2
$\begingroup$

This may seem like an elementary question, but I'm a bit confused right now about this. From the first and second laws of thermodynamics, and from the definition of enthalpy (per unit mass), we have the equation (as an example, and at constant pressure): $$ dh=c_p dT. $$ But I often come across this other form: $$ \Delta h=c_p\Delta T, $$ but from the sources I've seen, it's not made clear that these deltas represent incremental changes. That is the case? The second expression ought to be written $$ \Delta h = \int_{T_i}^{T_f}c_p dT, $$ right? In any case I'm not sure I understand that second form, because $c_P$ is measured at which temperature, $T_i$ or $T_i+\Delta T$?

$\endgroup$
1
  • 2
    $\begingroup$ Heat capacity can be temperature dependent $\endgroup$
    – K_inverse
    Sep 11 '18 at 0:25
2
$\begingroup$

For a perfect gas, $c_p$ is actually independent of temperature, so both equations are equivalent. Some real gases actually show behavior very close to temperature independence of $c_p$, e.g. ammonia.

In addition, because the coefficients of temperature dependence of $c_p$ of most gases are not that large, over a small temperature rise it is valid to approximate the first equation with the second form.

Or you may just be reading about some approximate or computational method.

$\endgroup$
1
$\begingroup$

This is the case when you take Heat Capacity as Constant.

But there are cases when the temperature dependence of heat capacities are to be taken into account. There are many empirical equations of heat capacities which relates it to temperature and they can be used for more accurate results.

For example:

enter image description here

where a,b,c and d are specie-dependent constants. Now you can replace Cp in the above equation with this expression and the integrate from T_1 to T_2

$\endgroup$
0
$\begingroup$

Such an approximation is valid in two cases:

  1. Assuming that there is no change in phase, as it incurs more heat loss or gain.
  2. When the value of C$^{p}$ is independent of the temperature rise.

These are the basic two cases taken , regardless of the system being a gas or liquid or solid.

$\endgroup$
5
  • $\begingroup$ I suppose that for an hydrogen gas for example, change in excitation or ionization levels is akin to change in phase? $\endgroup$ Sep 11 '18 at 0:48
  • $\begingroup$ Not true. Change in phase means melting or boiling. $\endgroup$ Sep 11 '18 at 1:15
  • $\begingroup$ But change in excitation or ionization levels incur heat loss or gain, no? $\endgroup$ Sep 11 '18 at 1:17
  • $\begingroup$ In thermodynamics, we don’t consider excitation or deexcitation, we consider it as a type of a macro system. And according to me, it won’t cause a comparable heat loss or gain as compared to phase change. $\endgroup$ Sep 11 '18 at 1:25
  • 1
    $\begingroup$ Well, I thinking about it, excitation levels are indeed irrelevant, but ionization levels? These are important latent heat releases, from what I recall. $\endgroup$ Sep 11 '18 at 3:31
0
$\begingroup$

Physicists regard perfect gases (aka ideal gases) as fluids for which the enthalpy (and internal energy) are functions only of temperature and for which the heat capacities are constant, independent of temperature.

Engineers, on the other hand, regard ideal gases as fluids which match the limiting behavior of real gases in the limit of low pressures. As such their enthalpy (and internal energy) are likewise functions only of temperature but their heat capacities are functions of temperature which match those of the specific gas under consideration in the low pressure limit.

So, using the engineering definition, we have $$\Delta h=\int_{T_i}^{T_f}{C_p(T)dT}$$ In practice, this will, of course, be more accurate in matching real gas behavior at low pressures.

$\endgroup$
3
  • $\begingroup$ Most sources I've seen treat perfect gases and ideal gases differently. Ideal gases may have temperature-varying $c_p$. $\endgroup$
    – Al Nejati
    Sep 11 '18 at 5:33
  • $\begingroup$ Thanks. Can you please cite some of those sources for my records. $\endgroup$ Sep 11 '18 at 12:08
  • $\begingroup$ Chester Miller: sure, check out Comprehensive Thermal Science and Engineering, page 297. The terminology is used somewhat inconsistently in the various literature, though. $\endgroup$
    – Al Nejati
    Sep 11 '18 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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