0
$\begingroup$

I'm reading a stat thermo text (Terrel Hill) about the freely jointed chain problem. It all goes well until I hit the thermodynamic function of state derived from the canonical partition function. The problem goes like the following:

Consider a linear polymer chain made up of M units, where M is large enough so that one chain can be considered a thermodynamic system. Each unit can exist in the states $i = 1, 2, . . . , n$ with partition functions $j_i(T)$ and lengths $l_i$. The total length of the chain is $l$. The system (chain) is characterized thermodynamically by $l, M, T$.

Because this is a freely jointed chain, the canonical ensemble partition function for the system follows the formalism of the mixture of ideal gas species. Therefore, the partition function $Q$ is then:

$$Q(l, M, T)=\sum_M M!\prod^n_{i=1}\frac{j_i^{M_i}}{M_i!} \tag{1}$$

where $M_i$ is the number of units with length $l_i$, and the sum is over all sets $M=\{M_1, M_2, ..., M_n\}$ consistent with the restrictions:

\begin{align*}\sum_{i=1}^nM_i &=M, \tag{2} \\ \sum_{i=1}^nl_iM_i &=l \tag{3}\end{align*}

If we choose $l$ as an independent variable, the text says that the appropriate thermodynamic equation is:

$$dA=-SdT+\tau dl+\mu dM \tag{4}$$

With $\tau$ being the force pulling on the chain. My question is:

Why "$+\tau dl$" rather than "$-\tau dl$"? Since we have $dA=-SdT -pdV + \mu dM$ for ideal gas.

$\endgroup$
1
$\begingroup$

It's a tensile force. You do work by pulling on the chain, so as to extend it. For the ideal gas, you do work by compressing it.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.