Heat depends on how a gas is heated
The heat needed to warm one mole of a substance by one kelvin is its molar heat capacity, \(C = Q/(n\,\Delta T)\), in J/(mol K). For a solid or a liquid it has one value, because they barely expand. A gas is different. Heat it in a rigid vessel and it cannot expand. Heat it under a free piston and it expands and lifts the piston. The two ways need different amounts of heat, so a gas has two molar heat capacities. This lesson uses \(R = 8.31\,\text{J/(mol K)}\).
In a rigid vessel the walls do not move, so the gas does no work. Every joule of heat goes into internal energy: \(Q = \Delta U = \tfrac f2 nR\,\Delta T\). Dividing by \(n\,\Delta T\) gives the molar heat capacity at constant volume, \(C_v = \tfrac f2 R\). The last lesson met these numbers in the steel bottles.
For a monatomic gas, \(C_v = \tfrac32 R = 12.5\,\text{J/(mol K)}\). For a diatomic gas near room temperature, \(C_v = \tfrac52 R = 20.8\,\text{J/(mol K)}\). For a non-linear molecule, \(C_v = 3R = 24.9\,\text{J/(mol K)}\). These values match the measured heat capacities of helium, nitrogen and steam closely.
One mole of helium in a rigid vessel is heated from \(300\,\text{K}\) to \(400\,\text{K}\). How much heat does it need, in joules? Take \(R = 8.31\,\text{J/(mol K)}\).
Two cylinders hold one mole each of nitrogen at \(300\,\text{K}\). One has its piston fixed; the other has a free, loaded piston. Identical burners give each \(600\,\text{J}\). Which gas ends hotter?
Press play in the figure above with nitrogen chosen. After \(600\,\text{J}\), the gas under the fixed piston is \(28.9\,\text{K}\) warmer. The gas under the free piston is only \(20.6\,\text{K}\) warmer, and its readout shows why: it has done \(171\,\text{J}\) of work lifting the loaded piston. The ratio of the two rises is \(28.9/20.6 = 1.4\). With helium chosen the rises are \(48.1\,\text{K}\) and \(28.9\,\text{K}\), a ratio of \(1.67\).
The work done by the free gas can be found from mechanics. The gas pushes the piston of area \(A\) with a steady force \(pA\). If the piston rises by \(\Delta h\), the work is \(pA\,\Delta h = p\,\Delta V\). At a fixed pressure, \(pV = nRT\) gives \(p\,\Delta V = nR\,\Delta T\). So the heat must supply the rise in internal energy and this work as well.
\[ Q_p = \Delta U + p\,\Delta V \]
Why. energy is conserved. The heat supplied at constant pressure raises the internal energy and pays for the work of lifting the piston.
\[ \Delta U = nC_v\,\Delta T \]
Why. the internal energy of an ideal gas depends on \(T\) alone, so the same \(\Delta T\) gives the same \(\Delta U\) whatever the process.
\[ p\,\Delta V = nR\,\Delta T \]
Why. \(pV = nRT\) at the start and \(p(V + \Delta V) = nR(T + \Delta T)\) at the end; subtract, with \(p\) fixed.
\[ nC_p\,\Delta T = nC_v\,\Delta T + nR\,\Delta T \]
Why. put steps 2 and 3 into step 1, and write the heat at constant pressure as \(Q_p = nC_p\,\Delta T\).
\[ C_p = C_v + R \]
Read it. divide by \(n\,\Delta T\).
For an ideal gas, the molar heat capacity at constant pressure exceeds that at constant volume by the gas constant: \(C_p - C_v = R\). The extra \(R\) per mole per kelvin pays for the work \(p\,\Delta V = nR\,\Delta T\) done by the expanding gas.
The ratio of the two heat capacities, \(\gamma = C_p/C_v\), is called the ratio of specific heats. With \(C_v = \tfrac f2 R\) and \(C_p = C_v + R\), it is \(\gamma = 1 + 2/f\). A monatomic gas has \(\gamma = 5/3 = 1.67\) and a diatomic gas \(7/5 = 1.40\). A non-linear molecule has \(4/3 = 1.33\). Measuring \(\gamma\) therefore counts the degrees of freedom of a molecule. The next chapter derives Mayer's relation again from the first law of thermodynamics.
Fig 5.2 shows both heat capacities as slopes. For a rise of \(100\,\text{K}\), nitrogen needs \(2078\,\text{J}\) at constant volume and \(2909\,\text{J}\) at constant pressure. The gap, \(831\,\text{J}\), is exactly \(nR\,\Delta T = 1 \times 8.31 \times 100\). At constant pressure a diatomic gas keeps \(C_v/C_p = 5/7\) of the heat as internal energy and spends \(2/7\) as work.
\(2\,\text{mol}\) of oxygen is heated by \(10\,\text{K}\). Find the heat needed at constant volume and at constant pressure, and the work done by the gas in the second case.
