0
$\begingroup$

The specific heat capacity at constant volume and the specific heat capacity at constant pressure are intensive properties defined for pure, simple compressible substances as partial derivatives of the functions u(T, υ) and h(T, p), respectively, $$c_v=\left ( \frac{\partial u}{\partial T} \right )_v$$ $$c_p=\left ( \frac{\partial h}{\partial T} \right )_p$$ Can someone explain why this is? The book from which I got this from doesn't derive these expressions.

$\endgroup$
4
  • 3
    $\begingroup$ Those can be regarded as definitions. Do you have a more specific question about the definitions? $\endgroup$
    – Buck Thorn
    Oct 5, 2020 at 14:25
  • $\begingroup$ @BuckThorn Yes. I was trying to find out how to obtain this expression through differentiation. $\endgroup$ Oct 5, 2020 at 14:41
  • $\begingroup$ Well, for $c_p$ for instance, start with the enthalpy function h(T) and take the derivative with respect to T while holding the pressure constant. It is not more mysterious than that. $\endgroup$
    – Buck Thorn
    Oct 5, 2020 at 15:09
  • $\begingroup$ @BuckThorn I'm just confused as to what h(T) would be. $\endgroup$ Oct 5, 2020 at 17:16

4 Answers 4

2
$\begingroup$

If a body absorbs a quantity of heat $q$ its temperature will normally rise by a value $\Delta T$. The average heat capacity over this temperature range is defined as $C_{av}\equiv q/\Delta T$. The instantaneous heat capacity at temperature $T$ is $C\equiv dq/dT$. This definition is not exact enough, however, until the path of heating is specified. From the first law $q$ could be replaced by $dU+dw$ so it is necessary to fix another parameter which is $V$ or $p$ giving rise to the definitions you quote.

Note (a) that at a phase change $C$ becomes infinity because heat is absorbed with no change in temperature. (b) As the heat capacity is the gradient of internal energy with temperature it describes how the internal energy increases. For example when $C$ becomes constant at high temperatures the molecules are absorbing heat a constant and maximum 'rate' with increase in $T$. (c) Using statistical mechanics the heat capacity can be calculated from a knowledge of a molecule's energy levels.

$\endgroup$
2
  • $\begingroup$ Can you explain what the function h(T,p) and u(T,v) would be mathematically? $\endgroup$ Oct 5, 2020 at 18:51
  • 1
    $\begingroup$ These functions are in any thermodynamics book, for example $H=U+pV, \Delta U=q+w$ $\endgroup$
    – porphyrin
    Oct 6, 2020 at 7:14
2
$\begingroup$

Consider a gas which was subjected to two different processes: isochoric and isobaric.

  1. Isochoric: If a certain amount of heat $dq$ was given to this gas, then this heat will be completely used up in raising it's temperature since $dw$ is zero. So, $$dq_v = dU$$ Also $$dq_v = nC_v dT$$ where $C_v$ is constant volume molar heat capacity. Therefore we can say that $$nC_v dT = dU\implies C_v =\frac{1}{n}\frac{dU}{dT}$$ So for 1 mole of gas $$C_v =\left(\frac{dU}{dT}\right)_v$$

  2. Isobaric : Now in this process a certain amount of heat $dq_p$ is given to the gas and this will raise it's temperature and a part will be used up in doing work against external pressure (if any). Change in Enthalpy $\Delta H$ as we know is defined as heat tranfered in an isobaric process. This process is isobaric so $$dq_p = dH$$ Again writing heat transer in terms of molar heat capacity $$dq_p = nC_p dT$$ Therefore $$nC_p dT = dH\implies C_p = \frac{1}{n}\frac{dH}{dT}$$ For 1 mole of this gas $$C_p = \left(\frac{dH}{dT}\right)_p$$

$\endgroup$
6
  • 1
    $\begingroup$ I don't think this is an answer because you have only illustrated the equations for an ideal gas instead of having a discussion on the actual quantities involved. $\endgroup$
    – Babu
    Oct 7, 2020 at 10:32
  • 1
    $\begingroup$ Actually these equations are applicable to real gases as well since I haven't used the ideal gas equation anywhere $\endgroup$
    – Eyy boss
    Oct 7, 2020 at 11:01
  • 2
    $\begingroup$ Oh you are right, I again misunderstood. On a lighter note, I changed places of isobaric and isochoric since you explain isochoric first $\endgroup$
    – Babu
    Oct 7, 2020 at 11:06
  • 1
    $\begingroup$ I think it'd be easier to undestand if you wrote from where you got each definition i.e: one where $ dq_v = nC_v dT$ and $dq_v = dU$ $\endgroup$
    – Babu
    Oct 7, 2020 at 11:09
  • 1
    $\begingroup$ I believe he already knows about this since he is talking about heat capacities in the question. $\endgroup$
    – Eyy boss
    Oct 7, 2020 at 11:18
1
$\begingroup$

Principally, definitions cannot be derived. They are chosen.

What you have described in the question are heat capacities without the adjective specific, being heat capacities of a whole system.

Note that specific heat capacities are related to the substance unit mass amount, mostly 1 kg, so I would expect the mass in denominator, like $$c_V = \frac 1m . \left(\frac{\partial U}{\partial T}\right)_V $$ $$c_p = \frac 1m . \left(\frac{\partial H}{\partial T}\right)_p $$

Similarly, molar heat capacities would have there the molar amount instead of mass.

$$C_V = \frac 1n . \left(\frac{\partial U}{\partial T}\right)_V $$ $$C_p = \frac 1n . \left(\frac{\partial H}{\partial T}\right)_p $$

$\endgroup$
1
$\begingroup$

The heat capacities arise when you define various differential forms of U and H. The differential forms define how the energy responds to changes in thermodynamic variables such as pressure, temperature and volume.

For instance, one form for U reads:

$$\begin{align}dU = \left(\frac{\partial U}{\partial T} \right)_V dT + \left(\frac{\partial U}{\partial V} \right)_T dV \end{align}$$

Here we can now use the following definition (symbolized with ":="):

$$C_V:= \left(\frac{\partial U}{\partial T} \right)_V$$

Think of "$C_V$" as shorthand for "$\left(\frac{\partial U}{\partial T} \right)_V$".

Use of shorthand forms leads to the following equation:

$$dU = C_V dT + \left[ \frac{\alpha T}{\kappa} -p\right] dV $$

Similarly

$$C_p:= \left(\frac{\partial H}{\partial T} \right)_p$$

Use of this and other shorthand forms leads to

$$\begin{align}dH = C_p dT + V\left[ 1-\alpha T\right] dp \end{align}$$

There is more to heat capacities than that, of course. For one thing, we can attach a physical interpretation to heat capacities in the context of particular experiments.

Regarding how a functional form of the enthalpy (or energy) may be obtained as a function of T, that is another can of worms. One strategy is to use the van't Hoff equation. Another is to use calorimetry.

Also, internal energies and enthalpies are not reported as absolute values. Values are reported as differences relative to a reference value (not "$H$", rather "$\Delta H$"). Once you define the enthalpy as a function of temperature at a particular value of the pressure, computing $C_p$ just involves taking the derivative wrt T. For $C_v$ we start from internal energies as functions of T at constant volume. Finally, as mentioned in another answer, you should keep in mind the difference between heat capacities as extensive properties, and molar or specific heat capacities, which are intensive properties.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.