Question 24.1: Adapted from Molecular Driving Forces - Statistical Thermody...

(a1) Because the number of moles, the temperature, and the system volume are constant, we can use the Canonical ensemble to solve this problem. Working on a per mole basis, underbars will not be carried in the solution of this problem. It is most convenient to compute the average energy, \langle E\rangle, or internal energy, U, as follows:

U=\langle E\rangle=\frac{\sum_{i=0}^{3} E_{i} e^{-E_{i} / R T}}{Q}                            (1)

where the Canonical partition function, Q, is given by:

Q=\sum_{i=0}^{3} e^{-E_{i} / R T}                           (2)

In Eqs. (1) and (2), we consider four distinguishable energy states, E_{0}, E_{1}, E_{2}, and E_{3} (see the energy ladder in the Problem Statement). At T = 300 K, it follows that:

\left.R T\right|_{T=300  K }=\left(1.98717 \times 10^{-3}  kCal /( molK )\right)(300  K )=0.5962  kCal / mol                         (3)

Using Eqs. (2) and (3), along with E_{0}=3  kCal / mol , E_{1}=E_{2}=8  kCal / mol, and E_{3}=11  kCal / mol, we obtain:

Q=e^{-(3) /(0.5962)}+2 e^{-(8) /(0.5962)}+e^{-(11) /(0.5962)}=0.006527                        (4)

In addition, the numerator in Eq. (1) is given by:

\sum_{i=0}^{3} E_{i} e^{E_{i} / R T}=\left[(3) e^{-(3) /(0.5962)}+2(8) e^{-(8) /(0.5962)}+(11) e^{-(11) /(0.5962)}\right] kCal / mol

\sum_{i=0}^{3} E_{i} e^{E_{i} / R T}=0.019596  kCal / mol                          (5)

Using Eqs. (5) and (4) in Eq. (1) yields the desired result:

U=\frac{0.019532  kCal / mol }{0.006506}=3.00229  kCal / mol                             (6)

(a2) The probability that the system will be in the ground state (0) having an energy E_{0}=3  kCal / mol is given by:

p_{0}=\frac{e^{-E_{0} / R T}}{Q}                             (7)

Using E_{0}=3  kCal / mol, along with Eqs. (3) and (4), in Eq. (7) yields:

p_{0}=\frac{e^{-3 / 0.5962}}{0.006506}=0.99954, \text { or } 99.954 \%, \text { a very high probability! }                         (8)

(a3) If each energy is increased by 2 kCal/mol, then, the possible energy states will be E_{0}=5  kCal / mol , E_{1}=E_{2}=10  kCal / mol \text {, and } E_{3}=13  kCal / mol. Therefore, the lowest possible energy state that the system can attain is now 5 kCal/mol. As a result, the system can never have E = 3 kCal/mol, and hence P (E = 3 kCal/mol) = 0!

(a4) As T \rightarrow \infty, the term e^{-E_{i} / R T} tends to 1. It then follows that:

p_{i}=\frac{e^{-E_{i} / R T}}{Q}=\frac{1}{Q}=\frac{1}{4}                              (9)

Equation (9) indicates that the four available energy states are likely to be equally populated. As a result, the ensemble-averaged energy, U, is equal to the average of E_{0}, E_{1}, E_{2}, \text { and } E_{3}. That is:

U=\langle E\rangle=\frac{[3+(2)(8)+11]  kCal / mol }{4}=\left(\frac{30}{4}\right)  kCal / mol =7.5  kCal / mol                         (10)

(a5) As T \rightarrow 0, the term e^{-E_{i} / R T} tends to zero. However, it tends to zero more rapidly for the higher-energy states than for the lower-energy states. As a result, the only energy state that is likely to be populated is the ground state (0) having energy E_{0}=3  kCal / mol. In other words, as T \rightarrow 0, the ensemble-averaged energy is equal to the energy of the ground state. In mathematical terms, one obtains:

