Thermodynamics - Equations

As we progress through Thermodynamics we will have more and more equations for your to learn and memorize. I will start putting the equations here as we learn them. Best for you to start memorizing them ASAP.You must have JavaScript enabled in your browser in order to see the formulas correctly.

Wed, 11/3

$\Delta E = E_{\rm f} - E_{\rm i}$         $\Delta E = q + w $ (First Law equation)

Internal Energy, $E$, is also shown as $U$ in many books and on Quest. Therefore:

$\Delta U = U_{\rm f} - U_{\rm i}$         $\Delta U = q + w$

Fri, 11/5

$w = -P_{\rm ext}\Delta V$ (must have constant pressure in order to use this formula)

Enthalpy is defined: $H = E + PV$ which means that (at constant pressure) $\Delta H = \Delta E + P\Delta V$, or rewritten to be $\Delta E = \Delta H - P\Delta V$. $\Delta E = q + w $ at constant pressure becomes, $\Delta E = q_P - P\Delta V $, which means that $\Delta H = q_P$
$\Delta E = q + w $ at constant volume becomes, $\Delta E = q_V$

For any substance not changing phase, $q = C_{\rm s}\, m\, \Delta T$.

$q_{\rm cal} = C_{\rm s, water} \cdot m_{\rm water} \cdot \Delta T$ (coffee cup calorimetry)

$q_{\rm cal} = C_{\rm s, water} \, m_{\rm water} \, \Delta T + C_{\rm hardware}\Delta T$ (bomb calorimetry)

Mon, 11/8

Work only comes from the expansion or compression of gases. This means that you can express work in terms of $\Delta n$ or in terms of $\Delta T$ because $\Delta(PV) = \Delta(nRT)$.

$w = -P\Delta V = -\Delta n R T$ where $\Delta n$ = (mol of gas products) - (mol of gas reactants).

Hess' Law: $\Delta H_{\rm rxn} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots $

Wed, 11/10

Depending on HOW you chose to expand your system and do work, you could calculate work via these equations:
$w = -P\Delta V$       $w = -\Delta nRT$       $w = -nR\Delta T$

You also have a definite difference in $C_V$ and $C_P$ for gases and the equation is
$C_P = C_V + R$

You also get 3 different values for $C_V$ and for $C_P$ depending on what form of ideal gas you have. This is explained best via the Equipartition Theorm which says to equally distribute the overall energy across ALL the degrees of freedom. The degrees of freedom are translational modes (3 total for x, y, and z directions), and rotational modes (rotations about the x, y, and z axes. A single atom is too small to have any significant rotational inertia and therefore only has the 3 modes of translational motion. A diatomic (or linear) molecule can have rotational inertia about 2 axes. A polyatomic non-linear molecule can have all 3 rotational modes occupied. Each mode of freedom has ${1\over2}RT$ (J/mol). This is what you sum up to get the overall motional contribution to the internal energy of an ideal gas.

FYI: Vibrational modes are only available at very very high temperatures. For this reason we do not quantify the vibrational modes of freedom in these ideal gases.

Fri, 11/12

You can use $\Delta H_{\rm f}^\circ$ for the rxn steps in Hess' Law.

$$\Delta H_{\rm rxn}^\circ =\sum{n\Delta H^\circ_{\rm f} (\rm products)} - \sum{n\Delta H^\circ_{\rm f} (\rm reactants)}$$

You can also get a good approximation of $\Delta H$ via the summation of bond energies. The concept is the same regardless of what book (or source) you read it from. Here are a few...

From Zumdahl, Chapter 13, section 8. The "D" is for the dissociation energy of the bond. $$\Delta H =\underbrace{\sum{D\:{\rm (bonds\;\;broken)}}}_{\rm energy\;\; required\uparrow} - \underbrace{\sum{D\:{\rm (bonds\;\;formed)}}}_{\rm energy\;\;released\downarrow}$$

From Aktins/Jones, they use "mean bond enthalpies", $\Delta H_{\rm B}$ $$\Delta H_{\rm rxn}^\circ =\sum{n\Delta H_{\rm B} (\rm reactants)} - \sum{n\Delta H_{\rm B} (\rm products)}$$ Of course when using bond energies or enthalpies, you should break only the bonds that need breaking and make the bonds that need making. You don't have to break every bond in the molecule if much of its structure is retained after the reaction.

