The first law of thermodynamics divides the internal energy change into contributions of heat and work.

$$\text dU=\omega_Q-\omega_W,$$

Here I chose the notation to emphasise that the two parts are in general not exact differentials, i.e. that there is no heat counting function.

What would be the proper way to express the heat capacity

$$C_V=\left(\frac{\text d U}{\text d T}\right)_V$$

in terms of $\omega_Q$?

  • 1
    $\begingroup$ Am I missing something or wouldn't that just be $C_V = \left(\partial Q/\partial T\right)_V$ ? $\endgroup$ – Alexander Mar 30 '12 at 18:29
  • $\begingroup$ @Alexander: Well yes, my problem is what $∂Q$ or even $\text d Q$ is supposed to mean, there is no function $Q$. So I'm searching for an answer in terms of the form $\omega_Q$. $\endgroup$ – Nikolaj-K Mar 30 '12 at 23:52
  • $\begingroup$ $\partial Q$ is the amount of thermal energy you put into your system. This is not a thermodynamic potential, therefore Q is not a function. It does not depend on anything. $\endgroup$ – Alexander Mar 31 '12 at 0:36
  • $\begingroup$ @Alexander: Can you express $∂Q$ in terms of $\omega_Q$? $\endgroup$ – Nikolaj-K Mar 31 '12 at 11:17

I don't blame you for being confused by this because the notation in which the first law of thermodynamics is often written ($\mathrm{d}U = \mathrm{d}Q - \mathrm{d}W$) is actually rather inconsistent. In your notation, the specific heat would be

$$C_V = \biggl(\frac{\omega_q}{\mathrm{d}T}\biggr)_V$$

It's just conventional to write $\omega_q$ as $\mathrm{d}Q$ (or $\partial Q$) to indicate that it is of infinitesimal magnitude.

As for a more in-depth explanation: the fundamental issue is that a differential quantity $\mathrm{d}x$ for some $x$ does not have to be representable as a change in some function. What it really represents is an "infinitesimal amount," which is some amount associated with an infinitesimal advance along the path that the system takes through its state space.

For example, with a thermodynamic system, the state space might be parametrized by coordinates $(N, P, V)$. Any thermodynamic process this system goes through can be represented as a one-parameter path through the state space. Now, imagine splitting up the path into segments of infinitesimal length. You could do this by dividing the time during which the process happens into infinitesimal increments $\mathrm{d}t$, but you don't have to use time; you can use any parameter that doesn't repeat values.

In a given segment, $\mathrm{d}X$ for any $X$ represents some sort of amount of $X$ associated with that segment. Sometimes $\mathrm{d}X$ is an actual change; for example, $\mathrm{d}P$ is the change in pressure from the beginning of the segment to the end, or $\mathrm{d}U$ is the change in internal energy from the beginning to the end. But other things cannot be expressed as a difference of initial and final amounts. For example, $\mathrm{d}Q$ (your $\omega_q$) is the amount of heat gained by the system during that segment - or more precisely, $\mathrm{d}Q$ is the amount of energy gained by the system through the mechanism of heat transfer. Similarly, $\mathrm{d}W$ is the amount of energy lost by the system through the mechanism of mechanical work.

Obviously, the mechanism by which the energy is transferred is meaningless when energy is not actually being transferred. So you can only make this division into work and heat for changes in your system's energy, not for its amount of energy. That's why there is no state function for heat or for work individually. However, if you ignore the division between the different ways in which energy is transferred, and only consider the total amount of energy, there's no problem: energy is conserved, so the initial amount of energy plus the change in energy equals the final amount of energy. So when you put together the change due to heat and the change due to work, you get something ($\mathrm{d}U$) that does correspond to a change in a state function.

By the way, this is the same sort of thing that goes on with conservative and nonconservative forces in Newtonian mechanics. Just as I described above for thermodynamical systems, you can track the path a mechanical system takes through physical space, split that path into small segments, and compute various quantities for each segment.

Sometimes those segments correspond to changes in some function of position. For example, if you compute the work done by gravity during the segment, $\mathrm{d}W_g = -mg\mathrm{d}y$, you find that it's path-independent, and you can then define a state function $U_g$ for it, $\mathrm{d}W_g = -\mathrm{d}U_g$. Potential energies are the state functions of Newtonian mechanics.

But if you compute the work done by friction during the segment, $\mathrm{d}W_f = -F_f \mathrm{d}s$, you can't express that as a change of any state function. Accordingly, friction is a nonconservative force.

In a sense, you could think of heat and work as the nonconservative "forces" of thermodynamics. Specifically, $\mathrm{d}Q$ and $\mathrm{d}W$ (your $\omega_{{w,q}}$) fill a similar role in thermodynamics as $\mathrm{d}W_f$ etc. fill in mechanics. Of course, $\mathrm{d}Q$ and $\mathrm{d}W$ happen to combine in such a way as to produce a state function; I don't know that that has any common equivalent in mechanics.

  • $\begingroup$ I've not used the $\omega$ notation, though while learning ThermoD (from the chemistry POV), I was extremely bugged that $q\cong\rm dq,w\cong \rm dw$, and stuff like $\int\frac{q}{T}=\Delta S$ is $\neq\infty$. :/ $\endgroup$ – Manishearth Mar 31 '12 at 16:05
  • $\begingroup$ This is a very nice description. I have also seen the notation $C_V = \left(\delta Q/\partial T\right)$ to emphasize that $Q$ is not a thermodynamic potential. $\endgroup$ – Alexander Mar 31 '12 at 17:23
  • $\begingroup$ Thanks for your explaination David. Wouldn't it be better to introduce an actual trajectory in state space like $X=X_T·e_T+X_V·e_V\equiv(X_T,X_V)$ and define $C_V:=i_{e_T}\omega_Q$? Because I've never seen someone using a notation like $\frac{\omega[X(\lambda)]}{\text d \lambda}$. $\endgroup$ – Nikolaj-K Mar 31 '12 at 19:47

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