Is it natural to answer that \(\Delta U = C_P\Delta T\), where \(C_P\) is the constant-pressure heat capacity? This answer certainly is natural for the student taking an introductory thermodynamics class, and it might be long afterwards. Such new practitioners of thermodynamics have likely learned in lectures and from their textbook(s) that closed systems can be manipulated in several possible ways:

• One might heat a system—such as an ideal gas—in a closed container (i.e., at constant volume) by exposing it to a succession of external heat reservoirs at increasing temperature or equivalently by applying a heat flux.
• Alternatively, one might add or extract energy from an ideal gas within a container (with a piston at one end) by slowly moving the piston to maintain a constant pressure within the container at some constant value. (In the simplest example, we simply maintain atmospheric pressure.)
• (Situations in which both the volume and the pressure change are of course also possible but may be less amenable to introductory analysis.)

Having been taught this framework, it’s natural for new practitioners to conclude that if certain variables are associated with constant-volume or constant-pressure conditions (as the heat capacities are), then one can apply the property with the corresponding subscript with impunity. Spoiler alert: they cannot ☹

The cruel part: Experienced thermodynamics practitioners will recognize immediately that \(\Delta U = C_V\Delta T\) for an ideal gas for any process.

Is this a cruel joke on all new thermodynamics students? I think it is—although certainly not deliberate—and I support this claim with evidence of how unexpected and frustrating this realization is to students by noting the many online posts with essentially the same question: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.

Repeatedly, new thermodynamics students optimistically internalize their new familiarity with constant-volume and constant-pressure processes only to be betrayed by the fact that even during a constant-pressure process (in fact, any process), the coupling between the energy and temperature of an ideal gas is the constant-volume heat capacity.

Here, I survey the following progressive approaches for understanding why this misunderstanding occurs and how to avoid it:

• Explanation 1: The heat capacity never coupled the internal energy to the temperature for arbitrary conditions.

• Explanation 2: Forget about any processes; the ideal gas is unusual in having one of its equations of state be \(U=CT+U_0\), where \(C\) is a constant.

• Explanation 3: For a single-phase closed system, we should expect state variables to depend on two variables. The ideal gas is unusual in that its internal energy always depends on only one.

• Explanation 4: The cruelty of \(dU=C_V\,dT\) for the ideal gas lies in its simplicity: The only material property that doesn't cancel out of the general relation unfortunately has “constant-volume” in its name.

I hope that at least one of the lines of reasoning below makes intuitive sense and provides a strategy to avoid the same sense of betrayal from other equations.

Explanation 1: The heat capacity never coupled the internal energy to the temperature for arbitrary conditions.

The equation \(dU=C_V\,dT\) might tempt us to think that \(C_V\) or \(C\) in general is some type of universal coupling constant between \(dU\) and \(dT\). Not true! Not true at all. What the heat capacity always does is to couple incoming heat transfer to an increase in temperature (when all we're doing is heating a system, under any conditions).

As a quick review, in equation form, recall that we have \(\Delta U=Q +W\) (i.e., the internal energy of a closed system increases via incoming heat \(Q\) and work \(W\)). In differential form, we have \(dU=T\,dS-P\,dV\), where we note that the conjugate variables of heat are the (intensive parameter) temperature \(T\) and the (extensive parameter) entropy \(S\); this pair implies that incoming heat transfer arises from a temperature difference that drives a transfer of entropy. On the work side, we’ve restricted outselves to pressure–volume work for simplicity; this particular type of work consists of a pressure difference that drives a decrease in volume, which is where the minus sign comes from.

Then, the heat capacity under a certain condition \(X\) can be defined as \(C_X=Q/\Delta T\) for small changes or, in differential form, \(C_X=T(\partial S/\partial T)_X\).