Mon, 11/15

Isothermal Expansion of an ideal gas. You stay on the isotherm throughout the entire expansion which means that $\Delta E = 0$ and $q = -w$. Also, the external pressure has to constantly match the dropping system pressure - this will result in a reversible process. So this is really isothermal reversible expansion. P is a function of volume ($P = nRT/V$) and so one can integrate ${\rm d}w = -nRT\;{\rm d}V/V$ to give:

$w = -nRT \ln \left({V_2\over V_1}\right)$         $q = nRT \ln \left({V_2\over V_1}\right)$

And, Boyle's Law is in effect so that $P_1/P_2 = V_2/V_1$ and therefore:

$w = -nRT \ln \left({P_1\over P_2}\right)$         $q = nRT \ln \left({P_1\over P_2}\right)$

Entropy is a measure of energy dispersal and the 2nd Law states that universal entropy must increase for any spontaneous process: $$\Delta S_{\rm univ} = \Delta S_{\rm sys} + \Delta S_{\rm surr}$$

Wed, 11/17

Entropy (S) is really a measure of the number of microstates (W) in the system:

$$S = k \ln W$$

Where k is the Boltzmann Constant (1.38 × 10^-23 J/K) which is the same as the universal gas constant, R, divided by Avogadro's number, N_A. It is easy to associate the number of available microstates with the available volume (V). For example, if you double the volume, you will double the number of available microstates. From this, we can consider state1 and then state2 and we get: $$ \Delta S = nR \ln \left({V_2\over V_1}\right) $$

That formula is much like the formula for heat (q) transfer during isothermal expansion and you can rightfully conclude that: $$ \Delta S = {q_{\rm rev}\over T} $$

Now the change in entropy is calculated via the heat flow in/out of the system divided by the temperature at which the transfer took place. Some other outcomes from this fact: $$ \Delta S = n C_v \ln \left({T_2\over T_1}\right) $$ $$ \Delta S = n C_p \ln \left({T_2\over T_1}\right) $$

Note that you use $C_v$ if done at constant volume, or $C_p$ if done at constant pressure. Also, remember that heat can flow and temperature can remain constant - like for phase changes. For phase changes $q_{\rm transition} = \Delta H_{\rm transition}$. This means that you can get entropy change via: $$ \Delta S_{\rm transition} = {\Delta H_{\rm transition}\over T_{\rm transition}}$$

"Transition" here means vaporization, or fusion, or sublimation, etc... and the temperature at transition is just the melting point (fusion) or the boiling point (vaporization).

Fri, 11/19

The Gibb's Free Energy (G) is another state function that is defined from 3 others: $G = H - TS$. When we hold temperature and pressure constant we get:

&DeltaG = ΔH − TΔS

We typically are getting the $\Delta X$'s in the above equation from Standard Thermodynamic Data Tables like you have in Appendix 4 in your textbook. If you are at standard state you add the superscript "°"'s to the terms:

&DeltaG° = ΔH° − TΔS°

Of course this means that the temperature must be 25°C or 298.15 K in the above equation. You can also show through a simple proof that $\Delta G$ (a system state function) is equal to $-T\Delta S_{\rm univ}$. So we can now use $\Delta G$ as our indicator of spontaneity for a reaction or process.

Definition of Thermodynamic equilibrium: $\Delta G = 0$

Also, we can assume that plain ol' $\Delta G$ (non-standard conditions) is approximately equal to the adjusted standard values for $\Delta H$ and $\Delta S$ via temperature:

&DeltaG ≈ ΔH° − TΔS°

If you assume you're at equilibrium, then $\Delta G = 0$, and you get the following:

$\Delta H = T\Delta S$        $\Delta S = \Delta H / T$        $T = \Delta H / \Delta S $

That is the same equation we got for $\Delta S$ back on Wed, 11/17, for phase transitions.