\left.U\right|_{T \rightarrow 0}=\left.\langle E\rangle\right|_{T \rightarrow 0}=\lim _{T \rightarrow 0} \frac{\sum_{i=0}^{3} E_{i} e^{-E_{i} / R T}}{\sum_{i=0}^{3} e^{-E_{i} / R T}}=\lim _{T \rightarrow 0} \frac{E_{0} e^{-E_{0} / R T}}{e^{-E_{0} / R T}}=E_{0}=3  kCal / mol                          (11)

(b1) As can be seen from the energy ladder in the Problem Statement, there are five conformations of the four-bead chain: the conformation with the bead-bead contact is taken to have energy \varepsilon=0 (the ground state). The other four conformations have no bead-bead contacts, and all have energy \varepsilon=\varepsilon_{0}, where \varepsilon_{0} is a constant. The average end-to-end distance of the four-bead chain is given by:

\langle d\rangle=\sum_{i=1}^{5} d_{i} p_{i}                           (12)

where according to the Problem Statement (in lattice units):

\begin{aligned}&d_{1}=1 \\&d_{2}=d_{3}=d_{4}=\sqrt{5} \\&d_{5}=3 \\&\varepsilon_{1}=0 \\&\varepsilon_{2}=\varepsilon_{3}=\varepsilon_{4}=\varepsilon_{5}=\varepsilon_{0}\end{aligned}                          (13)

In addition, the various probabilities are given by:

\begin{aligned}p_{1} &=\frac{e^{-\varepsilon_{1} / k_{B} T}}{q}=\frac{1}{q}, \quad p_{2}=p_{3}=p_{4}=p_{5}=\frac{e^{-\varepsilon_{0} / k_{B} T}}{q}, \text { and } q=\sum_{i=1}^{5} e^{-\varepsilon_{i} / k_{B} T} \\&=1+4 e^{-\varepsilon_{0} / k_{B} T}\end{aligned}                            (14)

Using Eqs. (13) and (14) in Eq. (12) yields:

\langle d\rangle=1\left(\frac{1}{q}\right)+\frac{3(\sqrt{5}) e^{-\varepsilon_{0} / k_{B} T}}{q}+\frac{3 e^{-\varepsilon_{0} / k_{B} T}}{q}                             (15)

Using q in Eq. (14) in Eq. (15) yields:

\langle d\rangle=\frac{1+9.71 e^{-\varepsilon_{0} / k_{B} T}}{1+4 e^{-\varepsilon_{0} / k_{B} T}}                         (16)

(b2) The maximum value of the end-to-end distance of the chain is obtained when T \rightarrow \infty \text { and } e^{-\varepsilon_{0} / k_{B} T} \rightarrow 1. In that case (in lattice units):

\lim _{T \rightarrow \infty}\langle d\rangle=\langle d\rangle_{\max }=\frac{1+9.71}{1+4}=\frac{10.71}{5} \cong 2.14                         (17)

(b3) We are asked to calculate at what temperature the end-to-end distance of the chain is equal to half of \langle d\rangle_{\max } given in Eq. (17). We therefore require that \langle d\rangle in Eq. (16) be equal to \frac{\langle d\rangle_{\max }}{2}. Specifically:

\frac{1+9.71 e^{-\varepsilon_{0} / k_{B} T}}{1+4 e^{-\varepsilon_{0} / k_{B} T}}=\frac{\langle d\rangle_{\max }}{2}=1.07                          (18)

1+9.71 e^{-\varepsilon_{0} / k_{B} T}=1.07\left(1+4 e^{-\varepsilon_{0} / k_{B} T}\right)

\begin{gathered}(9.71-1.07(4)) e^{-\varepsilon_{0} / k_{B} T}=5.43 e^{-\varepsilon_{0} / k_{B} T}=(1.07-1)=0.07 \\e^{-\varepsilon_{0} / k_{B} T}=\frac{0.07}{5.43}=0.01305 \\-\frac{\varepsilon_{0}}{k_{B} T}=\ln (0.01305)=-4.339\end{gathered}

T=0.2305\left(\frac{\varepsilon_{0}}{k_{B}}\right)=0.2305 T_{0}                         (19)