# How many types of specific heat can a gas have?

The number of specific heats that a gas can have is? Which of the following three options is correct?

1. Only one. 2. Only two. 3. Infinite

We know that a gas has two specific heats - $$C_p$$ (Specific heat at constant pressure) and $$C_v$$ (Specific heat at constant volume). I answered option 2 but the correct option according to the question paper solution is option 3.

So, why does a gas have infinite number of specific heats and how do I conceptualise this? Is each specific heat related to a unique thermodynamic process?

• Seems dumb to me also, and I would also have picked two. Any others are not really fundamental physical properties of the gas, related to the temperature derivatives of U and H. Any others are process related. Oct 30, 2022 at 15:31
• It looks like a homework dump. From "How do I ask homework questions on Physics Stack Exchange?": "The best way to produce a focused, specific question is to show your work. Explain what you've been able to figure out so far and how you did it. ... It's not enough to just show your work and ask where you went wrong.". In particular: "Don't just copy the exact problem from your homework assignment or textbook." Oct 31, 2022 at 22:18

The question may be aiming to expand one's scope beyond what's considered in introductory textbooks. This is the essence of scientific research.

The heat capacity $$C_X$$ at a condition of constant $$X$$ is $$C_X\equiv T\left(\frac{\partial S}{\partial T}\right)_X$$, with temperature $$T$$ and entropy $$S$$. We can interpret this as the heating required to obtain a certain temperature change (at constant $$X$$).

Introductory treatments often assume that a gas is enclosed in an an impermeable container (constant $$N$$); thus, we have the associated standard constant-volume and constant-pressure heat capacities

$$C_{V,N}\equiv T\left(\frac{\partial S}{\partial T}\right)_{V,N}\,\,\mathrm{and}$$

$$C_{P,N}\equiv T\left(\frac{\partial S}{\partial T}\right)_{P,N},$$

which for the ideal gas are conveniently equal to $$\frac{dU}{dT}$$ and $$\frac{dH}{dT}$$, respectively, with the internal energy $$U$$ and enthalpy $$H$$ potentials. (Note that no subscripts are required in this special case because they're irrelevant; for the ideal gas, the coefficients of $$dV$$ and $$dP$$ in the corresponding fundamental relations are zero.)

OK, enough review of introductory material. Consider now a gas at equilibrium with an adjacent material. (The system boundaries still include only the gas.) The corresponding heat capacities are now relevant:

$$C_{V,\mu}\equiv T\left(\frac{\partial S}{\partial T}\right)_{V,\mu}\,\,\mathrm{and}$$

$$C_{P,\mu}\equiv T\left(\frac{\partial S}{\partial T}\right)_{P,\mu},$$

with constant chemical potential $$\mu$$, as the number of gas molecules is no longer constant; the gas can diffuse into and/or react with the other material. (Note that this analysis requires care because the system entropy is now affected not just by heating but also by mass transfer, as mass carries its own entropy. One obtains an infinite heat capacity for some simple condensation models, for example, because no amount of cooling can lower the gas temperature while it's condensing—it just accelerates the condensation rate at the boiling temperature.)

Still broader, consider a magnetic gas; that is, a gas for which another type of work other than pressure–volume work is relevant: magnetic field–magnetization work, or $$B$$$$M$$ work.

We can define new potentials

$$\Phi\equiv U+MB=TS-PV+\mu N+MB;$$ $$\Psi\equiv\Phi-MB,$$

where we distinguish $$\Psi$$ from $$U$$ by the possibility of $$M$$$$B$$ work when the former is used. We might now heat a gas at constant magnetic field $$B$$ or constant magnetization $$M$$, respectively corresponding to heat capacities

$$C_{Y,B}\equiv T\left(\frac{\partial S}{\partial T}\right)_{Y,B}\,\,\mathrm{and}$$

$$C_{Y,M}\equiv T\left(\frac{\partial S}{\partial T}\right)_{Y,B},$$

where $$Y$$ refers to some combination of variables we already discussed ($$V$$, $$P$$, $$N$$, $$\mu$$) being held constant. Note that it can still be convenient to work in terms of a potential: $$C_{M}=\frac{\partial \Phi}{\partial T}$$ and $$C_{B}=\frac{\partial \Psi}{\partial T}$$, where the natural variables are being held constant. This is analogous to $$C_V=\frac{\partial U}{\partial T}$$ and $$C_P=\frac{\partial H}{\partial T}$$. (In other words, $$\Phi$$ and $$\Psi$$ are just as valid physical properties of the gas as the more familiar $$U$$ and $$H$$!) Recognizing such analogies and how they can be extended to an arbitrary degree is how one's thermodynamic muscles strengthen.

This gives eight possible heat capacities, with the possibility of extending the framework further with other work types. And we haven't even gotten into ways that functions of the thermodynamic variables might be held constant, rather than just the variables themselves. An infinite number of variations are possible, so the answer to the original question is (3).

• Nice answer, upvoted. Only a doubt: how can we justify the identification of $TdS$ at constant chemical potential with the corresponding heat transferred in a reversible transformation? Oct 30, 2022 at 18:11
• Good point. I added a couple of sentences noting that mass transfer introduces the nuance that the system entropy can vary even though no heating/cooling is occurring. Oct 30, 2022 at 19:06
• I do not understand the argument that a gas that is in contact with another material or radiation (magnetic field) is still "just the gas". Of course the mixed system has different properties, but it is not the same as "the gas". Oct 30, 2022 at 19:40
• The system boundaries always enclose only gas but may not be impermeable (termed an open system). I considered this possibility because the original question mentions only "a gas." Oct 30, 2022 at 20:15

Draw an indicator diagram (that is, a pV diagram where each point represents one state). $$C_p$$ is $$\delta Q/\delta T$$ for changes along a line at constant $$p$$, which is a horizonal line on the diagram. $$C_V$$ is $$\delta Q/\delta T$$ for changes along a line at constant $$V$$, which is a vertical line on the diagram.

But you can also draw any other line: it need not be either vertical or horizonal or even straight. It represents a path in state-space. There is a $$\delta Q/\delta T$$ for changes along that path. There are an infinite number of possible paths, and therefore an infinite number of heat capacities.

Having said that, you only need to consider two, and furthermore if you have the equation of state relating $$T$$ to the state variables, then you only need one because the other can be derived. The sense in which the word "need" is being used here is this: once you know $$C_V$$ for all states, and the equation of state, you can then derive all other thermodynamic information. That idea is a somewhat more advanced one but you will come to it if you are learning thermodynamics. It involves free energy, the second law, and natural variables.