19.7 The Specific Heats of Gases

Specific heats are given in Table 19.2 for solids and liquids. Gases are harder to characterize because the heat required to cause a specified temperature change depends on the process by which the gas changes state.

FIGURE 19.16 shows two isotherms on the pV diagram for a gas. Processes A and B, which start on the $T_{i}$ isotherm and end on the $T_{f}$ isotherm, have the same temperature change $Δ T = T_{f} - T_{i}$ . But process A, which takes place at constant volume, requires a different amount of heat than does process B, which occurs at constant pressure. The reason is that work is done in process B but not in process A. This is a situation that we are now equipped to analyze.

The figure illustrates a pressure versus volume graph.

FIGURE 19.16 Full Alternative Text

It is useful to define two different versions of the specific heat of gases, one for constant-volume (isochoric) processes and one for constant-pressure (isobaric) processes. We will define these as molar specific heats because we usually do gas calculations using moles instead of mass. The quantity of heat needed to change the temperature of $n$ moles of gas by $Δ T$ is

\begin{array}{l} Q = n C_{V} Δ T (temperature change at constant volume) \\ Q = n C_{P} Δ T (temperature change at constant pressure) \end{array}

(19.23)

where $C_{V}$ is the molar specific heat at constant volume and $C_{P}$ is the molar specific heat at constant pressure. TABLE 19.4 gives the values of $C_{V}$ and $C_{P}$ for a few common monatomic and diatomic gases. The units are $J / mol K$ . Molar specific heats do vary somewhat with temperature—we’ll look at an example in the next chapter—but the values in the table are adequate for temperatures from 200 K to 800 K.

TABLE 19.4 Molar specific heats of gases $(J / mol K)$ at $T = 0 ° C$

Gas	$C_{P}$	$C_{V}$	$C_{P}$ – $C_{V}$
Monatomic Gases
He	20.8	12.5	8.3
Ne	20.8	12.5	8.3
Ar	20.8	12.5	8.3
Diatomic Gases
$H_{2}$	28.7	20.4	8.3
$N_{2}$	29.1	20.8	8.3
$O_{2}$	29.2	20.9	8.3

NOTE Equations 19.23 apply to two specific ideal-gas processes. In a general gas process, for which neither $p$ nor $V$ is constant, we have no direct way to relate $Q$ to $Δ T$ . In that case, the heat must be found indirectly from the first law as $Q = Δ E_{th} - W$ .

EXAMPLE 19.7 Heating and cooling a gas

Three moles of $O_{2}$ gas are at $20.0 ° C$ . 600 J of heat energy are transferred to the gas at constant pressure, then 600 J are removed at constant volume. What is the final temperature? Show the process on a pV diagram.

MODEL $O_{2}$ is a diatomic ideal gas. The gas is heated as an isobaric process, then cooled as an isochoric process.

SOLVE The heat transferred during the constant-pressure process causes a temperature rise

Δ T = T_{2} - T_{1} = \frac{Q}{n C_{P}} = \frac{600 J}{(3.0 mol) (29.2 J / mol K)} = 6.8 ° C

where $C_{P}$ for oxygen was taken from Table 19.4. Heating leaves the gas at temperature $T_{2} = T_{1} + Δ T = 26.8 ° C$ . The temperature then falls as heat is removed during the constant-volume process:

Δ T = T_{3} - T_{2} = \frac{Q}{n C_{V}} = \frac{- 600 J}{(3.0 mol) (20.9 J / mol K)} = - 9.5 ° C

We used a negative value for $Q$ because heat energy is transferred from the gas to the environment. The final temperature of the gas is $T_{3} = T_{2} + Δ T = 17.3 ° C$ . FIGURE 19.17 shows the process on a pV diagram. The gas expands (moves horizontally on the diagram) as heat is added, then cools at constant volume (moves vertically on the diagram) as heat is removed.

FIGURE 19.17 Full Alternative Text

REVIEW The final temperature is lower than the initial temperature because $C_{P} > C_{V}$ .

EXAMPLE 19.8 Calorimetry with a gas and a solid

The interior volume of a 200 g hollow aluminum box is $800 {cm}^{3}$ . The box contains nitrogen gas at STP. A $20 {cm}^{3}$ block of copper at a temperature of $300 ° C$ is placed inside the box, then the box is sealed. What is the final temperature?

MODEL This example has three interacting systems: the aluminum box, the nitrogen gas, and the copper block. They must all come to a common final temperature $T_{f}$ .

VISUALIZE The box and gas have the same initial temperature: $T_{Al} = T_{N 2} = 0 ° C$ . The box doesn’t change size, so this is a constant-volume process. The final temperature is unknown.

SOLVE Although one of the systems is now a gas, the calorimetry equation $Q_{net} = Q_{Al} + Q_{N 2} + Q_{Cu} = 0$ is still appropriate. In this case,

\begin{array}{l} Q_{net} = & m_{Al} c_{Al} (T_{f} - T_{Al}) + n_{N 2} C_{V} (T_{f} - T_{N 2}) \\ + m_{Cu} c_{Cu} (T_{f} - T_{Cu}) = 0 \end{array}

Notice that we used masses and specific heats for the solids but moles and the molar specific heat for the gas. We used $C_{V}$ because this is a constant-volume process. Solving for $T_{f}$ gives

T_{f} = \frac{m_{Al} c_{Al} T_{Al} + n_{N 2} C_{V} T_{N 2} + m_{Cu} c_{Cu} T_{Cu}}{m_{Al} c_{Al} + n_{N 2} C_{V} + m_{Cu} c_{Cu}}

The specific heat values are found in Tables 19.2 and 19.4. The mass of the copper is

m_{Cu} = ρ_{C u} V_{Cu} = (8920 kg / {cm}^{3}) (20 \times 10^{- 6} m^{3}) = 0.178 kg

The number of moles of the gas is found from the ideal-gas law, using the initial conditions. Notice that inserting the copper block displaces $20 {cm}^{3}$ of gas; hence the gas volume is only $V = 780 {cm}^{3} = 7.80 \times 10^{- 4} m^{3}$ . Thus

n_{N 2} = \frac{p V}{R T} = 0.0348 mol

Computing the final temperature gives $T_{f} = 83 ° C$ .

19.7 The Specific Heats of Gases

FIGURE 19.16 Processes $A$ and $B$ have the same $Δ T$ and the same $Δ E_{th}$ , but they require different amounts of heat.

TABLE 19.4 Molar specific heats of gases $(J / mol K)$ at $T = 0 ° C$

EXAMPLE 19.7 Heating and cooling a gas

FIGURE 19.17 The pV diagram for Example 19.7.

EXAMPLE 19.8 Calorimetry with a gas and a solid