Thermodynamic differental relations problem

[knowlewj01]

Homework Statement

Assuming V is a function of P and T such that

[itex]V = V(P,T)[/itex]

express the differential changes in volume due to differential changes in Temperature and pressure, what is the fractional/relative change?

Homework Equations

The Attempt at a Solution

since V is a function of P and T:

[itex] dV = \frac{\partial V}{\partial T} dP + \frac{\partial V}{\partial P} dT[/itex]

so we can say:

[itex] \left[\frac{dV}{dP}\right]_T = \frac{\partial V}{\partial T}[/itex]

and

[itex] \left[\frac{dV}{dT}\right]_P = \frac{\partial V}{\partial P}[/itex]

is this correct or have i read the question wrong, I am not really sure what I'm doing.

 

[Mapes]

I'm not sure what you're doing either. Where does your first equation in the solution come from? Typically one writes

[tex]dV=\left(\frac{\partial V}{\partial P}\right)_T\,dP+\left(\frac{\partial V}{\partial T}\right)_P\,dT[/tex]

 

[knowlewj01]

Ah, i got it the wrong way around then. Makes slightly more sense now.

so we are looking for fractional change, ie. [itex] \frac{dV}{V}[/itex]

so if we say that:

[itex]dV=\left(\frac{\partial V}{\patial P}\right)_T dP + \left(\frac{\partial V}{\partial T}\right)_P dT[/itex]

then divide by V to get fractional change:

[itex]\frac{dV}{V}=\frac{1}{V}\left(\frac{\partial V}{\patial P}\right)_T dP + \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P dT[/itex]

now i notice that:

[itex] \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P[/itex] is the coeficcient of thermal expansion [itex]\alpha[/itex]

and
[itex] \frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T[/itex] is negative coeficcient of isothermal compression [itex]-\kappa_T[/itex]

so this can be re-written as:

[itex]\frac{dV}{V} = \alpha dP - \kappa_T dT[/itex]

i think this is right.

 

[Mapes]

Nice!

 

[knowlewj01]

Thanks for pointing that out, would have been scratching my head all night otherwise ;)