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Chapter 17: Problem 30

Several times it has been mentioned that \(q\) is a constant, but the expression for energy (as well as many other thermodynamic functions) contains the derivative of \(q\) (or the derivative of In \(q\) ). The derivatives of constants are zero. Why aren't thermodynamic state functions equal to zero, then?

Short Answer

Expert verified
Thermodynamic functions are influenced by variables other than constants, leading to non-zero changes.

Step by step solution

01

Understanding the role of q

The constant \( q \) refers to a particular configuration or state variable in thermodynamic functions. It is linked to partition functions or other constants under specific conditions.
02

Examining thermodynamic state functions

Thermodynamic state functions depend on many variables and are calculated using differentials, which capture changes in these variables. These functions, like energy, may include \( q \), but also depend on other variables that are not constant, such as temperature \( T \) or volume \( V \).
03

Differentiating expressions involving q

When deriving an expression where \( q \) is involved, such as internal energy \( U \), the derivative of \( q \) is zero since \( q \) is constant. However, the derivative of the entire function is not zero, as it involves changes in other variables that are not constants.
04

Connecting derivative constants to non-zero results

Although the derivative of a constant is zero, the overall thermodynamic derivation involves multiple terms. These terms capture non-constant factors leading to non-zero results for functions like energy, free energy, entropy, etc.
05

Concluding reason non-zero outputs occur

The derivative of a state function, like energy, reflects how it changes with respect to changes in non-constant variables. Constants remain unchanged, but their presence in the original expression ensures the formulation correctly reflects physical laws and constraints, meaning changes still occur.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

partition function
In thermodynamics, partition functions are fundamental to understanding how systems behave at different temperatures and energies. A partition function, often denoted as \( Z \), mathematically represents the sum of all possible states a system can occupy, weighted by their respective probabilities.
  • Each state has an associated energy level \( E_i \).
  • The partition function is written as \( Z = \, \sum_i e^{-E_i/k_BT} \), where \( k_B \) is the Boltzmann constant and \( T \) is the temperature.
The partition function links microscopic states to macroscopic observables like temperature and pressure. It plays a central role in calculating thermodynamic properties such as internal energy, free energy, and entropy.
thermodynamic derivatives
Thermodynamic derivatives are tools used to quantify how a system changes as its variables change. They offer insight into how properties like energy and entropy evolve with other parameters like temperature or pressure. Such derivatives help physicists and chemists understand reactions and processes in terms of changes in state variables.
In practice:
  • First derivatives can indicate stability properties of systems.
  • Second derivatives often relate to phase transitions.
For example, the derivative of internal energy with respect to temperature at constant volume (\( \left( \frac{\partial U}{\partial T} \right)_V \)) defines heat capacity at constant volume. These derivatives are crucial for deriving other thermodynamic equations and understanding material behavior.
thermodynamic variables
Thermodynamic variables are quantities that describe the state of a system in thermodynamic equilibrium. These variables fall into two main categories: intensive and extensive properties.
  • Intensive variables do not depend on the system size, such as temperature, pressure, and density.
  • Extensive variables depend on the system size, like volume, internal energy, and mass.
These variables are essential for computing state functions, as they help establish relationships between different physical properties. By analyzing them, we can derive equations of state that tie together combinations of thermodynamic variables, ultimately reflecting the behavior of systems under different conditions.
internal energy
Internal energy, represented by \( U \), is a crucial concept in thermodynamics and reflects the total energy stored within a system. It includes all kinetic and potential energy components at a microscopic level.
  • Kinetic energy stems from particle motion within the system.
  • Potential energy arises from interactions between particles.
Internal energy can change through heat transfer (Q) into or out of the system or through work (W) done on or by the system, forming the basis of the first law of thermodynamics: \(\Delta U = Q - W\). This law illustrates the conservation of energy and provides the groundwork for understanding energy transformations during processes.
differentials in thermodynamics
Differentials in thermodynamics are mathematical expressions that represent small changes in thermodynamic quantities. They help describe how these quantities respond to varying parameters, such as temperature or pressure. Differentials are key to understanding how thermodynamic processes evolve.
Differentials are often used in the context of state functions where:
  • A small change in a function, \(dX\), expresses the change with respect to its variables.
  • For instance, for a state function like internal energy, we might write:\( dU = \left( \frac{\partial U}{\partial V} \right)_T dV + \left( \frac{\partial U}{\partial T} \right)_V dT\), demonstrating changes with volume and temperature.
These calculations are essential for formulating precise expressions that govern energy exchanges and transformations in thermodynamic systems.

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Most popular questions from this chapter

Consider a system of five energy levels, each of which is doubly degenerate. The levels have energies of \(0,1 \times 10^{-21}\), \(2.5 \times 10^{-21}, 4 \times 10^{-21}\), and \(6 \times 10^{-21}\) J. Calculate the partition function of this system at \(50,100,200,300,500\), and \(1000 \mathrm{~K}\). Do you see a leveling off of the value of \(q\) as the temperature increases? What is the interpretation of the values of \(q\) ?

A common thought experiment is to suppose that all of the gas molecules in a room may instantaneously cluster in one corner of the room, killing everyone in the room. Explain in statistical thermodynamic terms why we don't need to worry about this ever happening.

What temperature is necessary to have twice as many atoms in the ground state as in the first excited state, at \(16.4 \mathrm{~cm}^{-1}\), of Catoms? What temperature is necessary to have equal populations in the ground state and the second excited state, at \(43.5 \mathrm{~cm}^{-1}\) ? What temperature is necessary to have equal populations in the first and second excited states? The degeneracies of the ground, first, and second excited states are 1,3 , and 5 , respectively. (Note that these equilibrium ratios would not be possible at any temperature if the degeneracies were equal.)

\(\mathrm{Ti}^{3+}\) has the following electronic energy levels: \begin{tabular}{ccc} \hline \(\boldsymbol{i}\) & degeneracy & \(\boldsymbol{E}\left(\mathbf{c m}^{-1}\right)\) \\ \hline 0 & 4 & 0 \\ 1 & 6 & \(384.3\) \\ 2 & 2 & \(80378.6\) \\ \hline \end{tabular} Calculate \(q\) by summing the three terms, assuming a temperature of \(2000 \mathrm{~K}\). Are additional terms needed to get a good value for \(q\) ?

Determine the average score on an exam two different ways and show that the same average score is obtained. The scores are \(78,44,74,92,85,50,74,80,80\), and 90 .
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