What Is the Equation of State Given Compressibility and Expansivity Relations?

In summary, the given equations are used to find the equation of state by integrating the partial derivatives of lnV with respect to P and T. The solution involves finding an arbitrary function f(V) and using it to express the natural logarithm of V in terms of P and T. This results in the equation V = V0 exp(aT^3/P), which is the desired equation of state. The hint given by the professor involves rewriting the original equations and using the concept of exact differentials to find the solution.
  • #1
NucEngMajor
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Homework Statement


Find the equation of state given that k = aT^(3) / P^2 (compressibility) and B = bT^(2) / P (expansivity) and the ratio, a/b?

Homework Equations


B = 1/v (DV /DT)Pressure constant ; k = -1/v (DV /DP)Temperature constant D= Partial derivative
dV = BVdT -kVdP (1)
ANSWER is V = V0exp(aT^(3)/P)

The Attempt at a Solution


a. Integrate (1) and obtain v = voexp (bT^(3)/3P) + 2aT^(3)/P^3) WRONG
b. Hint by Professor: rewrite as: (Let "D" = partial derivative): D/DP (lnV) = 1/V (DV/DP) then D/DP (lnV) + aT^(3)/P = 0. Write as D/DP(lnv + g(P,T)) = 0 where g is a function only of V. ==> lnV + g(P,T) = f(V) where V is an arbitrary function.

I don't understand "b" but following attempt "a" gave me the proper result for a very similar problem?
 
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  • #2
Your starting equations are:

$$\frac{\partial \ln V}{\partial P}=-a\frac{T^3}{P^2}$$
$$\frac{\partial \ln V}{\partial T}=b\frac{T^2}{P}$$
Together with $$d\ln V=\frac{\partial \ln V}{\partial P}dP+\frac{\partial \ln V}{\partial T}dT$$
The last equation implies that d lnV is an exact differential. What does that imply about the relationship between ##\frac{\partial}{\partial T}\left(\frac{\partial \ln V}{\partial P}\right)## and ##\frac{\partial}{\partial P}\left(\frac{\partial \ln V}{\partial T}\right)##?

Chet
 

Related to What Is the Equation of State Given Compressibility and Expansivity Relations?

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.

Why are partial derivatives important?

Partial derivatives are important in many areas of science and engineering, as they allow us to understand how multiple variables affect a system or process. They are particularly useful in optimization problems and in the study of multivariable functions.

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, while treating all other variables as constants. This can be done using the standard rules of differentiation.

What is the difference between a partial derivative and a regular derivative?

The main difference between a partial derivative and a regular derivative is that a partial derivative is taken with respect to only one variable, while a regular derivative is taken with respect to all variables in a function. Partial derivatives are also often used in multivariable calculus, while regular derivatives are used in single variable calculus.

How are partial derivatives used in real-world applications?

Partial derivatives are used in various fields such as physics, economics, and engineering to analyze and optimize complex systems. For example, in economics, partial derivatives are used to determine how changes in multiple variables affect the production of goods and services in a market. In physics, they are used to understand how different forces and variables affect the motion of objects. In engineering, they are used to optimize designs and processes by understanding the relationship between different variables.

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