# A thermodynamics polymer chain problem

• chemicaholic
In summary: I think the "partition function" is a red herring.In summary, the conversation discusses a one-dimensional polymer molecule and its energy state, as well as the effects of an electric field and temperature on the molecule. The energy of the molecule is determined by the distance between its ends and an electric field, and in the absence of charges involves maximizing entropy. The partition function is not relevant to the problem.
chemicaholic
A one-dimensional polymer molecule (rubber) is chain of N links of the same length a, the links can go either forward or backward but always stay parallel to the x axis. If one denotes the coordinates of the joints are ${x_0, x_1, . . . , x_N}$ , then $|x_n − x_{n+1}| = a$. The energy of the polymer does not depend on the angles at the joints.

Two electric charges $-q$ and $q$ are attached to the two ends of the molecule. An electric field $\mathcal{E}$ is turned on. The total energy of the molecule is the electrostatic energy,
$$E=−q\mathcal{E} (x_N −x_0)$$
Find the dependence of the temperature on the distance $L$ between the two ends of the molecule. Imagine the system interacts with a thermal bath, show that the rubber molecule contracts when the temperature increases.

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So I tried this problem with first setting up energy states, with $E_n = -qE(x_N - x_0) = -qE(-N+2n)a$, and getting the partition function $$Z = \frac{e^{-\beta q \mathcal{E} N a} - e^{\beta q \mathcal{E} N a}}{1-e^{2\beta q \mathcal{E} a}}$$
but I don't know how to proceed from here. In principle I should take $\langle L \rangle = \frac{1}{Z}\sum_{n=0}^N e^{-\beta E_n} |x_N - x_0|$, but the expression is extremely complicated and there is no way to answer the second question (that the polymer contracts when temperature increases).
Could anyone help me?

Last edited:
Finding the mean length of a polymer chain in the absence of charges involves maximizing the entropy. I imagine that your problem involves some combination of energy and entropy (Gibbs?).

## 1. What is a thermodynamics polymer chain problem?

A thermodynamics polymer chain problem refers to a problem in which the thermodynamic properties of a polymer chain, such as its entropy and free energy, are being studied and analyzed. This can involve examining the behavior of the chain under different conditions, such as temperature and pressure, and how it affects the overall system.

## 2. How is thermodynamics used to study polymer chains?

Thermodynamics is used to study polymer chains by providing a framework for understanding the behavior of the chains in different environments. It allows for the calculation of important properties, such as the entropy and free energy, which can then be used to predict the behavior of the polymer chain under various conditions.

## 3. What are some real-world applications of studying thermodynamics polymer chain problems?

Studying thermodynamics polymer chain problems has many practical applications, such as in the development of new materials and products. For example, understanding the thermodynamic properties of polymer chains can help in the design of more efficient and durable plastics for use in various industries.

## 4. How do you solve a thermodynamics polymer chain problem?

Solving a thermodynamics polymer chain problem involves applying the principles of thermodynamics, such as the laws of thermodynamics and the concept of equilibrium, to the specific problem at hand. This often involves using mathematical equations and calculations to determine the desired properties of the polymer chain.

## 5. What are some challenges in studying thermodynamics polymer chain problems?

One of the main challenges in studying thermodynamics polymer chain problems is the complexity of the systems involved. Polymer chains can have varying lengths, branching patterns, and chemical compositions, making it difficult to generalize their behavior. Additionally, the high temperatures and pressures often required for these studies can be challenging to control and measure accurately.

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