Deriving Results in Multivariable Calculus for Thermodynamics

In summary: K/∂V)_T]*(∂V/∂P)_T = [(∂K/∂V)_T]*(∂T/∂P)_VAgain, notice that we have replaced (∂V/∂P)_T with (∂T/∂P)_V using the fact that T is a function of V and P. Now, using the definition of (∂K/∂P)_V, we can write:(∂K/∂P)_V = (∂K/∂T)_V*(∂T/∂P)_VSubstituting this into
  • #1
Zatman
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0

Homework Statement


The thermodynamic state of a system may be defined by setting any 2 of the variables P, V and T to be constant because they are related by f(P,V,T)=0.

Given a thermodynamic function K(V,T), prove that:

(∂K/∂T)_P = (∂K/∂T)_V + [(∂K/∂V)_T]*[(∂V/∂T)_P]

and that

(∂K/∂P)_T = [(∂K/∂V)_T]*[∂V/∂P)_T]

2. The attempt at a solution
To be honest I have written down various differentials and tried to combine them in various ways to get these results but to no avail.

Obviously an expression for dK would be useful:

dK = [(∂K/∂V)_T]dV + [(∂K/∂T)_V]dT

Since f(P,V,T) = 0, I can say that p is a function of V and T, and then:

dP = [(∂P/∂V)_T]dV + [(∂P/∂T)_V]dT

Similarly:

dV = [(∂V/∂P)_T]dP + [(∂V/∂T)_P]dT
dT = [(∂T/∂V)_P]dV + [(∂T/∂P)_V]dP

But I cannot combine these to make the required results, and can't think of anything else I can write down.

I did write that since P is a function of V and T and K is a function of V and T then K is a function of P -- but I've just thought this isn't actually (necessarily) correct?

Any help/hints on this would be appreciated.
 
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  • #2





Thank you for your question. The first step in proving these results is to use the chain rule for partial derivatives. Recall that for a function f(x,y), the chain rule states that:

(∂f/∂x)_y = (∂f/∂u)(∂u/∂x) + (∂f/∂v)(∂v/∂x)

Where u and v are intermediate variables that depend on x and y. In our case, u could be P and v could be T, or vice versa.

Now, let's apply this to our problem. We start with the first result:

(∂K/∂T)_P = (∂K/∂T)_V + [(∂K/∂V)_T]*[(∂V/∂T)_P]

Using the chain rule, we can rewrite the second term on the right-hand side as:

[(∂K/∂V)_T]*[(∂V/∂T)_P] = [(∂K/∂V)_T]*(∂P/∂T)_V

Notice that we have replaced (∂V/∂T)_P with (∂P/∂T)_V using the fact that P is a function of V and T. Now, using the definition of (∂K/∂T)_V, we can write:

(∂K/∂T)_V = (∂K/∂P)_V*(∂P/∂T)_V

Substituting this into our original expression, we get:

(∂K/∂T)_P = (∂K/∂P)_V*(∂P/∂T)_V + [(∂K/∂V)_T]*(∂P/∂T)_V

Factorizing (∂P/∂T)_V, we get the desired result:

(∂K/∂T)_P = [(∂K/∂P)_V + (∂K/∂V)_T]*(∂P/∂T)_V

Similarly, we can prove the second result:

(∂K/∂P)_T = [(∂K/∂V)_T]*(∂V/∂P)_T

Using the chain rule,
 

1. What is multivariable calculus and how is it related to thermodynamics?

Multivariable calculus is a branch of mathematics that deals with functions of more than one variable. It is used in thermodynamics to analyze and understand changes in temperature, pressure, and volume of a system as it undergoes physical or chemical processes.

2. How is the process of deriving results in multivariable calculus applied in thermodynamics?

In thermodynamics, multivariable calculus is used to derive equations and relationships between different thermodynamic variables, such as temperature, pressure, and volume. These derivations help to understand and predict the behavior of a system under different conditions.

3. What are some common techniques used in deriving results in multivariable calculus for thermodynamics?

Some common techniques used in deriving results in multivariable calculus for thermodynamics include partial derivatives, total differentials, and vector calculus. These techniques allow for the analysis of multiple variables and their relationships in a system.

4. How do the results derived from multivariable calculus help in understanding thermodynamic processes?

The results derived from multivariable calculus provide a quantitative understanding of how a system changes and evolves under different thermodynamic processes. They can also help in predicting the behavior of a system in response to changes in variables such as temperature, pressure, and volume.

5. Can multivariable calculus be used to solve real-world thermodynamics problems?

Yes, multivariable calculus is an essential tool in solving real-world thermodynamics problems. It allows for the analysis and prediction of complex systems and their behavior, making it a valuable tool for engineers and scientists working in the field of thermodynamics.

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