3.2 Internal energy, enthalpy, and specific heats

Internal energy, enthalpy, and specific heats

The First Law of Thermodynamics is fundamentally about energy conservation: energy can't be created or destroyed, only transferred or converted. Internal energy, enthalpy, and specific heats are the quantities that let you track where energy goes and how much the temperature changes during a process. Mastering these concepts is essential for analyzing everything from piston-cylinder devices to turbines and heat exchangers.

Defining and explaining key concepts

Internal energy ($U$) represents the total microscopic energy stored within a system. This includes the kinetic and potential energies of all the particles, as well as chemical and nuclear energies. You can't measure $U$ directly, but you can measure changes in it, which is what matters in thermodynamics.

• Kinetic energy contributions come from particle motion: translational, rotational, and vibrational
• Potential energy contributions come from intermolecular forces, chemical bonds, and nuclear forces

Enthalpy ($H$) is a state function defined as:

$H = U + PV$

Because it bundles together internal energy and the pressure-volume product, enthalpy is especially useful for constant-pressure processes (which are extremely common in engineering). As a state function, enthalpy depends only on the initial and final states of the system, not on the path taken between them.

Specific heat is the amount of heat required to raise the temperature of a unit mass of a substance by one degree. It tells you how readily a substance stores thermal energy.

• Specific heat at constant volume ($C_v$): heat capacity measured while holding volume constant. All the added heat goes into raising internal energy.
• Specific heat at constant pressure ($C_p$): heat capacity measured while holding pressure constant. Some of the added heat goes into doing expansion work, so $C_p$ is always greater than $C_v$.
• For an ideal gas, the two are related by:

$C_p - C_v = R$

where $R$ is the specific gas constant (not the universal gas constant, unless you're working on a per-mole basis). On a molar basis, $\bar{C}_p - \bar{C}_v = R_u$, where $R_u = 8.314 \text{ kJ/(kmol·K)}$.

Relationship between specific heats and molecular structure

The number of ways a molecule can store energy (its degrees of freedom) directly determines its specific heat.

• Monatomic gases (helium, neon) only have three translational degrees of freedom, giving them the lowest specific heats. For a monatomic ideal gas: $\bar{C}_v = \frac{3}{2}R_u$ and $\bar{C}_p = \frac{5}{2}R_u$.
• Diatomic gases (nitrogen, oxygen) add rotational modes, and at higher temperatures, vibrational modes. This means more energy storage per degree of temperature rise, so their specific heats are higher.
• Polyatomic gases (carbon dioxide, methane) have even more rotational and vibrational modes, giving them the highest specific heats among gases.

Solids generally have lower specific heats than liquids because particle motion is more restricted. Also, substances with higher molecular weights tend to have lower specific heats per unit mass, since the same amount of heat is spread across heavier molecules.

Internal energy and enthalpy changes

First law of thermodynamics and heat-work interactions

The first law in its most common form:

$\Delta U = Q - W$

where $Q$ is heat added to the system and $W$ is work done by the system. (Watch the sign convention your course uses; some textbooks define $W$ as work done on the system, flipping the sign.)

Here's how this plays out in common processes:

• Constant pressure process: The change in enthalpy equals the heat transfer: $\Delta H = Q_p$. This is why enthalpy is so useful for constant-pressure analysis. The system does expansion work against the surroundings, and the enthalpy change accounts for both the internal energy change and that work.
• Adiabatic process ($Q = 0$): All energy exchange happens through work.
  • Adiabatic expansion: the system does work, so $\Delta U = -W$ and internal energy decreases (temperature drops)
  • Adiabatic compression: work is done on the system, so internal energy increases (temperature rises)
• Isothermal process (constant temperature): For an ideal gas, $\Delta U = 0$ because internal energy depends only on temperature. That means $Q = W$: whatever work the system does must be supplied as heat, and vice versa.

Reversible and irreversible processes

A reversible process is an idealized, quasi-static process that can be completely undone with no net effect on the system or surroundings. Think of it as the theoretical best case.