Now, if one mistakenly considers the heat capacity to always couple internal energy with temperature, then one might see the equation $$dU=C_V\,dT\;\mathrm{(under\; the\; condition\; of\; constant\; volume)}$$ and conclude incorrectly that the following analogous relationship holds by replacing \(V\) with \(P\): $$\color{red}{\mathrm{(WRONG!)}}\quad dU=C_P\,dT\;\mathrm{(under\; the\; condition\; of\; constant\; pressure)}\quad\color{red}{\mathrm{(WRONG!)}}.$$ Instead, the constant-volume heat capacity \(C_V=T(\partial S/\partial T)_V\) equals \(\left(\partial U/\partial T\right)_V\) only because \(dU = T\,dS-P\,dV\) for a simple closed system simplifies to \(dU=T\,dS\) when the volume is held constant. The appropriate replacement equation is actually $$dH=C_P\,dT\;\mathrm{(under\; the\; condition\; of\; constant\; pressure)},$$ where \(H=U+PV\) is the enthalpy, which is the internal energy of a system—the energy required to create it—plus the work required to move the atmosphere out of the way. (I discuss how \(H\), \(U\), \(C_P\), and \(C_V\) all fit together in the context of an ideal gas in the next section.) Recognizing and understanding why the equation marked as WRONG! above is wrong is a source of major frustration for the new thermodynamics practitioner.

A well-recognized problematic aspect of thermodynamics is that multiple distinct parameters all have units of energy. In “Student understanding of the first law of thermodynamics: Relating work to the adiabatic compression of an ideal gas,” a study of the understanding of themodynamic concepts by undergrads (Am J Phys 70(2) (2002)), Loverude et al. noted that

Many students seemed to confuse quantities associated with processes (such as heat transfer and work) with those associated with states (such as the temperature and internal energy). For some this confusion was reflected in a tendency to refer to the “change in heat” or the “change in work.” … [T]he inability to distinguish between two seemingly similar concepts was often at a sufficiently deep level that it precluded students from correctly applying the first law. … We found that many students had…a tendency to confuse heat, temperature, and internal energy. Some students admitted confusion on this point, even wondering why both \(\Delta U\) and \(Q\) are present in the algebraic form of the first law, because both seem to describe the “heat in an object.” … Although some students treated heat transfer as a process that can produce a change in internal energy, many others treated heat as essentially the same thing as the internal energy.

If you don’t understand the difference between \(\Delta U\) (a system’s increase in internal energy) and \(Q\) (the heat transfer into a system), then the cruelest equation will continue to challenge you.

The strength of this explanation moving forward is that we don’t need to know anything about any specific material system; we just need to know how the heat capacity is defined to know not to switch its subscript inappropriately. The weakness is that we still don’t know how to treat the ideal gas.

Explanation 2: Forget about any processes; the ideal gas is unusual in having one of its equations of state be \(U=CT+U_0\), where \(C\) is a constant.

Once you understand that the equation of state \(U\propto T\) always applies to an ideal gas, it’s straightforward to see that the type of any applied process is immaterial. (Here, I’m referring to the calorically perfect gas, with constant heat capacity. In addition, the presence of \(U_0\) is not strictly necessary—we could define \(U_0=0\), as I do for the rest of this note—but is shown to remind us that energy can be defined only relative to a reference state.) Of course, \(C\) turns out to be \(C_V\), but in this line of reasoning, the subscript is unimportant.

The essential consequence of having an equation of state of this form is that the internal energy of the ideal gas depends only on temperature. The ideal gas is a unique state of matter in which the energy is stored only as the kinetic energy of molecules interacting solely through elastic collisions. Under this idealization, it simply doesn’t matter whether we compress the system (as long as we maintain a constant temperature). Given that this equation of state applies at any single temperature, processes—and whether an associated process is conducted at constant volume or constant pressure—are irrelevant.

As introduced earlier, the equation \(dH=d(U+PV)=C_P\,dT\) works because the other equation of state for an ideal gas is \(PV=nRT\) (where \(n\) is the number of moles of gas and \(R\) is the universal gas constant). Under this condition, for the enthalpy \(H=U+nRT\), the differential form (for a closed system) is \(dH=dU+nR\,dT\), and thus \(dU=(C_P-nR)dT=C_V\,dT\) even for a constant-pressure process.

The advantage of this explanation is the distinction it provides between an equation of state (e.g., \(U\propto T\) for an ideal gas at any \(T\)) and a differential relationship (e.g., \(dU\propto dT\) under a certain condition and over a certain range of \(T\)); the cruelest equation arises when the finite-difference or differential form of the former equation is confused with the latter equation. This explanation also incorporates information specific to the ideal gas. However, there’s at least one more insight available in thermodynamics.

Explanation 3: For a single-phase closed system, we should expect state variables to depend on two variables. The ideal gas is unusual in that its internal energy always depends on only one.

