The Molar Specific Heats of an Ideal Gas

In this section, we want to derive from molecular considerations an expression for the internal energy E_int of an ideal gas. In other words, we want an expression for the energy associated with the random motions of the atoms or molecules in the gas. We shall then use that expression to derive the molar specific heats of an ideal gas.

Internal Energy E_int

Let us first assume that our ideal gas is a monatomic gas (which has individual atoms rather than molecules), such as helium, neon, or argon. Let us also assume that the internal energy E_int of our ideal gas is simply the sum of the translational kinetic energies of its atoms. (As explained by quantum theory, individual atoms do not have rotational kinetic energy.)

The average translational kinetic energy of a single atom depends only on the gas temperature and is given by Eq. 19-24 as . A sample of n moles of such a gas contains nN_A atoms. The internal energy E_int of the sample is then

Using Eq. 19-7 (k = R/N_A), we can rewrite this as

Thus,

 The internal energy E_int of an ideal gas is a function of the gas temperature only; it does not depend on any other variable.

With Eq. 19-38 in hand, we are now able to derive an expression for the molar specific heat of an ideal gas. Actually, we shall derive two expressions. One is for the case in which the volume of the gas remains constant as energy is transferred to or from it as heat. The other is for the case in which the pressure of the gas remains constant as energy is transferred to or from it as heat. The symbols for these two molar specific heats are C_V and C_p, respectively. (By convention, the capital letter C is used in both cases, even though C_V and C_p represent types of specific heat and not heat capacities.)

Fig. 19-8 (a) The temperature of an ideal gas is raised from T to T + ΔT in a constant-volume process. Heat is added, but no work is done. (b) The process on a p–V diagram.

Molar Specific Heat at Constant Volume

Figure 19-8a shows n moles of an ideal gas at pressure p and temperature T, confined to a cylinder of fixed volume V. This initial state i of the gas is marked on the p–V diagram of Fig. 19-8b. Suppose now that you add a small amount of energy to the gas as heat Q by slowly turning up the temperature of the thermal reservoir. The gas temperature rises a small amount to T + ΔT, and its pressure rises to p + Δp, bringing the gas to final state f.

In such experiments, we would find that the heat Q is related to the temperature change ΔT by

where C_V is a constant called the molar specific heat at constant volume. Substituting this expression for Q into the first law of thermodynamics as given by Eq. 18-26 (ΔE_int = Q − W) yields

With the volume held constant, the gas cannot expand and thus cannot do any work. Therefore, W = 0, and Eq. 19-40 gives us

From Eq. 19-38 we know that , so the change in internal energy must be

Substituting this result into Eq. 19-41 yields

As Table 19-2 shows, this prediction of the kinetic theory (for ideal gases) agrees very well with experiment for real monatomic gases, the case that we have assumed. The (predicted and) experimental values of C_V for diatomic gases (which have molecules with two atoms) and polyatomic gases (which have molecules with more than two atoms) are greater than those for monatomic gases for reasons that will be suggested in Section 19-9.

We can now generalize Eq. 19-38 for the internal energy of any ideal gas by substituting C_v for  we get

This equation applies not only to an ideal monatomic gas but also to diatomic and polyatomic ideal gases, provided the appropriate value of C_V is used. Just as with Eq. 19-38, we see that the internal energy of a gas depends on the temperature of the gas but not on its pressure or density.

When an ideal gas that is confined to a container undergoes a temperature change ΔT, then from either Eq. 19-41 or Eq. 19-44 we can write the resulting change in its internal energy as

This equation tells us:

 A change in the internal energy E_int of a confined ideal gas depends on the change in the gas temperature only; it does not depend on what type of process produces the change in the temperature.

As examples, consider the three paths between the two isotherms in the p–V diagram of Fig. 19-9. Path 1 represents a constant-volume process. Path 2 represents a constant-pressure process (that we are about to examine). Path 3 represents a process in which no heat is exchanged with the system’s environment (we discuss this in Section 19-11). Although the values of heat Q and work W associated with these three paths differ, as do p_f and V_f, the values of Δ E_int associated with the three paths are identical and are all given by Eq. 19-45, because they all involve the same temperature change ΔT. Therefore, no matter what path is actually taken between T and T + ΔT, we can always use path 1 and Eq. 19-45 to compute Δ E_int easily.

