David Bauer Physics & Astronomy · UCLA

Physics 5B Chapter 6

Internal Energy and Heat Capacity

Internal energy of a gas

Internal energy of a monatomic ideal gas

The average kinetic energy allows us to calculate the internal energy of the ideal gas. We will start by considering a monatomic ideal gas, in which the gas particles are single atoms. The noble gases (helium, neon, argon, etc.) can be modeled as monatomic ideal gases.

In the ideal gas model, we neglect potential energy between molecules. For a monatomic ideal gas, all of the internal energy is random translational kinetic energy:

\[E_{\mathrm{int}} = K_{\mathrm{tot}}.\]

For $N$ particles, the total kinetic energy is $N$ times the average kinetic energy of a single particle,

\[K_{\mathrm{tot}}=NK_{\mathrm{av}}.\]

Multiplying the average kinetic energy $K_{\mathrm{av}} = \tfrac{3}{2}k_BT$ by the number of particles $N$, we get the total internal energy of a monatomic ideal gas:

\[\boxed{E_{\mathrm{int}}=\frac{3}{2}Nk_BT}\quad \text{or}\quad \boxed{E_{\mathrm{int}}=\frac{3}{2}nRT.}\]

For a fixed amount of a monatomic ideal gas, this means that the internal energy changes by

\[\Delta E_{\mathrm{int}}=\frac{3}{2}nR\,\Delta T\]

when the temperature changes by $\Delta T$. This is consistent with our earlier statement that the internal energy is a state function, since it only depends on the state variable $T$.

Example: Warming a helium balloon

Lucia fills a balloon with $10.0\ \mathrm{L}$ of helium, a monatomic ideal gas, at $101.3\ \mathrm{kPa}$ and $20.0^\circ\mathrm{C}$. She carries it outside, where the sun warms the helium to $40.0^\circ\mathrm{C}$, and the balloon expands as it warms. Assume the pressure of the helium stays constant.

  1. What is the change in the internal energy of the helium?
  2. How much work is done on the helium?
  3. How much heat flows into the helium?
Solution

1. Converting the givens to SI units, with temperatures in kelvin,

\[p = 1.013\times 10^{5}\ \mathrm{Pa}, \qquad V_i = 1.00\times 10^{-2}\ \mathrm{m^3}, \qquad T_i = 293\ \mathrm{K}, \qquad T_f = 313\ \mathrm{K}.\]

The pressure is atmospheric pressure and stays constant. The internal energy of a monatomic ideal gas changes by

\[\Delta E_{\mathrm{int}} = \frac{3}{2}nR\,\Delta T.\]

We could find $n$ from the initial state, $n = pV_i/RT_i$. Instead, we can avoid computing $n$ by using the ideal gas law, $pV = nRT$, directly. At constant pressure, a change in temperature only changes the volume, so $p\,\Delta V = nR\,\Delta T$, and

\[\Delta E_{\mathrm{int}} = \frac{3}{2}p\,\Delta V.\]

To find $\Delta V$, note that $n$ and $p$ stay fixed, so the ideal gas law gives $V_i/T_i = V_f/T_f$, or

\[V_f = \frac{T_f}{T_i}V_i = \frac{313\ \mathrm{K}}{293\ \mathrm{K}}\,(10.0\ \mathrm{L}) = 10.68\ \mathrm{L}.\]

The helium’s volume increases by $\Delta V = V_f - V_i = 0.683\ \mathrm{L} = 6.83\times 10^{-4}\ \mathrm{m^3}$. Then

\[\Delta E_{\mathrm{int}} = \frac{3}{2}\left(1.013\times 10^{5}\ \mathrm{Pa}\right)\left(6.83\times 10^{-4}\ \mathrm{m^3}\right) = 103.7\ \mathrm{J} \approx 104\ \mathrm{J}.\]

2. The pressure stays constant, so the work done on the helium is

\[W = -p\,\Delta V = -\left(1.013\times 10^{5}\ \mathrm{Pa}\right)\left(6.83\times 10^{-4}\ \mathrm{m^3}\right) = -69.1\ \mathrm{J}.\]

The work is negative because the helium expands, pushing the surrounding air outward.