Consider our closed system of a single component with only \(P\,dV\) work (i.e., no electrical or magnetic work and no strain energy other than that applied by compressive equitriaxial stress, also known as hydrostatic stress, also known as pressure). To express \(U\), \(T\), \(S\), \(P\), or \(V\) or some combination of these, we’ve got to pick two variables. We can remember this number by referring back to \(dU=T\,dS-P\,dV\), in which \(dU\) is expressed in terms of two conjugate pairs (each with an intensive variable, \(T\) and \(P\), and an extensive variable, \(S\) and \(V\); the symmetry is broken only by the minus sign that arises because a compressive stress tends to reduce the volume of the system it acts on). Never forget, though, that this equation is just another way to write $$dU=\left(\frac{\partial U}{\partial S}\right)_V\,dS+\left(\frac{\partial U}{\partial V}\right)_S\,dV.$$ We just happen to know \((\partial U/\partial S)_V\) and \(-(\partial U/\partial V)_S\) by their more common names, the temperature \(T\) and pressure \(P\), respectively.

But we could have more generally written $$dU=\left(\frac{\partial U}{\partial T}\right)_X\,dT+\left(\frac{\partial U}{\partial X}\right)_T\,dX$$ where \(X\) is either \(V\), \(P\), or \(S\). And once we understand that the absence of a \(dV\) or \(dP\) or \(dS\) term in such an expression \(dU=C\,dT\) is anomalous, then we can move forward to understanding that the ideal gas is an idealized state of matter in which the coefficients of the \(dV\), \(dP\), and \(dS\) terms in such an expression are all zero. Therefore, it doesn’t matter at all what parameter we’re holding constant in any process; the temperature \(T\) is the only arbiter of whether the internal energy of an ideal gas changes.

The advantage of this explanation is that it reminds us that state variables depend on a certain number of independent variables and that something unusual is happening when one of the dependencies disappears. Under this view, \(\Delta U=C_V\Delta T\) is useful as a simplified equation in an introductory class but ultimately misleads students by increasing the mystery about which parameters depend on which other parameters in the more general case.

Having fun with it

Thermodynamics is arguably the hardest subject we ask engineering students to comprehend, in part because the state variables of temperature and entropy have no general meaning at the single-particle level; they are so-called ensemble parameters that describe not energy levels but distributions of energy levels. (It’s only in the case of the ideal gas that the total energy straightforwardly scales linearly with temperature.) In addition, we have the fact that heat and work along with energy, enthalpy, and the various other potentials share the same units but are crucially different. We first encounter the concept that material properties are second derivatives of energy; for example, the bulk thermal expansion coefficient is defined as \(\alpha=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P=\frac{1}{V}\left(\frac{\partial^2 G}{\partial P\partial T}\right)_{T,P}\), where \(G=U-TS+PV\) is the Gibbs free energy. And then there’s entropy, which one could spend a lifetime understanding.

In graduate thermodynamics, when I remarked on this challenge, the response I got was, “John, you’ve just got to play around and have fun with it,” which was the last thing I wanted to hear. But if we consider what’s outlined in the review above, we’ll find that we can play around with it, at least by defining the internal energy \(U\) (or the enthalpy \(H=U+PV\)) in differential terms for a variety of state functions and by obtaining results systematically. Here, I show one example, and I provide the results for other cases as a work-checking resource for manipulating thermodynamic expressions.

Consider, for example, if we wish to express \(dU\) in terms of material properties and changes in other state variables (say, \(T\) and \(V\) to start; I summarize other possible examples below). In this derivation, I’ll note each technique and tool applied to obtain the final expression.

