19.7.1 $C_{P}$ and $C_{V}$

You may have noticed two curious features in Table 19.4. First, the molar specific heats of monatomic gases are all alike. And the molar specific heats of diatomic gases, while different from monatomic gases, are again very nearly alike. We saw a similar feature in Table 19.2 for the molar specific heats of solids. Second, the difference $C_{P} - C_{V} = 8.3 J / mol K$ is the same in every case. And, most puzzling of all, the value of $C_{P} - C_{V}$ appears to be equal to the universal gas constant $R$ ! Why should this be?

The relationship between $C_{V}$ and $C_{P}$ hinges on one crucial idea: $Δ E_{th},$ the change in the thermal energy of a gas, is the same for any two processes that have the same $Δ T$ . The thermal energy of a gas is associated with temperature, so any process that changes the gas temperature from $T_{i}$ to $T_{f}$ has the same $Δ E_{th}$ as any other process that goes from $T_{i}$ to $T_{f}$ . Furthermore, the first law $Δ E_{th} = Q + W$ tells us that a gas cannot distinguish between heat and work. The system’s thermal energy changes in response to energy added to or removed from the system, but the response of the gas is the same whether you heat the system, do work on the system, or do some combination of both. Thus any two processes that change the thermal energy of the gas by $Δ E_{t h}$ will cause the same temperature change $Δ T .$

With that in mind, look back at Figure 19.16. Both gas processes have the same $Δ T$ , so both have the same value of $Δ E_{th}$ . Process A is an isochoric process in which no work is done (the piston doesn’t move), so the first law for this process is

(Δ E_{th})_{A} = W + Q = 0 + Q_{const vol} = n C_{V} Δ T

(19.24)

Process B is an isobaric process. You learned earlier that the work done on the gas during an isobaric process is $W = - p Δ V$ . Thus

(Δ E_{th})_{B} = W + Q = - p Δ V + Q_{const press} = - p Δ V + n C_{P} Δ T

(19.25)

$(Δ E_{th})_{B} = (Δ E_{th})_{A}$ because both have the same $Δ T$ , so we can equate the right sides of Equations 19.24 and 19.25:

- p Δ V + n C_{P} Δ T = n C_{V} Δ T

(19.26)

For the final step, we can use the ideal-gas law $p V = n R T$ to relate $Δ V$ and $Δ T$ during process B. For any gas process,

Δ (p V) = Δ (n R T)

(19.27)

A photo of a jet engine. — Designing a jet engine requires a thorough understanding of thermodynamics. The *compressor* of a jet engine does work to adiabatically compress the incoming air, which substantially increases its pressure and temperature. Fuel is sprayed into this hot air and ignited, further increasing the temperature. The expanding hot gas is exhausted from the engine and, by Newton’s third law, generates thrust to push the plane forward. Part of the exhaust gas spins a *turbine* that runs the compressor at the front end and also generates the electric power needed onboard. The overall efficiency of a jet engine—the ratio of the work done by thrust to the chemical energy of the fuel—is about 35%.

For a constant-pressure process, where $p$ is constant, Equation 19.27 becomes

p Δ V = n R Δ T

(19.28)

Substituting this expression for $p Δ V$ into Equation 19.26 gives

- n R Δ T + n C_{P} Δ T = n C_{V} Δ T

(19.29)

The $n Δ T$ cancels, and we are left with

C_{P} = C_{V} + R

(19.30)

This result, which applies to all ideal gases, is exactly what we see in the data of Table 19.4.

But that’s not the only conclusion we can draw. Equation 19.24 found that $Δ E_{th} = n C_{V} Δ T$ for a constant-volume process. But we had just noted that $Δ E_{th}$ is the same for all gas processes that have the same $Δ T$ . Consequently, this expression for $Δ E_{th}$ is equally true for any other process. That is

Δ E_{th} = n C_{V} Δ T (any ideal-gas process)

(19.31)

Compare this result to Equations 19.23. We first made a distinction between constant-volume and constant-pressure processes, but now we’re saying that Equation 19.31 is true for any process. Are we contradicting ourselves? No, the difference lies in what you need to calculate.

The change in thermal energy when the temperature changes by $Δ T$ is the same for any process. That’s Equation 19.31.
The heat required to bring about the temperature change depends on what the process is. That’s Equations 19.23. An isobaric process requires more heat than an isochoric process that produces the same $Δ T$ .

The reason for the difference is seen by writing the first law as $Q = Δ E_{th} - W$ . In an isochoric process, where $W = 0$ , all the heat input is used to increase the gas temperature. But in an isobaric process, some of the energy that enters the system as heat then leaves the system as work $(W < 0)$ done by the expanding gas. Thus more heat is needed to produce the same $Δ T$ .