6
$\begingroup$

The definition of enthalpy as a function of heat capacity and temperature change. $ \Delta H = C_p \Delta T $.

Does it only apply at constant pressure? In my discussions on this board and also with links elsewhere, it looks like this equation applies during the Carnot Cycle, where there is no constant pressure.

Why is Cv used in the adiabatic expansion of Carnot Cycle to calculate internal energy

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/Thermodynamic_Cycles/Carnot_Cycle

I generally think this equation does not need constant pressure to apply. However today, when I was going through Atkins Physical Chemistry, it specifically says:

$ \Delta H = C_p \Delta T $ (at constant pressure)

equation 2B.6b. I think the constant pressure part is wrong, but this is a book with 10 editions with 5 star reviews on amazon.

http://www.amazon.com/gp/product/1429290196/

$\endgroup$
2
  • $\begingroup$ Please find the relevant derivation of $\Delta H= C_p \Delta T$ here: aerostudents.com/files/aerodynamicsA/enthalpySpecificHeat.pdf $\endgroup$
    – Gert
    Commented Aug 31, 2015 at 21:22
  • $\begingroup$ Here is my derivation which I think explains why $\Delta H = C_p\Delta T$ for the adiabatic portion of the Carnot cycle. For an ideal gas, (3/2)NkdT = dq + PdV. (3/2)NkdT + d(PV) = dq - PdV + d(PV). (3/2) NkdT + NkdT = dq + -PdV + PdV + VdP. (5/2)NkdT = dq + VdP. For an adiabat, (5/2)NkdT = VdP. For an adiabat and ideal gas: (5/2)NkdT = dH. CpdT = dH. $\endgroup$
    – Ted Yu
    Commented Sep 10, 2015 at 1:20

3 Answers 3

3
$\begingroup$

Certainly, you agree $dH=C_P dT$ if we're at constant pressure. If $H$ and $C_P$ don't actually depend on pressure, then you can use this equation regardless of whether pressure changes. However, to determine $C_P$ and $H$ you first need an equation of state (such as $PV=NkT$). Without knowing the specific equation of state (aka, if your gas is ideal or not), you can only say $dH=C_P dT$ at constant pressure. However, in the specific case of the ideal gas, you know that both $H$ and $C_P$ don't depend on $P$, and so you can apply the equation more generally.

$\endgroup$
8
  • $\begingroup$ I think you meant $dH=C_pdT$. I understand what you are trying to say, but many people who (like me) used this book for their PChem class. When reading it as it is written, we explicitly took this to mean it only applies at constant pressure for all cases (even including ideal gases). Many fail to see that the Cp definition comes from $Cp = dq/dT$ at constant P. And this is not the same as $Cp = dH/dT$. Look at the other answer. $\endgroup$
    – Ted Yu
    Commented Sep 1, 2015 at 1:15
  • $\begingroup$ Edited! Force of habit, from writing so many $PdV$s. $\endgroup$ Commented Sep 1, 2015 at 2:59
  • $\begingroup$ @TedYu At constant pressure, it is true that $\delta Q/\delta T=\partial H/\partial T$ for any gas, anytime. Just use the chain rule, and the fact that $dP=0$. $\endgroup$ Commented Sep 1, 2015 at 3:05
  • $\begingroup$ @TedYu and the point is, this equation is always true as stated. If you have a constant pressure, the equation is factually correct no matter what. In the case of an ideal gas you can say a little more, but that isn't generalizable. $\endgroup$ Commented Sep 1, 2015 at 3:06
  • 1
    $\begingroup$ @TedYu That's fair, but I think it's important when studying statistical physics to make sure you separate things which are generally true and things that are only true for ideal gases. Otherwise, you could develop bad intuition for many physical processes. $\endgroup$ Commented Sep 1, 2015 at 19:14
2
$\begingroup$

Enthalpy is defined as $H=U+pV$ where $U$ is the internal energy. This definition has nothing to do with pressure being constant or not. If the system under consideration has only pressure $p$ and thermal $T$ internal variables and interaction with its environment (simple homogeneous system) then for constant external pressure $dp=0$, and we have $dH = dU + pdV + Vdp = dU + pdV=\delta Q$ therefore $dH = C_p dT$

$\endgroup$
8
  • 1
    $\begingroup$ Then, since H doesn't depend on pressure for an ideal gas, we have that Cp =dH/dT even if pressure is changing. So dH=CpdT. $\endgroup$
    – Ted Yu
    Commented Aug 31, 2015 at 22:26
  • $\begingroup$ The constant external pressure is not required. $\endgroup$
    – Ted Yu
    Commented Aug 31, 2015 at 22:27
  • $\begingroup$ The heat capacity $C_p$ is function of both $p$ and $T$, $C_p=C_p(p,T)$ but the equation $C_p=\frac {\partial H}{ \partial T}$ holds only if $p$ is kept constant while $T$ is varying $\endgroup$
    – hyportnex
    Commented Aug 31, 2015 at 22:32
  • $\begingroup$ If that's the case, why can this equation be used to determine $\Delta H$ during the adiabatic portion of the Carnot cycle? chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/… $\endgroup$
    – Ted Yu
    Commented Sep 1, 2015 at 1:05
  • 2
    $\begingroup$ I actually think Atkins stating that it applies for only constant pressure is wrong. Just because Cp is defined as $dq/dT$ at constant pressure does NOT mean that it is by definition also $dH/dT$ at constant pressure. My point is that Atkins is wrong. $\endgroup$
    – Ted Yu
    Commented Sep 1, 2015 at 1:39
0
$\begingroup$

The simple answer to this question is logic. Look the simple thing is if pressure is made constant, than the supplied energy is the sum of two energies. Supplied Energy= Work done +Change in Internal Energy #Remember the above case is calculated by making pressure constant, with volume constant the case is different So, enthalpy is the total heat of the system. Change in enthalpy indicates change in total energy of the system.
So, it does not mean that it is applied, or it is the condition, it is the calculation of value of the enthalpy assuming a condition using pressure is constant. Similar can be the cause with u=cvdT in this if volume is made constant. So, assuming this case the energy supplied will be spent in the internal energy of the system. Hope this helps Good Luck

$\endgroup$

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.