As noted above, for a closed system with only simple \(P\,dV\) work, we have two independent variables; therefore, we can write $$dU=\left(\frac{\partial U}{\partial T}\right)_V\,dT+\left(\frac{\partial U}{\partial V}\right)_T\,dV.$$ Now apply the partial derivatives to \(dU=T\,dS-P\,dV\), dropping any terms with two differentials as negligible and noting that the derivative of any constant term is zero and that the derivative of any parameter with respect to itself is unity: $$dU=\left(T\left. \frac{\partial S}{\partial T}\right|_V\right)dT+ \left(T\left. \frac{\partial S}{\partial V}\right|_T-P\right)dV.$$ We immediately recognize the left-hand term as \(C_V\,dT\). The right-hand term can be refined by applying a Maxwell relation: since \(S=(\partial A/\partial T)_V\), where \(A\) is the Helmholz energy \(A=U-TS\) (implying that \(dA=-S\,dT-P\,dV\)), let’s differentiate \(S\) again and—since it doesn’t matter in which order partial derivatives are applied—we have $$\left(\frac{\partial S}{\partial V}\right)_T=-\left(\frac{\partial^2 A}{\partial T\partial V}\right)_{V,T}=-\left(\frac{\partial^2 A}{\partial V\partial T}\right)_{T,V}=\left(\frac{\partial P}{\partial T}\right)_V.$$ This form still isn’t suitably convenient, so let’s apply the triple product rule to turn \((\partial P/\partial T)_V\) into \(-(\partial P/\partial V)_T(\partial V/\partial T)_P\). Now, \(-(\partial P/\partial V)_T\) is simply the isothermal bulk modulus \(K\) divided by \(V\), and \((\partial V/\partial T)_P\) is simply the isobaric volumetric thermal expansion coefficient \(\alpha\) multiplied by \(V\). Therefore, the full expression for \(dU\) in terms of \(T\) and \(V\), conveniently expressed in terms of material properties (for any material), is $$dU=C_V\,dT+(\alpha T K-P)\,dV.$$ Furthermore, for the special case of an ideal gas, one can quickly verify using \(PV=nRT\) that \(K=P\) and \(\alpha=1/T\) for an ideal gas and therefore that the coefficient of the right-hand term disappears. Insight!

We’re familiar with the idea that \(dV\) is zero for a constant-volume process. But for an ideal gas, it doesn’t matter if \(dV\) is zero; the associated coefficient is always zero.

I suggest that if you derive at least one of the related examples in the next section (e.g., \(dU\) in terms of \(dT\) and \(dS\)), then you’ll come much closer to understanding how to use (and not to misuse or misinterpret) the relationship \(dU=C_V\,dT\).

Expressing anything in terms of anything

Energized by this insight, let’s continue having fun. For example, what if we want a more general equation relating \(dU\) (or \(dH\)) and \(dT\) for any material? If we wish to express the internal energy in terms of the constant-volume heat capacity as dependent on changes in volume (and temperature), we have, as just derived, \begin{align}dU&=\left(T\left. \frac{\partial S}{\partial T}\right|_V\right)dT+\left(T\left.\frac{\partial P}{\partial T}\right|_V-P\right)dV\\ &=C_V\,dT+\left(\alpha TK-P\right)dV.\end{align}

How about the internal energy and constant-pressure heat capacity for changes in volume? That’s just $$dU=\left(C_P-\alpha^2 TV K\right)dT-\left(\alpha TK-P\right)dV$$ where we’ve used the classic result that \(C_P-C_V=\alpha^2 TV K\), the derivation of which is an excellent exercise.

Internal energy for changes in pressure? That’s $$dU=\left(C_V+\alpha^2 TVK-PV\alpha\right)dT-V\left(\frac{P}{K}-\alpha T\right)dP.$$

And of course we can express the internal energy in terms of changes in temperature and entropy. That’s $$dU=\left(\frac{PC_V}{T\alpha K}\right)dT+\left(T-\frac{P}{\alpha K}\right)dS.$$

For enthalpy and changes in pressure (and temperature), we have $$dH=C_P\,dT+V\left(1-\alpha T\right)dP.$$ Note how at constant pressure, we obtain the relatively simple \(dH=C_P\,dT\) for any material at constant pressure. This finding doesn’t imply that \(dH=C_V\,dT\) at constant volume any more than the incorrect deduction marked as WRONG! above, again because the heat capacity never coupled \(dH\) to anything; again, it couples the input heat transfer (i.e., \(T\,dS\)) to a rise in system temperature. Note also that the coefficient of the \(dP\) term is zero only for an ideal gas; thus, \(dH=C_P\,dT\) holds for an ideal gas for any process or even for no process: at any state, \(H=C_PT\) for an ideal gas (i.e., \(H=C_PT+H_0\) with the reference enthalpy \(H_0\) set to zero).

For enthalpy and changes in volume, we have $$dH=\left(C_V+VK\alpha\right)dT+K\left(\alpha T-1\right)dV.$$

And for changes in entropy, we have $$dH=\left(\frac{C_P}{\alpha T}\right)dT+\left(T-1/\alpha\right)dS.$$

Did you notice how the coefficients of all terms other than \(dT\) in this section are zero for an ideal gas? Let’s sum this up as follows.

