CHEM2524Chemical Physics

    08-01 · Week 9

    Soft matter and the kT yardstick

    Soft matter is anything that the random jostling of heat is big enough to change, and kT is the yardstick that tells you what thermal energy can and cannot do.

    • Say what makes a material soft matter, in terms of thermal fluctuations.
    • Use the Boltzmann distribution and the population ratio to judge how many molecules reach a higher state.
    • Quote the size of kT at 0, 25 and 100 degrees C from the lecture table.
    • Explain why polymer chains bend but do not break, and why colloids stay suspended.

    15 min readJump to the quiz ↓

    Soft matter is anything that heat alone can push around

    Intuition

    Imagine a crowd shuffling in a station. Nobody is shoving hard, but the crowd keeps drifting and reshaping itself. Now imagine a crowd of people locked into a rigid grid. A little shuffling changes nothing. Soft matter is the shuffling crowd. The random jostling of heat is big enough to change its shape.

    The lecture’s definition is short. When a system can be altered by stresses of the size that might be found in thermal fluctuations, it is called soft matter. So the question becomes: how big are thermal fluctuations, and what can they do?

    Gels, polymers, colloids and the liquid crystals in later lectures are all examples. Their parts are held together loosely enough that ordinary heat can rearrange them.

    The Boltzmann distribution turns an energy gap into a head count

    Intuition

    Think of a hill and a crowd of walkers who each get a random push from the weather. Lots of walkers are standing in the valley. Fewer reach a low ledge. Almost nobody reaches a very high ledge. The higher the ledge, the fewer walkers. How high is “too high” depends on how strong the typical push is. That push is kT.

    The Boltzmann distribution says the probability pip_i that part of an equilibrium system is in state ii depends on that state’s energy and the temperature.

    pi∝e−εi/kTp_i \propto e^{-\varepsilon_i/kT}

    Here εi\varepsilon_i is the energy of the state (in J), kk is Boltzmann’s constant (1.38×10−231.38\times10^{-23} J K⁻¹) and TT is the temperature (in K).

    If a molecule has two states, with energies εi\varepsilon_i and εj\varepsilon_j, the ratio of the numbers of molecules in each state is

    NiNj=pipj=e−(εi−εj)/kT\frac{N_i}{N_j} = \frac{p_i}{p_j} = e^{-(\varepsilon_i-\varepsilon_j)/kT}

    If εi−εj≫kT\varepsilon_i-\varepsilon_j \gg kT, very few molecules are ever thermally excited into the higher state. So very little change or alteration is possible. That is the lecture’s test for whether heat can do something.

    The lecture’s table shows how fast the ratio collapses as the gap grows, measured in units of kTkT.

    (εi−εj)/kT(\varepsilon_i-\varepsilon_j)/kTNi/NjN_i/N_j
    0.01.0
    0.50.607
    1.00.368
    3.00.05
    10.00.000045
    50.00.00000000000000000000019

    Try it: worked example

    Two states are 3.0 kT apart in energy. Roughly what fraction of the molecules in the lower state are found in the higher state, according to the lecture's table?

    Show answer and steps

    AnswerNi/Nj = 0.05, so about 5 molecules in the higher state for every 100 in the lower one.

    1. The gap is (εi−εj)/kT=3.0(\varepsilon_i-\varepsilon_j)/kT = 3.0.
    2. Put it into the ratio: Ni/Nj=e−3.0N_i/N_j = e^{-3.0}.
    3. The lecture’s table lists this as 0.050.05.
    4. So the higher state is thinly populated but not empty. A gap of 10 kT10\,kT gives 0.0000450.000045, which is almost nothing.

    kT is a few kJ per mole at everyday temperatures

    Everything above is measured against kTkT, so you need its size. The lecture gives these constants: k=1.38×10−23k = 1.38\times10^{-23} J K⁻¹, e=1.6×10−19e = 1.6\times10^{-19} C, NA=6.02×1023N_A = 6.02\times10^{23} mol⁻¹ and cal = 4.18 J. Its table of thermal energy:

    TT (°C)TT (K)kTkT (J)kTkT (eV)kTkT (J/mol)kTkT (kJ/mol)kTkT (kcal/mol)
    02733.8×10⁻²¹0.02423002.30.54
    252984.1×10⁻²¹0.02625002.50.59
    1003735.1×10⁻²¹0.03231003.10.74

    Exam detail

    Know the population ratio, the idea that a gap much bigger than kTkT means almost no change, and that kTkT is about 2.5 kJ/mol near room temperature. The slide table is the source for these numbers.

    Polymers and colloids sit right at the kT scale

    Polymers. A polymer is a long chain of repeating units. Compare two energies with kTkT of about 2.5 kJ/mol. The C–C bond energy is 348 kJ/mol, far above kTkT, so chains do not break thermally. C–C bond torsion (twisting about the bond) is about 12 kJ/mol, a little above kTkT, so chains can bend and twist. Their interactions with solvents or other polymer molecules are at a similar scale. So solubility, swelling, mixing and crystallisation are all possible, and can change with temperature alone, even before any chemical modification.

    Colloids. A colloid is a heterogeneous mixture where the dispersed particles are between the size of a solution and a suspension. Think of milk or paint. The tiny particles of a stable colloid do not settle under gravity if their gravitational potential energy is about kTkT. Heat keeps knocking them around. For stability, the energy barrier to particles combining and clumping together must be much bigger than kTkT. At high volume fractions, polymeric colloids can form a gel, and strongly change the viscosity of the solution.

    Pitfall

    Do not read “soft” as “weak bonds everywhere”. The same polymer has a bond energy of 348 kJ/mol (strong, far above kTkT) and a torsion energy of about 12 kJ/mol (soft, near kTkT). The softness comes from the motions that sit near kTkT, not from every bond.

    Recall

    Why can a molecule's chain twist freely at room temperature while its backbone bonds never break?

    Twisting costs about 12 kJ/mol, which is close enough to kTkT (about 2.5 kJ/mol) for heat to supply it. Breaking a C–C bond costs 348 kJ/mol, so almost no molecule ever gets that much energy.

    Check yourself

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