What is the implicit differentiation of the van der Waals equation?

In summary: When calculating the derivative of a function with respect to one of these variables, you would usually only need the derivative of the function with respect to the other two. However, in this case it looks like you might need the derivative of the function with respect to all three.Can you clarify what you're trying to do?
  • #1
ABearon
5
1
Member warned that homework problems should be posted in a homework forum section
Summary:: van der waals

I have a pretty good understanding of implicit differentiation. However I'm stuck on a homework problem and could use some help.

[P + (an^2)/V^2][V - nb] = nRT a,n,b,R are constants

My professor wants me to take the implicit differentiation of P wrt T. She then says solve for P then take the derivative to check your answer. I understand the chain rule and product rule. Could someone show the steps to solving using these rules. I'm used to having 3 intermediate variables as functions of t and seeing it this way is confusing me.
 
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  • #2
Do [itex]\frac{d}{dT} [/itex] on both sides.
 
  • #3
Svein said:
Do [itex]\frac{d}{dT} [/itex] on both sides.

I got {nR - [P+(an^2)/V^2][dV/dT]}/(V-nb) - d/dt[(an^2)/V^2]
 
  • #4
ABearon said:
I got {nR - [P+(an^2)/V^2][dV/dT]}/(V-nb) - d/dt[(an^2)/V^2]
Using Latex might help:

https://www.physicsforums.com/help/latexhelp/

Although, to be honest, I have no idea what you are trying to do. How does differentiating an equation check the answer? Check it in what way?
 
  • #5
ABearon said:
She then says solve for P then take the derivative to check your answer.
PeroK said:
Although, to be honest, I have no idea what you are trying to do. How does differentiating an equation check the answer? Check it in what way?
I believe what is meant here is to do these operations:
  1. Find the derivative dP/dT using implicit differentiation.
  2. Solve for P, and then differentiate normally to find dP/dT.
Both techniques should result in the same derivative.
 
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  • #6
@ABearon, you have posted several threads in sections other than the homework sections. In the future, please post homework questions in one of the forum sections under Homework & Coursework.
 
  • #7
Mark44 said:
I believe what is meant here is to do these operations:
  1. Find the derivative dP/dT using implicit differentiation.
  2. Solve for P, and then differentiate normally to find dP/dT.
Both techniques should result in the same derivative.

I guess the objective is to find ##\frac{dP}{dT}##. That seems like an obvious thing to say in the OP.
 
  • #8
PeroK said:
I guess the objective is to find ##\frac{dP}{dT}##. That seems like an obvious thing to say in the OP.
From post #1:
ABearon said:
My professor wants me to take the implicit differentiation of P wrt T.
A better way to say this probably would be "Find dP/dT using implicit differentiation."
 
  • #9
ABearon said:
My professor wants me to take the implicit differentiation of P wrt T. She then says solve for P then take the derivative to check your answer. I understand the chain rule and product rule. Could someone show the steps to solving using these rules. I'm used to having 3 intermediate variables as functions of t and seeing it this way is confusing me.

Wait. Are you differentiating with respect to temperature (T) or time (t)? Don't keep switching back and forth between them, choose one symbol. it's very confusing.

ABearon said:
I got {nR - [P+(an^2)/V^2][dV/dT]}/(V-nb) - d/dt[(an^2)/V^2]

Adjusting the formatting to what I think that says based on the brackets, I think that line says this:
$$\frac{nR - \frac {P + an^2} {V^2} \frac {dV}{dT}}{V - nb} - \frac {d}{d?} \frac {an^2}{V^2}$$

Since V is apparently treated as a function of t or T, whatever you're differentiating with respect to, then you can apply the Chain Rule to that last term.

I'm not saying whether this is a correct differentiation or not, just what I think you wrote. Actually I can see one thing clearly wrong with it at a glance: Implicit differentiation of an equation should give you an equation.
 
  • #10
I'm a bit puzzled by the question. In thermodynamics it's important to say which quantities are kept constant when calculating derivatives. For a gas you have three independent variables (e.g., ##n##, ##V##, and ##T##).
 

What is Implicit Differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not written explicitly in terms of a single variable. It is commonly used when a function is written in the form of an equation rather than a function of a single variable.

Why is Implicit Differentiation useful?

Implicit differentiation allows us to find the slope or rate of change of a function at any point, even when the function is not written explicitly in terms of a single variable. This is useful in many real-world applications, such as physics, engineering, and economics.

What is the process for Implicit Differentiation?

The process for implicit differentiation involves treating the dependent variable as a function of the independent variable and using the chain rule to find the derivative. This involves differentiating both sides of the equation with respect to the independent variable, isolating the derivative, and simplifying the result.

When is Implicit Differentiation used?

Implicit differentiation is commonly used in situations where the dependent variable cannot be easily isolated and expressed explicitly in terms of the independent variable. This includes equations involving multiple variables, trigonometric functions, and implicit curves.

What are some common applications of Implicit Differentiation?

Implicit differentiation is used in a variety of fields, including physics, engineering, economics, and biology. It can be used to find the rate of change of physical quantities, optimize functions, and solve related rates problems. It is also useful in curve sketching and optimization problems in economics and biology.

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