Explanation 4: Few people are confused by the general \(dU\) equation of state’s use of the constant-volume heat capacity \(C_V\), the constant-temperature bulk modulus \(K\), and the constant-pressure thermal expansion coefficient \(\alpha\); it’s clear that these are just material properties. But the simplified version for the ideal gas, \(dU=C_VdT\), is too simple for its own good, causing new learners to conflate the name of the constant-volume heat capacity with a constraint that acts on the entire equation.

Derive all of the above equations on your own and you’ll be in good shape in terms of your familiarity with the chain rule, the triple product rule, and Maxwell relations.

Before I present a final twist, what have the authors of well-known thermodynamics textbooks done, and what are the best practices for teaching the new practitioners?

When I took graduate thermodynamics, we used Callen’s Thermodynamics and an Introduction to Thermostatistics (1985) supplemented with Zemansky and Dittman’s Heat and Thermodynamics (1997) to get up to speed. I thought I’d survey how various authors of thermodynamics textbooks addressed the point raised in this note. (I also looked back at my recitation notes from MIT 3.20, Materials at Equilibrium and am happy to note that—at least at the time—the equation \(\Delta U=C_V\Delta T\) was carefully derived with a dedicated discussion of why \(C_V\) is suitable for any process involving an ideal gas. Similar recitation notes are here.)

• Callen simply states that an equation of state of an ideal gas is \(U\propto T\) without getting into any parameters defined in terms of constant-volume or constant-pressure processes. However, the practitioner who later notices that the mediating parameter is \(C_V\) may still be confused.

• Zemansky and Dittman take care to caution that

  The term “heat capacity” implies that a substance can hold heat, which is completely false. Heat is not a function of the thermodynamic state of a system; internal energy is!

  However, their statement that

  ...when systematic experiments were performed to measure the capability of a substance to store internal energy, heat rather than work was used, and the results came to be known as the heat capacity of the sample. [boldface emphasis mine]

  is somewhat misleading; as I note above, the heat capacity is fundamentally defined as \(T(\partial S/\partial T)_X\) at some condition of constant \(X\), which is not equivalent to \((dU/dT)_X\) unless \(X\) is the volume for a closed system. In fact, a gas may store energy when we heat it at constant pressure, but plenty of energy is not being stored but rather doing compressive work on the rest of the atmosphere (or whatever is maintaining the constant pressure).

  To be sure, Zemansky and Dittman subsequently rigorously define the heat capacity as the heat transfer to achieve a certain temperature difference. However, the adjacent mention of the internal energy when explaining the heat capacity contributes somewhat to \(dU=C_VdT\) being the cruelest equation for beginner thermodynamics practitioners.

• In Thermodynamics, an Advanced Treatment (1967), Guggenheim introduces entropy before the heat capacity and begins by defining \(C_P=T(\partial S/\partial T)_P\). Presenting the constant-pressure case first lets one define \(C_V=T(\partial S/\partial T)_V\) and only then show that this is \((\partial U/\partial T)_V\); thus, the introduction of the internal energy can be shown to be a special feature (i.e., one applicable only under the special circumstance of \(dV=0\)), not an automatic one. Nice!

  Unfortunately, as its title suggests, Guggenheim’s book is far from introductory; for example, he never considers the case of an ideal gas, instead starting his discussion with a “slightly imperfect” gas, i.e., one that follows the equation of state \(PV/RT=1+B_2/V\), where an ideal gas would have \(B_2=0\).

• Kubo, in his Thermodynamics (1976), discusses the equations of state of an ideal gas, noting that

  The internal gas is independent of the volume…The form of \(U(T)\) is determined empirically or theoretically by statistical mechanics. In the normal range of temperature, \(U=nC_VT\) is a good approximation for more pure gases…

  This description is straightforward but sidesteps somewhat the question of why a constant-volume material property ends up being used in any arbitrary process.

• Fermi, in his Thermodynamics (1956), discusses the constant-volume and constant-pressure heat capacities and then turns to the case of the ideal gas, reviewing experiments indicating that the internal energy of such an idealized state depends only on its temperature. Notably, Fermi comments

  Since \(U\) depends only on \(T\), it is not necessary to specify that the volume is to be kept constant…for an ideal gas, we may write \(C_V=dU/dT\). [boldface emphasis mine]

  This is the clear statement we’ve been looking for that may avert confusion in the attentive reader. Bravo, Fermi!

