Chapter 2: Problem 34
The thermodynamic equation of state \((\partial U / \partial V) \mathrm{T}=T(\partial p / \partial T)_{v}-p\) was quoted in the chapter. Derive its partner from it and the general relations between partial differentials.
Short Answer
Expert verified
\((\partial T / \partial U)_p = - (\partial V / \partial T)_p / (T(\partial p / \partial T)_V - p)\)
Step by step solution
01
Leverage the Reciprocal Relation
According to the general relations for partial derivatives, if we have a function of three variables, say, f(x, y, z), then the reciprocal relation holds: \((\partial x / \partial y)_z (\partial y / \partial z)_x (\partial z / \partial x)_y = -1\). This relation helps to relate different partial derivatives of state functions.
02
Identify the Function and Variables
We need to see the internal energy U as a function of entropy S, volume V, and the number of particles N; U = U(S, V, N). However, since number of particles N is constant for most thermodynamics problems and is not included in our equation, we will only focus on S, V. We're given \((\partial U / \partial V)_T\), so T takes the implicit role of the constant variable, akin to z in the reciprocal relation we mentioned.
03
Apply the Reciprocal Relation to Thermodynamic Variables
Now we can write the reciprocal relation for the thermodynamic variables: \((\partial U / \partial V)_T (\partial V / \partial T)_p (\partial T / \partial U)_p = -1\). Since we're interested in the relationship that includes the derivative \((\partial T / \partial U)_p\), we need to express this in terms of known derivatives.
04
Rearrange to Obtain the Sought Partial Derivative
We rearrange the reciprocal relation to solve for \((\partial T / \partial U)_p\) which gives: \((\partial T / \partial U)_p = - (\partial V / \partial T)_p / (\partial U / \partial V)_T\). We can plug in the given thermodynamic equation of state \((\partial U / \partial V)_T = T(\partial p / \partial T)_V - p\) into the expression.
05
Substitute Given Information
Substituting, we get \((\partial T / \partial U)_p = - (\partial V / \partial T)_p / (T(\partial p / \partial T)_V - p)\). This is essentially the desired partner relation we need to derive using the general relations between partial differentials.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Thermodynamic Equation of State
In thermodynamics, the equation of state is a fundamental principle that describes the relationship between different state variables, typically pressure (p), volume (V), temperature (T), and internal energy (U). These variables are interdependent, and the equation of state provides a mathematical link between them, relating one variable to the others under certain conditions, such as constant temperature.
One common form of the thermodynamic equation of state connects the partial derivative of internal energy with respect to volume at constant temperature \(\left(\frac{\partial U}{\partial V}\right)_T\) to other measurable properties like pressure and temperature. According to the exercise, we're given the equation \(\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p\), which indicates that the change in internal energy with volume, when temperature is fixed, is tied to how pressure changes with temperature and the actual pressure value itself.
Understanding this equation provides insight into the physical behavior of a system. For example, if we know how pressure changes with temperature at a constant volume, we can infer how the internal energy of the system will vary as we adjust its volume at a fixed temperature. Such equations are vital in predicting and explaining the outcomes of various processes in physical and chemical systems.
One common form of the thermodynamic equation of state connects the partial derivative of internal energy with respect to volume at constant temperature \(\left(\frac{\partial U}{\partial V}\right)_T\) to other measurable properties like pressure and temperature. According to the exercise, we're given the equation \(\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p\), which indicates that the change in internal energy with volume, when temperature is fixed, is tied to how pressure changes with temperature and the actual pressure value itself.
Understanding this equation provides insight into the physical behavior of a system. For example, if we know how pressure changes with temperature at a constant volume, we can infer how the internal energy of the system will vary as we adjust its volume at a fixed temperature. Such equations are vital in predicting and explaining the outcomes of various processes in physical and chemical systems.
Reciprocal Relation in Thermodynamics
The reciprocal relations play a crucial role in thermodynamics, especially when dealing with partial derivatives of state functions like internal energy, entropy, and pressure. These relational expressions are a form of mathematical symmetry that comes from the exact differential of a state function.
In the context of the exercise provided, the reciprocal relation states that for any function of three variables, the product of the cyclic permutations of their partial derivatives taken with one variable held constant in each case is equal to -1: \(\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y = -1\). When applied to thermodynamic variables, this relation helps derive new expressions or 'partner relations' that may not be immediately obvious from the original equations.
For example, knowing \(\left(\frac{\partial U}{\partial V}\right)_T\), we can derive a related expression involving \(\left(\frac{\partial T}{\partial U}\right)_p\) by employing the reciprocal relation on the appropriate variables. This ability to generate connected derivatives is invaluable for constructing the thermodynamic web of relations that allows for a deeper understanding of the system under study.
In the context of the exercise provided, the reciprocal relation states that for any function of three variables, the product of the cyclic permutations of their partial derivatives taken with one variable held constant in each case is equal to -1: \(\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y = -1\). When applied to thermodynamic variables, this relation helps derive new expressions or 'partner relations' that may not be immediately obvious from the original equations.
For example, knowing \(\left(\frac{\partial U}{\partial V}\right)_T\), we can derive a related expression involving \(\left(\frac{\partial T}{\partial U}\right)_p\) by employing the reciprocal relation on the appropriate variables. This ability to generate connected derivatives is invaluable for constructing the thermodynamic web of relations that allows for a deeper understanding of the system under study.
Partial Derivative Relations
Partial derivative relations in thermodynamics allow us to express the sensitivity of one state variable with respect to another while keeping a third variable constant. These relationships often come up when we need to describe how a system's properties might change under specific conditions.
In solving the exercise, we encountered the relationship \(\left(\frac{\partial T}{\partial U}\right)_p\), where we are interested in how temperature changes with internal energy at a constant pressure. This is not a directly measurable quantity, but with knowledge of other partial derivatives, thermodynamics allows us to deduce it using the reciprocal relation. We isolate the desired partial derivative from the reciprocal relationship and then substitute in the known variables as done in step 4 and 5 of the exercise solution.
The use of such partial derivative relations extends beyond textbook examples and holds critical practical value in predicting the response of a system when subjected to different processes. A firm grasp of these concepts is essential for any student looking to apply thermodynamic principles effectively in real-world situations, including engineering, weather forecasting, and material science.
In solving the exercise, we encountered the relationship \(\left(\frac{\partial T}{\partial U}\right)_p\), where we are interested in how temperature changes with internal energy at a constant pressure. This is not a directly measurable quantity, but with knowledge of other partial derivatives, thermodynamics allows us to deduce it using the reciprocal relation. We isolate the desired partial derivative from the reciprocal relationship and then substitute in the known variables as done in step 4 and 5 of the exercise solution.
The use of such partial derivative relations extends beyond textbook examples and holds critical practical value in predicting the response of a system when subjected to different processes. A firm grasp of these concepts is essential for any student looking to apply thermodynamic principles effectively in real-world situations, including engineering, weather forecasting, and material science.