3. The first law of thermodynamics relates the heat to the results of parts 1 and 2:

\[\Delta E_{\mathrm{int}} = Q + W.\]

Solving for $Q$,

\[Q = \Delta E_{\mathrm{int}} - W = 103.7\ \mathrm{J} - (-69.1\ \mathrm{J}) = 173\ \mathrm{J}.\]

The heat from the sun increases the helium’s internal energy by $104\ \mathrm{J}$, with the other $69\ \mathrm{J}$ leaving as work the helium does pushing the surrounding air.

Specific heats of gases

Specific heats of gases

In the chapter on specific heat, we introduced the concept of specific heat, which measures how much heat is required to raise the temperature of a substance by a given amount:

\[Q = m c \Delta T.\]

We specifically excluded gases from that discussion. One reason to treat gases differently is that the specific heat depends on the process by which the gas is heated (e.g., constant volume vs. constant pressure).

The first law of thermodynamics tells us that

\[Q = \Delta E_{\mathrm{int}} - W.\]

While the internal energy change depends only on the temperature change, the heat and work transfer depend on the process.

Molar specific heats

When working with gases, it is common to use the molar specific heat, which measures the heat required to raise the temperature of one mole of the gas by a given amount:

\[Q = n C \Delta T,\]

where $n$ is the number of moles and $C$ is the molar specific heat.

The molar specific heat at constant volume $C_V$ is defined by

\[\boxed{Q_V = n C_V \Delta T.}\]

Similarly, the molar specific heat at constant pressure is defined by

\[\boxed{Q_P = n C_P \Delta T.}\]

Specific heat at constant volume

If a gas is heated or cooled in a rigid container at constant volume, no work is done on the gas, so the heat added goes entirely into changing the internal energy:

\[Q_V = \Delta E_{\mathrm{int}} = \frac{3}{2} n R \Delta T.\]

Comparing this to the definition of $C_V$, we find that for a monatomic ideal gas

\[C_V = \frac{3}{2} R \approx 12.47\ \mathrm{J/(mol\cdot K)}.\]

Although the heat transfer $Q$ depends on the process, the internal energy change for any process is

\[\Delta E_{\mathrm{int}} = nC_V\Delta T\]

because $E_{\mathrm{int}}$ is a state function.

Specific heat at constant pressure

If a gas is instead heated or cooled at constant pressure, the volume of the gas will change. In this case, work is done on the gas by the environment, and a different amount of heat $Q_P$ is required to change the temperature of the gas.

For example, consider the scenario in which we heat a gas in a flexible container (like a balloon) such that the gas stays at atmospheric pressure. The ideal gas law tells us that

\[V = \frac{n R T}{p},\]

so the volume of the gas will increase as the temperature increases.

Since the gas expands, energy is transferred out of the gas as work. This means that not all of the energy added to the gas as heat ends up increasing the internal energy. Compared to a constant volume process, we would need to add more heat to the gas to produce the same temperature increase, to account for the energy “lost” as work.

A before-and-after schematic of a gas heated at constant pressure in a balloon. On the left is a small balloon, a light cyan circle with blue dots for the gas molecules, labeled below V sub 1, T sub 1. Three orange arrows point in from the left toward the balloon, labeled heat in. A dark grey process arrow runs from the small balloon to the right. On the right is the same balloon after heating, holding the same molecules but drawn larger, labeled below V sub 2 greater than V sub 1, T sub 2 greater than T sub 1. Three purple arrows point outward from its right side, labeled work out: the expanding gas pushes the surrounding air back, so the work done on the gas, minus p times the change in volume, is negative. The pressure stays at atmospheric pressure throughout. Because energy leaves the gas as work while the balloon expands, only part of the heat added is left to raise the internal energy and hence the temperature, so heating a gas at constant pressure takes more heat than heating the same gas at constant volume.

