How Do I Take Partial Derivatives of the Dieterici Equation of State?

In summary, the person is having difficulty with the partial derivatives of the equation of state. They need to take the derivative of T with respect to v and then separate it out from the other variables.
  • #1
cashmerelc
18
0
I have to find the expansivity of a substance obeying the Dieterici equation of state using the cyclical relation.

I understand what I need to do, but I'm having a problem with the partial derivatives of the equation of state. I was wondering if anyone could refresh me on how to take a partial derivative of this equation:

P=(RT/v-b)e^(-a/RTv)

Taking the partial derivative of P with respect to T is simply the product rule, correct?

But when I have to take the derivative of T wrt v, or the der. of v wrt P, I have to solve the equation for T and v respectively, right? That is where I'm having the problem, with the variables in the exponent of e.

Any help someone could give me would be greatly appreciated.
 
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  • #2
this is in wrong section, someone will move this thread.

also show the work you have done, so we can see if you have done right and what it is that you don't understand.
 
  • #3
Well, that's what I'm having trouble with.

I have to take dT/dv keeping P constant.

So I need to get all of the Ts on one side of the equation. The problem I'm having deals with the T in the exponent. I know to solve for a variable in an exponent of e, you take the ln of both sides. But in this problem, there is a T not in the exponent also, so you'd end up with a T and a ln |T| on one side. So there's got to be another way to do it unless I'm missing something obvious.

And sorry about posting in the wrong thread.
 
  • #4
cashmerelc said:
Well, that's what I'm having trouble with.

I have to take dT/dv keeping P constant.

So I need to get all of the Ts on one side of the equation. The problem I'm having deals with the T in the exponent. I know to solve for a variable in an exponent of e, you take the ln of both sides. But in this problem, there is a T not in the exponent also, so you'd end up with a T and a ln |T| on one side. So there's got to be another way to do it unless I'm missing something obvious.

And sorry about posting in the wrong thread.

Can you write the equation and how you obtained it?

The rules of this forum is that you show work done before getting help.
 
  • #5
You don't have to solve for T or v. You could just use "implicit differentiation".
 
  • #6
This is the original equation, which is given:

P=(RT/v-b)e^(-a/RTv)

I have to find the partial derivative of T with respect to v. This is what I'm saying my problem is. I am trying to separate T out so I can write the equation as a function of T.

Here is what I have...

T= (v-b)P/Re^(-a/RTv)

So there is still a T on both sides of the equation.. So even if I get them on the same side, I have this:

Te^(-a/RTv)=(v-b)P/R

I can't figure out how to get the T out of the exponent without messing up the other T.
 
  • #7
/facepalm

I always forget implicit differentiation. Thanks a million, HallsofIvy!
 

1. What is a partial derivative?

A partial derivative is a mathematical concept that measures the rate of change of a function with respect to one of its variables, while holding all other variables constant.

2. Why are partial derivatives important?

Partial derivatives are important in many fields of science, including physics, engineering, and economics. They allow us to analyze the behavior of complex systems and make predictions about how they will change over time.

3. How do you calculate a partial derivative?

To calculate a partial derivative, you take the derivative of a function with respect to one of its variables, treating all other variables as constants. This can be done using the standard rules of differentiation, such as the power rule and the chain rule.

4. What is the relationship between partial derivatives and gradients?

A gradient is a vector that points in the direction of maximum increase of a function. The components of this vector are the partial derivatives of the function with respect to each of its variables. In other words, the gradient is a generalization of the concept of a partial derivative.

5. How are partial derivatives used in optimization?

Partial derivatives are used in optimization to find the minimum or maximum value of a function. By setting the partial derivatives of a function equal to zero, we can solve for the critical points, which represent the optimal values of the function for a given set of constraints.

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