What is the Derivative of P/T with respect to T?

[IngridR]

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 environment, 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!

 

[Simon Bridge]

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.

 

[IngridR]

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

 

[Simon Bridge]

A free expansion
http://en.wikipedia.org/wiki/Free_expansion

... or an adiabatic free expansion?
http://en.wikipedia.org/wiki/Adiabatic_process#Adiabatic_free_expansion_of_a_gas

... what is it you are trying to calculate?

 

[IngridR]

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)

 

[Simon Bridge]

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 $$\frac{d}{dT}\frac{P}{T}$$