Using the first law and the work $W = - p\Delta V$ for a constant pressure process, the heat transfer into a gas in a constant pressure process is

\[Q_P = \Delta E_{\mathrm{int}} + p\Delta V.\]

If the pressure is constant, then the ideal gas law tells us that

\[p \Delta V = n R \Delta T,\]

so

\[Q_P = \Delta E_{\mathrm{int}} + n R\Delta T.\]

As we noted earlier, the internal energy change for any process is $\Delta E_{\mathrm{int}} = n C_V \Delta T$. Plugging $\Delta E_{\mathrm{int}}$ into the equation for $Q_P$, we get

\[Q_P = n C_V \Delta T + n R \Delta T = n (C_V + R) \Delta T.\]

Comparing this to the definition of $C_P$, we see that the constant pressure specific heat is related to the constant volume specific heat by

\[\boxed{C_P = C_V + R.}\]

This is a general relation for any ideal gas. For monatomic gases, we find

\[C_P = \frac{5}{2}R \approx 20.79\ \mathrm{J/(mol\cdot K)}.\]

Specific heats of monatomic gases

The following figure shows measured values of $C_V$ and $C_P$ for real gases at $25^\circ\mathrm{C}$ and $1\ \mathrm{atm}$. Note that both panels are zoomed in, and the measured values are within about one percent of the prediction.

A two-panel plot of the measured molar specific heats of the five noble gases helium, neon, argon, krypton, and xenon, in that order along the horizontal axis. The vertical axis is molar specific heat C in joules per mole per kelvin, and it is broken, with a double slash marking the break: the lower panel is zoomed on the narrow range 12.30 to 12.60 and shows the measured constant-volume specific heat C sub V, while the upper panel is zoomed on 20.60 to 21.10 and shows the measured constant-pressure specific heat C sub P. In each panel a dashed red horizontal line marks the ideal monatomic gas prediction: three halves R, or 12.47, in the lower panel, and five halves R, or 20.79, in the upper panel. The blue measured points lie on or just above the prediction line in both panels. Helium and neon sit essentially on the line, and the points rise slightly and steadily from argon to krypton to xenon, with xenon the highest, about 0.06 above the line for C sub V and about 0.23 above it for C sub P. The data stay within half a percent of the prediction for C sub V and about one percent for C sub P, so the ideal-gas prediction describes all five gases well, and the small upward drift with increasing molar mass is the departure from ideal behaviour of the heavier, more strongly interacting atoms.

Data: Air Liquide Gas Encyclopedia.

The equipartition theorem

The equipartition theorem

The expression $E_{\mathrm{int}} = \tfrac{3}{2} N k_B T$ for the internal energy of a monatomic ideal gas is a special case of a general principle known as the equipartition theorem. The equipartition theorem says that each quadratic energy term in the microscopic energy of a system contributes $\frac{1}{2} k_B T$ to the thermal energy when we average over all microscopic states.

By “quadratic energy term,” we mean an energy that is proportional to the square of a microscopic variable. For example, the kinetic energy of a single particle

\[K = \tfrac{1}{2} m v_x^2 + \tfrac{1}{2} m v_y^2 + \tfrac{1}{2} m v_z^2\]

has three quadratic energy terms, one for each component of the velocity.

Diatomic molecules

Although a monatomic ideal gas only has translational kinetic energy, more complex molecules can have additional types of energy, such as rotational and vibrational energy. These other types of energy also have quadratic energy terms, leading to different expressions for the thermal energy and specific heat of the gas.

Let’s consider a diatomic molecule, which has the following additional types of motion:

  • Rotational motion about the two axes perpendicular to the line connecting the two atoms.
  • Vibrational motion of the two atoms along the line connecting them.

Rotational energy of a diatomic molecule

The molecule can rotate about either of the two axes perpendicular to the bond, and each rotation contributes a quadratic energy term to the total energy.

Two schematic diagrams of the same diatomic molecule, each showing rotation about one of the two axes perpendicular to the bond. In both panels the molecule is drawn as two shaded blue atoms of equal size joined by a thick horizontal bond, and both axes pass through the centre of the bond. In the left panel, captioned Axis 1, the axis lies in the page, drawn as a dashed grey line running vertically through the centre of the molecule. The atoms swing out of the page, so the circular path they sweep out is seen almost edge-on, drawn as a dotted flattened ellipse through both atoms, and a green arrow runs along the near, lower half of that path from the left atom to the right atom. In the right panel, captioned Axis 2, the axis points out of the page and is drawn as a grey circled dot at the centre of the bond; the path of the atoms is a dotted circle through both atoms lying in the page, and a green arrow runs counterclockwise along the top of that circle, so the molecule spins within the page like a propeller. Each rotation has its own moment of inertia and angular velocity and contributes a quadratic energy term, one half I omega squared, so the two rotations together add two degrees of freedom.