Fig. 19-9 Three paths representing three different processes that take an ideal gas from an initial state i at temperature T to some final state f at temperature T + ΔT. The change ΔE_int in the internal energy of the gas is the same for these three processes and for any others that result in the same change of temperature.

Molar Specific Heat at Constant Pressure

We now assume that the temperature of our ideal gas is increased by the same small amount ΔT as previously but now the necessary energy (heat Q) is added with the gas under constant pressure. An experiment for doing this is shown in Fig. 19-10a; the p–V diagram for the process is plotted in Fig. 19-10b. From such experiments we find that the heat Q is related to the temperature change ΔT by

where C_p is a constant called the molar specific heat at constant pressure. This C_p is greater than the molar specific heat at constant volume C_V, because energy must now be supplied not only to raise the temperature of the gas but also for the gas to do work—that is, to lift the weighted piston of Fig. 19-10a.

Fig. 19-10 (a) The temperature of an ideal gas is raised from T to T + ΔT in a constant-pressure process. Heat is added and work is done in lifting the loaded piston. (b) The process on a p–V diagram. The work p ΔV is given by the shaded area.

To relate molar specific heats C_p and C_V, we start with the first law of thermodynamics (Eq. 19-26):

We next replace each term in Eq. 19-47. For ΔE_int, we substitute from Eq. 19-45. For Q, we substitute from Eq. 19-46. To replace W, we first note that since the pressure remains constant, Eq. 19-16 tells us that W = p ΔV. Then we note that, using the ideal gas equation (pV = nRT), we can write

Making these substitutions in Eq. 19-47 and then dividing through by n ΔT, we find

C_V = C_p − R

and then

This prediction of kinetic theory agrees well with experiment, not only for monatomic gases but for gases in general, as long as their density is low enough so that we may treat them as ideal.

CHECKPOINT 4    The figure here shows five paths traversed by a gas on a p–V diagram. Rank the paths according to the change in internal energy of the gas, greatest first.

Sample Problem 19-7

A bubble of 5.00 mol of helium is submerged at a certain depth in liquid water when the water (and thus the helium) undergoes a temperature increase ΔT of 20.0 C° at constant pressure. As a result, the bubble expands. The helium is monatomic and ideal.

(a) How much energy is added to the helium as heat during the increase and expansion?

Solution: One Key Idea here is that the heat Q is related to the temperature change ΔT by a molar specific heat of the gas. Because the pressure p is held constant during the addition of energy, we use the molar specific heat at constant pressure C_p and Eq. 19-46,

to find Q. To evaluate C_p we go to Eq. 19-49, which tells us that for any ideal gas, C_p = C_V + R. Then from Eq. 19-43, we know that for any monatomic gas (like the helium here), . Thus, Eq. 19-50 gives us

(b) What is the change ΔE_int in the internal energy of the helium during the temperature increase?

Solution: Because the bubble expands, this is not a constant-volume process. However, the helium is nonetheless confined (to the bubble). Thus, a Key Idea here is that the change ΔE_int is the same as would occur in a constant-volume process with the same temperature change ΔT. We can easily find the constant-volume change ΔE_int with Eq. 19-45:

(c) How much work W is done by the helium as it expands against the pressure of the surrounding water during the temperature increase?

Solution: One Key Idea here is that the work done by any gas expanding against the pressure from its environment is given by Eq. 19-11, which tells us to integrate p dV. When the pressure is constant (as here), we can simplify that to W = p ΔV. When the gas is ideal (as here), we can use the ideal gas law (Eq. 19-5) to write p ΔV = nR ΔT. We end up with

Because we happen to know Q and ΔE_int, we can work this problem another way: The Key Idea now is that we can account for the energy changes of the gas with the first law of thermodynamics, writing

Note that during the temperature increase, only a portion (1250 J) of the energy (2080 J) that is transferred to the helium as heat goes to increasing the internal energy of the helium and thus the temperature of the helium. The rest (831 J) is transferred out of the helium as work that the helium does during the expansion. If the water were frozen, it would not allow that expansion. Then the same temperature increase of 20.0 C° would require only 1250 J of heat, because no work would be done by the helium.