Partial Derivative of Van der Waals Equation

In summary, we are given the Van Der Waals equation with constants n,a,R, and b. We are asked to find the derivative of p with respect to v and the derivative of p with respect to T without further differentiation. The question asks for guidance in solving this type of problem and the difference between implicit and explicit functions. A sample problem is provided for reference.
  • #1
Scharles
16
0
Given that the Van Der Waals equation is (p + (an^2)/v^2)(v-nb)=nRT where n,a,R and b are constants...

How to we find the derivative of p wrt v ?

How to find the derivative of p wrt T without further differentiation ??

Can anyone teach me how to do this question ?

Sincerly thanks~
 
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  • #2
Scharles said:
Given that the Van Der Waals equation is (p + (an^2)/v^2)(v-nb)=nRT where n,a,R and b are constants...

How to we find the derivative of p wrt v ?

How to find the derivative of p wrt T without further differentiation ??

Can anyone teach me how to do this question ?

Sincerely thanks~
What have you tried?

Where are you stuck ?
 
  • #3
i have no idea on how to solving this...
please kindly teach me how to start on solving this sort of question...
 
  • #4
Do you know the difference between an implicit function and an explicit one ? What do you know about the derivatives for explicit functions ? How about implicit ones ?
 
  • #5
Scharles said:
i have no idea on how to solving this...
please kindly teach me how to start on solving this sort of question...

If I had an equation:

[tex]y(x)+x=k[/tex]

and I wanted to take the derivative of y with respect to x, I'd get:

[tex]y'(x)+1=0[/tex]

Ok, not too bad.

Suppose I had:

[tex]y(x)+\frac{1}{x^2}=k[/tex]

still not too bad if I want the derivative of y with respect to x. That's:

[tex]y'-2x^{-3}=0[/tex]

How about:

[tex](y(x)+\frac{c}{x^2})(x-k)=a[/tex]

That's still not too bad cus' I'd use the chain rule this time:

[tex](y(x)+\frac{c}{x^2}) \frac{d}{dx} (x-k)+(x-k)\frac{d}{dx}(y(x)+\frac{c}{x^2})=0[/tex]

and that's:

[tex](y(x)+\frac{c}{x^2})(1)+(x-k)(y'(x)-2cx^{-3})=0[/tex]

ok, now you do one but instead of y(x), I'll say:

[tex](p(v)+\frac{k}{v^2})(v-c)=a[/tex]

and I want to take the derivative of p with respect to v. Do that one, then do yours with all the other parameters.
 
Last edited:

Related to Partial Derivative of Van der Waals Equation

1. What is the Van der Waals equation?

The Van der Waals equation is an equation of state that describes the behavior of real gases, taking into account the intermolecular forces between the gas molecules. It is an improvement over the ideal gas law, which assumes that gas molecules have no volume and do not interact with each other.

2. What is a partial derivative?

A partial derivative is a mathematical concept used in multivariable calculus to describe how a function changes with respect to one of its variables while holding the other variables constant. In the context of the Van der Waals equation, we take the partial derivative to understand how the equation changes when one of its parameters, such as temperature or pressure, is varied.

3. Why do we need to take the partial derivative of the Van der Waals equation?

By taking the partial derivative of the Van der Waals equation, we can better understand how the equation behaves under different conditions. This allows us to make predictions about the behavior of real gases and how they deviate from ideal gas behavior.

4. How is the partial derivative of the Van der Waals equation calculated?

The partial derivative of the Van der Waals equation is calculated using standard calculus techniques. We take the derivative of the equation with respect to the variable we are interested in, while treating all other variables as constants. This gives us an equation that describes the rate of change of the Van der Waals equation with respect to that variable.

5. What is the significance of the partial derivative of the Van der Waals equation?

The partial derivative of the Van der Waals equation allows us to understand the behavior of real gases and how they differ from ideal gases. It also helps us to make predictions about the behavior of gases under different conditions, such as varying temperature or pressure. This is important in many fields, including chemistry, physics, and engineering.

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