Vibrational energy of a diatomic molecule

The two atoms can also vibrate along the line connecting them, with the bond acting as a spring. This gives two quadratic energy terms: kinetic energy and elastic potential energy.

\[E_{\mathrm{vib}} = \frac{1}{2} m v_s^2 + \frac{1}{2} k_s s^2.\]

Three uncaptioned snapshots of the same diatomic molecule vibrating along the line joining its two atoms. In every snapshot the molecule is two shaded blue atoms joined by a coiled spring, standing for the bond. On the left the bond is stretched: the atoms sit farther apart than their equilibrium separation, the spring is drawn with long, open coils, two dashed grey lines mark the equilibrium positions of the atoms, and a short arrow labeled s runs outward from the right-hand equilibrium line to the atom, measuring how far it has moved. At this turning point the atoms are momentarily at rest, so all of the vibrational energy is potential. In the middle the atoms sit at their equilibrium separation, and green velocity arrows point outward from each atom, showing the atoms passing through equilibrium at their greatest speed, so all of the vibrational energy is kinetic. On the right the bond is compressed: the atoms sit closer together than equilibrium, the spring is drawn with tight coils, the dashed equilibrium lines are again marked, now outside the atoms, and the arrow labeled s runs inward from the right-hand equilibrium line to the atom. The atoms are once more momentarily at rest, so the energy is again all potential. The energy shuttles back and forth between the kinetic term, one half m v sub s squared, and the potential term, one half k sub s s squared, which is why vibration contributes two quadratic energy terms rather than one.

Specific heat of a diatomic gas

When all of these motions are active, a diatomic molecule has a total of seven quadratic energy terms: three from translational motion, two from rotational motion, and two from vibrational motion. The equipartition theorem would then predict an internal energy $E_{\mathrm{int}} = \tfrac{7}{2}Nk_B T$.

However, the vibrational motion requires higher energies to become active. At everyday temperatures, the vibrational motion is not active, so a diatomic ideal gas has only five quadratic energy terms, and the internal energy is

\[E_{\mathrm{int}} = \frac{5}{2} N k_B T = \frac{5}{2}nRT,\]

so the constant volume specific heat for a diatomic gas is

\[C_V = \frac{5}{2}R.\]

Rotation becomes active at about $100\ \mathrm{K}$, and vibration only becomes active above about $1000\ \mathrm{K}$. The following figure shows the theoretical $C_V$ for $\mathrm{H_2}$ gas.

A graph of the molar specific heat at constant volume of hydrogen gas, divided by the gas constant R, against temperature. The temperature axis is logarithmic, running from 10 kelvin to 10000 kelvin with ticks at each factor of ten; the vertical axis runs from 0 to about 4, with labelled dashed guide lines at three halves, five halves, and seven halves. The single blue curve climbs from left to right as a staircase with three plateaus. From 10 to about 25 kelvin it sits flat at three halves, where only the three translational degrees of freedom are active. Between about 30 and 80 kelvin it rises steeply to five halves, slightly overshooting that value near 70 kelvin before settling, as the two rotational degrees of freedom become active. It then runs flat at five halves across a broad range that includes room temperature, 300 kelvin, which is marked by a vertical dashed line: this is the value five halves R quoted for a diatomic gas at everyday temperatures. Above about 1000 kelvin the curve rises again toward seven halves as the two vibrational degrees of freedom switch on, reaching about 3.47 at 10000 kelvin. The last part of that rise, above roughly 3000 kelvin, is drawn as a dashed line because hydrogen molecules dissociate at those temperatures, so the seven halves plateau is never actually reached by real hydrogen. The staircase shape shows that a degree of freedom contributes its one half k B T only once the temperature is high enough, and is frozen out below that.