Does your textbook take a particularly clear or a particularly misleading approach to analyzing the ideal energy of an ideal gas? Send the information to me, and I’ll include it in this survey.

So: what are the best practices for professors, lecturers, recitation leaders, and tutors?

• Emphasize the difference between a change in internal energy \(\Delta U\) and the incoming heat transfer \(Q\). Emphasize that multiple parameters exist that have the same units but that are not identical. (This point addresses Loverude et al.’s finding that students tend to confuse multiple parameters that each have units of energy.)

• Emphasize that in the equation \(dU=C_V \,dT\), \(C_V\) is simply a constant whose subscript has no bearing on the process being applied—at least for an ideal gas. Alternatively, we might consider an equivalent expression for an ideal gas: \(dU=\left(C_P-nR\right)dT\). In this case as well, neither \(C_P\) nor \(R\) directly couples temperatures to energies; they’re simply acting as constants in this differential form of an equation of state.

• Emphasize the unique nature of the ideal gas. Put simply, the equation \(dU\propto dT\) would generally require another \(dV\), \(dP\), or \(dS\) term, but in the case of the ideal gas, the coefficients of these terms are all zero. This conclusion is most fully understood when self-derived, as described earlier.

• Emphasize the existence of independent thermodynamic degrees of freedom for a certain type of system and that we can correctly express the differential form only by including all independent parameters. For a closed, single-phase system, for example, we’re free to write $$dU=\left(\frac{\partial U}{\partial S}\right)_V\,dS+\left(\frac{\partial U}{\partial V}\right)_S\,dV$$ for the (natural) variables \(S\) and \(V\) or $$dU=\left(\frac{\partial U}{\partial S}\right)_P\,dS+\left(\frac{\partial U}{\partial P}\right)_S\,dP$$ for the variables \(S\) and \(P\) or $$dU=\left(\frac{\partial U}{\partial T}\right)_P\,dT+\left(\frac{\partial U}{\partial P}\right)_T\,dP$$ for the variables \(T\) and \(P\), and so on. (Some of these formulations may be more useful than others, of course.)

  If we allowed the system to be open, we’d add the term \(\mu\,dN\) to the fundamental relation \(dU=T\,dS-P\,dV\), where \(\mu\) is the chemical potential and \(N\) is the amount of stuff in the system, which would give an additional independent parameter. (To avoid confusion, I must note that for an open system, \(T\,dS\) no longer exactly matches the incoming heat transfer—we must acknowledge any entropy brought in by new matter—and \(-P\,dV\) no longer exactly matches the total incoming work, as we separate so-called flow work. This point is addressed in Knuiman and Barneveld’s “On the relation between the fundamental equation of thermodynamics and the energy balance equation in the context of closed and open systems”.)

  If we wished to incorporate surface considerations, we’d add the term \(\sigma\,dA\), where \(\sigma\) is the surface tension and \(A\) is the surface area, which would give yet another independent parameter. And so on. The fundamental concept hasn’t changed: each new term consists of a conjugate pair of terms, one intensive and one extensive, that multiply to produce units of energy.

A final twist

Feeling confident at this point? Consider a final twist: As we heat a typical room by a temperature difference \(\Delta T\), how does the internal energy of the air in the room change? (Assume that the air acts as an ideal gas.)

Sorry, it’s not \(\Delta U=C_V\Delta T\) (and it’s certainly not \(\Delta U=C_P\Delta T\)). It’s zero, and this answer tricks plenty of experienced thermodynamics practitioners (see the short note by Emden, “Why do we have winter heating?” Nature 141 908 (1938), and the related discussion by Kreuzer and Payne, “Thermodynamics of heating a room” Am J Phys 71 74 (2011)). Up to this point, we’ve assumed a closed system. But a real room operates at a constant pressure—heated air expands and exits the room around the door and window openings. As a result, the heat capacity \(C_V\) isn’t constant because the amount of air is changing. From the ideal gas law, we have \(n=PV/RT\); thus, the internal energy of an ideal gas in a room remains a constant \(U=C_VT=c_VPV/R\), where \(c_V=C_V/n\) is the molar heat capacity. Clearly, \(T\) appears nowhere in this equation. We’re left with the curious conclusion that by heating a typical room, any energy gained by the air is inevitably dumped outside!