Oxygen is diatomic: \(C_v = \tfrac52 R = 20.775\) and \(C_p = \tfrac72 R = 29.085\,\text{J/(mol K)}\).
Constant volume: \(Q_v = 2 \times 20.775 \times 10 = 416\,\text{J}\). Constant pressure: \(Q_p = 2 \times 29.085 \times 10 = 582\,\text{J}\).
The work is \(p\,\Delta V = nR\,\Delta T = 2 \times 8.31 \times 10 = 166\,\text{J}\), which is \(Q_p - Q_v\).
A diatomic ideal gas is heated at constant pressure. What fraction of the heat supplied becomes work done by the gas?
A gas has \(\gamma = 1.4\). How many degrees of freedom does each of its molecules have?
Mixtures of gases
In a mixture, each gas keeps its own internal energy, and the energies add. Warm \(n_1\) moles of one gas and \(n_2\) of another together by \(\Delta T\) at constant volume. The heat is \((n_1C_{v1} + n_2C_{v2})\,\Delta T\). So the mixture behaves as one gas with a mole-weighted average \(C_v\). Its \(C_p\) is \(C_v + R\) as before, because the mixture still obeys \(pV = nRT\).
Mix \(1\,\text{mol}\) of helium with \(1\,\text{mol}\) of oxygen. Then \(C_v = (\tfrac32 R + \tfrac52 R)/2 = 2R\), \(C_p = 3R\) and \(\gamma = 3R/2R = 1.5\). The mixture falls between the \(1.67\) of helium and the \(1.40\) of oxygen. Note that \(\gamma\) itself is not averaged. The heat capacities are averaged, and \(\gamma\) follows from them.
The equivalent molar mass of a mixture is its total mass divided by its total moles, \(M_{\text{mix}} = (n_1M_1 + n_2M_2)/(n_1 + n_2)\). For \(2\,\text{mol}\) of helium and \(1\,\text{mol}\) of nitrogen, \(M = (2 \times 4 + 1 \times 28)/3 = 12\,\text{g/mol}\). Per kilogram, light gases need much more heat. Helium has \(c_v = C_v/M = 12.465/0.004\), which is \(3116\,\text{J/(kg K)}\). Nitrogen has \(742\,\text{J/(kg K)}\). A kilogram of helium holds seven times as many molecules, and each needs its share of energy.
\(2\,\text{mol}\) of helium is mixed with \(1\,\text{mol}\) of nitrogen. Find \(\gamma\) for the mixture.
Solids: the law of Dulong and Petit
Equipartition works for solids too. An atom in a crystal is held in place by its neighbours, as if by springs. It vibrates about its place in three directions. Each direction has a kinetic energy term and a potential energy term, so each atom has \(f = 6\). One mole therefore holds \(U = 6 \times \tfrac12 RT = 3RT\). A solid barely expands, so the work is negligible and \(C = 3R = 24.9\,\text{J/(mol K)}\).
At room temperature and above, the molar heat capacity of most solid elements is close to \(3R\), about \(25\,\text{J/(mol K)}\), whatever the element.
The law explains why specific heats per kilogram differ so much among metals. Divide \(3R\) by the molar mass. Copper (\(63.5\,\text{g/mol}\)) gets \(393\,\text{J/(kg K)}\), against a measured \(385\). Aluminium (\(27\,\text{g/mol}\)) gets \(923\), against \(900\), and lead (\(207\,\text{g/mol}\)) gets \(120\), against \(128\). A light metal has more atoms in each kilogram, so it needs more heat. The law fails at low temperatures and for hard, light solids such as diamond, whose vibrations are partly frozen, like the rotations of cold hydrogen.
Use the law of Dulong and Petit to estimate the specific heat of iron (\(M = 56\,\text{g/mol}\)) in J/(kg K). Take \(R = 8.31\,\text{J/(mol K)}\).
In 1819 two French physicists, Pierre Dulong and Alexis Petit, measured the specific heats of thirteen solid elements. They noticed that the specific heat times the atomic mass came out nearly the same for all of them. Chemists soon used the rule the other way: from a measured specific heat they could estimate an unknown atomic mass. For decades the law of Dulong and Petit helped to fix the atomic masses of new elements.
- Using \(C_v\) for a gas heated under a free piston.A free piston keeps the pressure fixed, so use \(C_p = C_v + R\). A fixed piston or a rigid vessel uses \(C_v\).
- Averaging the values of \(\gamma\) for a mixture.Average \(C_v\) by moles, add \(R\) for \(C_p\), then divide. One mole of helium with one of oxygen gives \(1.5\), not \((1.67 + 1.40)/2\).
- Expecting every solid to have \(3R\).The law holds at room temperature for most elements. Light, hard solids such as diamond, and all solids at low temperature, have less.