Specific Heat Capacity

Why two specific heats

When the temperature of a gas is increased, even a small rise causes a considerable change in volume and in pressure. Therefore two specific heats are defined for gases: the specific heat at constant volume $C_{V}$ and the specific heat at constant pressure $C_{P}$. Mayer's relation connects the two.

Mayer's relation — derivation

Consider one mole of an ideal gas enclosed in a cylinder by a light, frictionless, airtight piston. Let $P$, $V$ and $T$ be the pressure, volume and temperature of the gas.

i. If the gas is heated so that its temperature rises by $dT$ while the volume remains constant, the heat supplied $dQ_{1}$ is used entirely to increase the internal energy $dE$ of the gas, because with the volume constant no work is done in moving the piston.

$$dQ_{1} = dE = C_{V}\,dT \qquad ....(1)$$

where $C_{V}$ is the molar specific heat at constant volume.

ii. If instead the gas is heated through the same rise in temperature at constant pressure, the volume increases by $dV$. The heat supplied is now used both to increase the internal energy of the gas and to move the piston backwards to allow the gas to expand, the work done being $dW = P\,dV$.

$$dQ_{2} = dE + dW = C_{P}\,dT \qquad ....(2)$$

where $C_{P}$ is the molar specific heat at constant pressure.

iii. Since the internal energy of an ideal gas depends only on its temperature, $dE = C_{V}\,dT$ in both cases. From equations (1) and (2),

$$C_{P}\,dT = C_{V}\,dT + dW$$

$$\therefore \left(C_{P} - C_{V}\right)dT = P\,dV \qquad ....(3)$$

iv. For one mole of gas, $PV = RT$. At constant pressure, differentiating, $P\,dV = R\,dT$.

v. Substituting in equation (3),

$$\left(C_{P} - C_{V}\right)dT = R\,dT$$

$$\therefore \ C_{P} - C_{V} = R$$

This is Mayer's relation between $C_{P}$ and $C_{V}$.

The form with the mechanical equivalent of heat

The relation above assumes that heat energy and mechanical work are measured in the same units. Generally heat supplied is measured in calories and work done in joules, and the relation is then written

$$C_{P} - C_{V} = \dfrac{R}{J}$$

where $J$ is the mechanical equivalent of heat.

Also $C_{P} = M_{0}S_{P}$ and $C_{V} = M_{0}S_{V}$, where $M_{0}$ is the molar mass of the gas and $S_{P}$, $S_{V}$ are the respective principal specific heats (per unit mass). Then $M_{0}S_{P} - M_{0}S_{V} = \dfrac{R}{J}$, so

$$S_{P} - S_{V} = \dfrac{R}{M_{0}J}$$

This is the form used, for instance, to find $J$ for hydrogen from $S_{P} = 3400$ cal kg$^{-1}$ K$^{-1}$, $S_{V} = 2400$ cal kg$^{-1}$ K$^{-1}$, $R = 8300$ J kg$^{-1}$ K$^{-1}$ and $M_{0} = 2$, giving $J = 4.15$ J cal$^{-1}$.

Note also that $C_{P} > C_{V}$ always, because at constant pressure part of the heat supplied is spent as work in moving the piston.

Specific heats from the law of equipartition

(a) Monatomic gases. For a monatomic gas held at constant temperature $T$ and containing $N_{A}$ atoms, each atom has only 3 translational degrees of freedom, so the average energy per atom is $\dfrac{3}{2}k_{B}T$ and the total internal energy per mole is

$$E = \dfrac{3}{2}N_{A}k_{B}T$$

$$\therefore C_{V} = \dfrac{dE}{dT} = \dfrac{3}{2}N_{A}k_{B} = \dfrac{3}{2}R$$

Using Mayer's relation, $C_{P} = \dfrac{5}{2}R$, and

$$\gamma = \dfrac{C_{P}}{C_{V}} = \dfrac{5}{3}$$

(b) Diatomic gases, rigid rotator. For a gas of diatomic molecules such as O$_2$, N$_2$, CO, HCl, treated as a rigid rotator, each molecule has three translational and two rotational degrees of freedom, so the internal energy of one mole is

$$E = \dfrac{3}{2}N_{A}k_{B}T + \dfrac{2}{2}N_{A}k_{B}T = \dfrac{5}{2}N_{A}k_{B}T$$

$$\therefore C_{V} = \dfrac{5}{2}N_{A}k_{B} = \dfrac{5}{2}R, \qquad C_{P} = \dfrac{7}{2}R, \qquad \gamma = \dfrac{7}{5}$$

(c) Diatomic gases, non-rigid vibrating molecules. Vibrational motion contributes $k_{B}T$ per molecule, so the internal energy per mole is

$$E = \dfrac{3}{2}N_{A}k_{B}T + \dfrac{2}{2}N_{A}k_{B}T + \dfrac{2}{2}N_{A}k_{B}T = \dfrac{7}{2}N_{A}k_{B}T$$

$$\therefore C_{V} = \dfrac{7}{2}R, \qquad C_{P} = \dfrac{9}{2}R, \qquad \gamma = \dfrac{9}{7}$$

(d) Polyatomic gases. Gases whose molecules contain more than two atoms are polyatomic, for example ammonia, in which each molecule has one N atom and three H atoms. Each molecule has three translational degrees of freedom, each contributing $\dfrac{1}{2}k_{B}T$. Only linear molecules have two degrees of freedom for rotation; all other polyatomic molecules can rotate about three mutually perpendicular axes through their centre of mass and so have three rotational degrees of freedom, each again contributing $\dfrac{1}{2}k_{B}T$. Polyatomic molecules have more than one mode of vibrational motion; the number $f$ of vibrational degrees of freedom depends on the geometric structure of the molecule, that is, on the arrangement of its atoms. Each such mode contributes $2 \times \dfrac{1}{2}k_{B}T$ from its kinetic and potential terms. Hence for one mole,

$$E = \dfrac{3}{2}N_{A}k_{B}T + \dfrac{3}{2}N_{A}k_{B}T + f\,N_{A}k_{B}T = \left(3 + f\right)N_{A}k_{B}T$$

$$\therefore C_{V} = \left(3 + f\right)R, \qquad C_{P} = \left(4 + f\right)R, \qquad \gamma = \dfrac{4 + f}{3 + f}$$

Summary table

Working with a given difference and ratio

If the difference $C_{P} - C_{V}$ and the ratio $\gamma = C_{P}/C_{V}$ are given, write $C_{P} = \gamma C_{V}$ so that $\left(\gamma - 1\right)C_{V} = C_{P} - C_{V}$, giving

$$C_{V} = \dfrac{C_{P} - C_{V}}{\gamma - 1}, \qquad C_{P} = C_{V} + \left(C_{P} - C_{V}\right)$$

For example, with a difference of 9000 J kg$^{-1}$ K$^{-1}$ and $\gamma = 1.5$: $C_{V} = 9000/0.5 = 1.8 \times 10^{4}$ and $C_{P} = 2.7 \times 10^{4}$ J kg$^{-1}$ K$^{-1}$.

Standard results

$$C_{P} - C_{V} = R, \qquad C_{P} - C_{V} = \dfrac{R}{J}, \qquad S_{P} - S_{V} = \dfrac{R}{M_{0}J}$$

$$C_{V} = \dfrac{f}{2}R, \qquad C_{P} = \left(\dfrac{f}{2} + 1\right)R, \qquad \gamma = 1 + \dfrac{2}{f}$$

where $f$ is the number of quadratic terms per molecule.