How to find the thermal compressibility of a gas

In summary: This suggests that the thermal compressibility of the gas undergoing this process tends to a constant value at very high temperature.
  • #1
PKM
49
16

Homework Statement

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

Homework Equations


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

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 [itex]PV = RT[/itex], to substitute [itex]RT[/itex] for [itex]PV[/itex]. 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?
 
Physics news on Phys.org
  • #2
What did you get for ##\kappa##? I'll tell you if I confirm.
 
  • #3
Chestermiller said:
What did you get for ##\kappa##? I'll tell you if I confirm.
I proceeded in two ways:
First, in the equation [itex]
PV^\frac{5}{3} e^\frac{-PV}{E_0} = c_1[/itex], I put [itex]PV=RT[/itex]. Then partially differentiated it w.r.t. [itex]P[/itex]. It yields [tex]\frac{1}{\sqrt[3] {V}}\frac{dV}{dP} = const[/tex]
Secondly, I differentiated the equation
[itex]
PV^\frac{5}{3} e^\frac{-PV}{E_0} = c_1[/itex] w.r.t. [itex]P[/itex]. The result appears to be [tex]-\frac{1}{V}\frac{\delta V}{\delta P} = \kappa = \frac{1-E_0}{P}+\frac{5}{3V}[/tex]
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?
 
  • #4
PKM said:
I proceeded in two ways:
First, in the equation [itex]
PV^\frac{5}{3} e^\frac{-PV}{E_0} = c_1[/itex], I put [itex]PV=RT[/itex]. Then partially differentiated it w.r.t. [itex]P[/itex]. It yields [tex]\frac{1}{\sqrt[3] {V}}\frac{dV}{dP} = const[/tex]
Secondly, I differentiated the equation
[itex]
PV^\frac{5}{3} e^\frac{-PV}{E_0} = c_1[/itex] w.r.t. [itex]P[/itex]. The result appears to be [tex]-\frac{1}{V}\frac{\delta V}{\delta P} = \kappa = \frac{1-E_0}{P}+\frac{5}{3V}[/tex]
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##.
 

Related to How to find the thermal compressibility of a gas

1. What is thermal compressibility of a gas?

Thermal compressibility is a measure of how much a gas will decrease in volume when its temperature is increased while keeping the pressure constant. It is a property that describes the response of a gas to changes in temperature.

2. How is thermal compressibility calculated?

The thermal compressibility of a gas can be calculated by dividing the change in volume by the change in temperature at constant pressure. This can be represented by the equation: β = -1/V(∂V/∂T)P, where β is the thermal compressibility, V is the volume, T is the temperature, and P is the pressure.

3. What factors affect the thermal compressibility of a gas?

The thermal compressibility of a gas is affected by its molecular structure and the strength of intermolecular forces, as well as the temperature and pressure of the gas. Gases with weaker intermolecular forces typically have higher thermal compressibility.

4. How is thermal compressibility related to other properties of gases?

Thermal compressibility is related to other properties of gases such as specific heat capacity, thermal expansion coefficient, and adiabatic compressibility. These properties are all dependent on the molecular structure and behavior of gases under different conditions.

5. Why is thermal compressibility important in scientific research?

Thermal compressibility is an important property in scientific research as it helps to understand the behavior of gases under different conditions. It is particularly useful in studying the relationship between temperature, pressure, and volume of gases, which has applications in fields such as thermodynamics, chemistry, and engineering.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
916
  • Introductory Physics Homework Help
Replies
33
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
261
  • Introductory Physics Homework Help
Replies
1
Views
639
  • Introductory Physics Homework Help
Replies
3
Views
343
  • Introductory Physics Homework Help
Replies
2
Views
856
  • Introductory Physics Homework Help
Replies
3
Views
929
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
2
Views
819
Back
Top