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

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Does it only use at constant pressure? In mine discussions ~ above this board and likewise with web links elsewhere, that looks prefer this equation uses during the Carnot Cycle, whereby there is no consistent pressure.

Why is Cv provided in the adiabatic development of Carnot Cycle come calculate internal energy

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

I usually think this equation does not need constant pressure come apply. Yet today, when I to be going v Atkins physics Chemistry, it particularly says:

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

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

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

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edited Apr 13 "17 at 12:39

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asked Aug 31 "15 at 20:13

Ted YuTed Yu
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Certainly, friend agree $dH=C_P dT$ if we"re at constant pressure. If $H$ and also $C_P$ don"t actually depend on pressure, then you have the right to use this equation regardless of whether pressure changes. However, to recognize $C_P$ and $H$ you first need an equation of state (such as $PV=NkT$). Without learning the particular equation the state (aka, if her gas is right or not), you have the right to only say $dH=C_P dT$ at consistent pressure. However, in the details case the the ideal gas, you know that both $H$ and also $C_P$ don"t rely on $P$, and so friend can use the equation an ext generally.

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edited Sep 1 "15 in ~ 2:59
answered Aug 31 "15 at 23:14

Jahan ClaesJahan Claes
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Enthalpy is characterized as $H=U+pV$ whereby $U$ is the interior energy. This an interpretation has nothing to execute with pressure being constant or not. If the system under factor to consider has just pressure $p$ and also thermal $T$ inner variables and interaction v its setting (simple homogeneous system) then for continuous external push $dp=0$, and we have $dH = dU + pdV + Vdp = dU + pdV=\delta Q$ because of this $dH = C_p dT$

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answer Aug 31 "15 in ~ 21:28

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