constant value. Participants are encouraged to verify the calculations and provide insights on any potential errors. The exploration of this problem emphasizes the relationship between thermal compressibility and temperature in gas behavior.
PKM
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Homework Statement

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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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