Partial Derivative Notation in Chemistry Thermodynamics

[danago]

I am currently doing a class on chemical thermodynamics which involves a fair amount of calculus. So far it is going well, however i have a very quick question about the notation being used for partial derivatives.

If there is some property of a mixture, K=K(T,P), then the differential change in that propety is given by:

$$dK = \left(\frac{\partial K}{\partial T}\right)_P dT + \left(\frac{\partial K}{\partial P}\right)_T dP$$

Where the subscripts T and P imply that they are being held constant. My question is -- Does the partial derivative not already imply everything except for one variable is held constant? Would $$\frac{\partial K}{\partial T}$$, by definition, be the change in K when ONLY T changes, without having to specift that P is held constant?

I guess what i am asking is -- is there is a difference between $$\frac{\partial K}{\partial T}$$ and $$\left(\frac{\partial K}{\partial T}\right)_P$$ that i was not made aware of in my first year calculus courses?

 

[Mark44]

I agree with you. The P and T subscripts seem redundant to me, for exactly the same reason you gave.

 

[danago]

Thanks Mark for clearing that up. I find it a bit strange that the book does it, because it really just makes equations look messier than they should.

 

[Mark44]

I'm open to someone who can give a justification for those subscripts.

 

[D H]

As a justification, suppose $u=y/x$ and $v=ux$. Is $\partial v/\partial x=u$ just because you only see that one occurrence of $x$ directly in the equation for $v$? Of course not. That $u$ in the equation for $v$ is not truly an independent variable -- and neither are most of the variables involved in statistical physics.

 

[danago]

As a justification, suppose $u=y/x$ and $v=ux$. Is $\partial v/\partial x=u$ just because you only see that one occurrence of $x$ directly in the equation for $v$? Of course not. That $u$ in the equation for $v$ is not truly an independent variable -- and neither are most of the variables involved in statistical physics.

So are you saying that $$\partial v/\partial x \ne u$$, but $$\left(\partial v/\partial x\right)_u = u$$? Have i understood you correctly?

 

[D H]

Exactly.

 

[danago]

Alright

Thanks for shedding some light on that