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.


1 Answer 1


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.


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