Lecture 1

Readings

http://umdberg.pbworks.com/w/page/73084109/Thermodynamic%20equilibrium%20and%20equipartition%20%282013%29

http://umdberg.pbworks.com/w/page/104122763/Example%3A%20Degrees%20of%20freedom

Thermodynamic Equilibrium

• When an object is in equilibrium with its surroundings, it is still exchanging energy with its surroundings

  • But equal amounts of energy go in each direction.
  • Object -> Surroundings = Surroundings -> Object

• Systems naturally move toward equilibrium

• As a system ( system = multiple objects, or object plus its surroundings ) moves toward equilibirum, there is a net direction of energy flow, and this flow can be used

  • Like water flowing down a slope

    • You can put in a water wheel and generate mechanical energy by the spinning of the wheel
    • But water in a still lake won't work

Equipartition of Energy

• Always applied to molecular scale, so not a baseball

• When a system is in thermodynamic equilibrium, each degree of freedom will have thee same amount of energy, on average.

• Degrees of Freedom: A molecule in 3-D space can move in 3 possible directions

  • Think x,y,z axes

    • So 3 degrees of freedom for moion

• Kinetic energy will be evenly divided among these 3 directions

• Notice the statement says on average.

  • So average kinetic energy due to motion of molecules in x-direction is the same as the average kinetic energy due to motion of molecules in the y-direction.

• But molecules have a distribution of kinetic energies:

Degrees of Freedom

• Basically a counting game.

  • Add up which of the following motions can occur for your molecule

    • Translational kinetic energy - velocity in a direction

      • x,y, and z if 3-D motion possible

    • Rotational kinetic Energy = spin about an axis

      • Spin about x,y, and z.

        • But watch out for symmetry, see below\

    • Vibrational Energy

      • Kinetic Energy of Spring
      • Potential Energy of Spring

• Each option above adds $\frac{1}{2}\ast k_B\ast T$$\frac{1}{2}*k_B*T$ to molecule's average energy at a given temperature

• Example: N molecules of ideal gas have x,y,z translational, nothing else.

  • So 3 degrees of freedom
  • $E_{gastotal}=\frac{3}{2}\ast N\ast k_B\ast T$$E_{gas\ total} = \frac{3}{2} * N * k_B * T$ = average energy per molecule of $\frac{3}{2}\ast k_B\ast T$$\frac{3}{2}*k_B * T$ times "N" molecules

• If, when a molecule rotates, it doesn't "change locations" ( due to symmetry ), that rotation doesn't count , Like spinning a rod

Energy Bar Charts

• Prepare for kinesin problem
• Consider the formation of a bond $\mathrm{H}-\mathrm{H}$$\ce{H-H}$
• Before, have 2 free atoms, far apart, each with very little kinetic energy:

• After, the Hydrogens are bonded:

• We'll discuss why energy is below the line and negative, in the next section
• Where did energy go?
• In the first picture, total energy was positive
• In the second picture, total energy is negative.
• But total energy can't be created or destroyed.
• There must be some positive energy in the second picture.
• So after picture should look like:

• Other energy could be:

  • Motion (kinetic energy) pf H=H molecule
  • Or it could be motion of surrounding molecules = heat released by reaction

Chemical Bonds

• Define the zero energy to be when the 2 atoms are far apart

• When bond exists, net energy is negative

• Need to add energy to break a bond

  • Need to bring net energy back up to at least zero

• Need to remove energy to form a bond

  • Unbound atoms/molecules have zero or more energy on this scale

Basic Problems

1. How many degrees of freedom fro 25 molecules of $\mathrm{N}{\phantom{X}}_2$$\ce{N2}$ at a low temperature?

• 3 Translation + 2 Rotational = 5 total Degrees of Freedom per molecule
• No vibrational because of low temperature
• = $5\ast 25=125for{N}_2$$5 * 25 = 125\ for\ N_2$

2. How many degrees of freedom for 25 molecules of $\mathrm{N}{\phantom{X}}_2$$\ce{N2}$ at high temperature?

3 Translation + 2 Rotational + 2 Vibrational = 7 total Degrees of Freedom per molecule

= $7\ast 25=175for{N}_2$$7 * 25 = 175\ for\ N_2$

3. If you added the same amount of heat energy (say 10 Joules) to both 25 molecules of $\mathrm{N}{\phantom{X}}_2$$\ce{N2}$ at low temperature and 25 molecules of $N_2$$N_2$ at high temperature, how would the change in temperature compare?

• More degrees means more places to put the energy

$$<E>=DegreesofFreedom\ast \frac{k_B\ast T}{2}$$

$$\therefore T=\frac{2\ast <E>}{DegreesofFreedom\ast k_B}$$

At Low Temperature:

$$\mathrm{\Delta }T=\frac{\mathrm{\Delta }E\ast 2}{5\ast 25\ast k_B}$$

At High Temperature:

$$\mathrm{\Delta }T=\frac{\mathrm{\Delta }E\ast 2}{7\ast 25\ast k_B}$$

• $\mathrm{\Delta }E$$\Delta E$ is the same

  • So at high temperatures, $\mathrm{\Delta }T$$\Delta T$ is smaller