• Examples: infinitely slow frictionless expansion, heat transfer across an infinitesimally small temperature difference

An irreversible process leaves a permanent change somewhere. Nearly all real processes are irreversible.

• Causes of irreversibility: friction, rapid expansion/compression, turbulence, heat transfer across large temperature differences, mixing
• Irreversible processes always produce less useful work (or require more input work) than their reversible counterparts

Calculating internal energy and enthalpy changes

Using specific heat values

For processes where specific heats can be treated as constant over the temperature range:

Internal energy change:

$\Delta U = m \cdot C_v \cdot \Delta T$

This works for any ideal gas process (not just constant-volume ones), because $U$ for an ideal gas depends only on temperature.

Enthalpy change:

$\Delta H = m \cdot C_p \cdot \Delta T$

Similarly, this works for any ideal gas process, since $H$ for an ideal gas also depends only on temperature. For solids and liquids, $C_p \approx C_v$, so you can use either one.

When specific heats vary significantly with temperature (large $\Delta T$), you should use tabulated values or integrate:

$\Delta h = \int_{T_1}^{T_2} C_p(T) \, dT$

Most thermodynamics textbooks provide ideal-gas property tables for this purpose.

Applying Hess's law

Hess's law states that the total enthalpy change for a reaction is independent of the path from reactants to products. This follows directly from enthalpy being a state function.

To apply Hess's law:

1. Write the target reaction you need the enthalpy change for.
2. Find a set of reactions with known enthalpy changes that, when combined, give you the target reaction.
3. If you reverse a reaction, flip the sign of its $\Delta H$.
4. If you multiply a reaction by a factor, multiply its $\Delta H$ by the same factor.
5. Sum all the individual $\Delta H$ values to get the total enthalpy change.

This technique is most commonly used in chemistry-adjacent thermo problems (combustion, formation reactions), but the underlying principle applies whenever you need to find enthalpy changes for processes that are hard to measure directly.

Temperature and pressure effects on internal energy and enthalpy

Ideal gas behavior

For an ideal gas, internal energy is a function of temperature only. Pressure and volume changes at constant temperature do not affect $U$. For a monatomic ideal gas:

$U = \frac{3}{2}nR_uT$

Enthalpy for an ideal gas is also a function of temperature only. This is a point that trips people up. Even though $H = U + PV$ and $PV = nR_uT$, both $U$ and $PV$ depend only on $T$ for an ideal gas, so $H$ does too. That means:

$\left(\frac{\partial H}{\partial P}\right)_T = 0 \quad \text{(ideal gas)}$

This is a key distinction from real gases.

Real gas behavior and the Joule-Thomson effect

Real gases deviate from ideal behavior because of intermolecular forces and finite molecular size. These deviations mean that $U$ and $H$ depend on pressure (or specific volume) in addition to temperature.

• Attractive intermolecular forces (van der Waals forces) tend to lower the internal energy and enthalpy compared to an ideal gas at the same conditions.
• Finite molecular size reduces available volume, effectively increasing pressure effects on enthalpy.

The Joule-Thomson effect describes the temperature change when a real gas undergoes a throttling process (adiabatic expansion through a valve or porous plug, with no work or heat transfer). The key quantity is the Joule-Thomson coefficient:

$\mu_{JT} = \left(\frac{\partial T}{\partial P}\right)_H$

• If $\mu_{JT} > 0$: the gas cools upon expansion (pressure drops, temperature drops). Most gases at room temperature behave this way, including nitrogen, oxygen, and carbon dioxide.
• If $\mu_{JT} < 0$: the gas warms upon expansion. Hydrogen and helium at room temperature are the classic examples.
• The inversion temperature is where $\mu_{JT} = 0$. Above this temperature, a gas that normally cools will instead warm upon throttling.

The Joule-Thomson effect is the working principle behind many industrial cooling and gas liquefaction processes. For an ideal gas, $\mu_{JT} = 0$ (no temperature change during throttling), which is another way to see that ideal gas enthalpy doesn't depend on pressure.