One-dimensional polymer (Gibbs canonical ensemble)

Let’s consider a polymer that is formed by an horizontal linear chain of $N$ disc-shaped monomers. Each monomer can either adopt either a vertical alignment (with length $l_1$ and energy $E_1$) or an horizontal alignment (with length $l_2$ and energy $E_2$) (obviously $l_2>l_1$). The chain is also subject to a tension $T$.

I want to compute the average energy $<E>$ and average length $<L>$ of the chain using the Gibbs canonical ensemble, whose partition function is given by:

$$Z_G=sum_Ssum_i exp[beta(E_i+pV_S)]$$

I began by calculating the denegeracy of energies $g(n$) of the system having $n$ monomers aligned horizontally, which is a basic problem of combinatorics:

$$g(n)={N choose n}$$

We can exchange the sum in the partition function to:

$$sum_Ssum_irightarrowsum_{{ n }}g(n)$$

So the partition function is reduced to,

$$Z_G=sum_{{ n }} {N choose n} exp[beta(E_i+pV_S)]$$

If I can find the form of the partition function, the calculation of $<E>$ and $<L>$ is straightforward as:

$$<E>=frac{sum_{{ n }}E_i {N choose n} exp[beta(E_i+pV_S)]}{sum_{{ n }} {N choose n} exp[beta(E_i+pV_S)]}$$

$$<L>=frac{sum_{{ n }}L_i {N choose n} exp[beta(E_i+pV_S)]}{sum_{{ n }} {N choose n} exp[beta(E_i+pV_S)]}$$

However, what I don’t understand is how to express the term $E_i+pV_S$ in terms of the energies of the monomers, their lengths, and the total tension (a general way to express this would be useful for more general problems).

For example, I thought about substituting $pV_Srightarrow TL_S$, with $L_S=l_1+l_2$ but I’m not even sure if this is valid. I found a similar problems online but they don’t have a solution.