Finding the Joule Coefficient

In summary, the conversation is about a problem with a demonstration involving a system containing gas in a container with no heat exchange or work with the environment. The goal is to find the Joule coefficient, which is equal to -(T^2/Cv)(d/dT)(P/T) constant V. The conversation includes discussions about a free expansion and an adiabatic free expansion, as well as calculations to find the value of the Joule coefficient. The final step in the calculation involves finding the value of d/dT(P/T).
  • #1
IngridR
3
0

Homework Statement



Hi, i have a little problem with a demostration, I hope you can help me.

Homework Equations


this said that we have a system, a gas is containing in a recipe, there's no heat exchange neither work with the enviroment, only an expansion v to 2v, we have to find that

(dT/dv)u=-(T^2/Cv)(d/dT)(P/T)v

The Attempt at a Solution



I start with
dU=(dU/dT)vdT+(dU/dv)t dU=0
(dT/dv)u=-(dU/dv)(dT/dU)vdv

(dT/du)v=1/Cv

(dT/du)=-(1/cv)(T(dP/dT)-P)

I don't know how continue!
 
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  • #2
Welcome to PF;

Have I understood you correctly:
An ideal(?) gas in a container (recipe?!), no heat may enter or leave the container(?), and no work is done on or by the gas(?) ... yet there is an expansion? How can this be?

It sounds like you are trying to describe an adiabatic expansion.
 
  • #3
Simon Bridge said:
Welcome to PF;

Have I understood you correctly:
An ideal(?) gas in a container (recipe?!), no heat may enter or leave the container(?), and no work is done on or by the gas(?) ... yet there is an expansion? How can this be?

It sounds like you are trying to describe an adiabatic expansion.


yes! Sorry i have some problems lol! In fact it's a free expansion U=0 Q=0 and w=0
 
  • #5
A free expansion

I try to find the Joule coefficient (dT/dv) constant U.
I must find that it's equal to -(T^2/Cv)d/dT(P/T) constant V, but
I found that it's equal to (-1/Cv)(T(dP/dT)v-P)
 
  • #6
Oh you mean - like in the title ?!

All right: you got
$$\left.\frac{dT}{dv}\right|_U = -\frac{1}{C_v}\left(T\left.\frac{dP}{dT}\right|_v -P \right)$$

You need to get from there to:$$\left.\frac{dT}{dv}\right|_U = -\frac{T^2}{C_v}\left. \frac{d}{dT}\frac{P}{T}\right|_v$$

... it looks like you are almost there since you expression rearranges as:

$$\left.\frac{dT}{dv}\right|_U = -\frac{T^2}{C_v}\left(\frac{1}{T}\left.\frac{dP}{dT}\right|_v -\frac{P}{T^2} \right)$$

So what is $$\frac{d}{dT}\frac{P}{T}$$
 
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What is the Joule Coefficient?

The Joule Coefficient, also known as the Joule-Thomson Coefficient, is a physical property of a substance that measures the change in temperature of the substance when it experiences a change in pressure at constant enthalpy (heat content).

How is the Joule Coefficient calculated?

The Joule Coefficient is calculated by taking the derivative of the change in temperature with respect to the change in pressure at constant enthalpy. This can be represented mathematically as dT/dP|H, where T is temperature, P is pressure, and H is enthalpy.

What is the significance of the Joule Coefficient?

The Joule Coefficient is an important property for understanding the thermodynamic behavior of a substance. It is used to determine the conditions at which a substance will undergo a phase change, such as from liquid to gas, and can also be used in the design of refrigeration and cooling systems.

How does the Joule Coefficient vary between different substances?

The Joule Coefficient can vary greatly between different substances. It is influenced by factors such as the intermolecular forces between particles, the molecular structure of the substance, and the temperature and pressure conditions. Substances with weaker intermolecular forces tend to have a larger Joule Coefficient.

What are some real-world applications of the Joule Coefficient?

The Joule Coefficient is used in various industries, such as in the design of refrigeration systems, natural gas processing, and cryogenic technology. It is also relevant in fields such as chemical engineering, thermodynamics, and material science.

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