Enthalpy: Difference between revisions

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(Added a couple of references)
m (Slight tidy)
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:<math>\left.H\right.=U+pV</math>
:<math>\left.H\right.=U+pV</math>


where <math>U</math>  is the [[internal energy]], <math>p</math> is the [[pressure]], <math>V</math> is the volume and ''(-pV)'' is a ''conjugate pair''. The differential of this function is
where <math>U</math>  is the [[internal energy]], <math>p</math> is the [[pressure]], <math>V</math> is the volume.
<math>pV</math> is a ''conjugate pair''. The differential of this function is


:<math>\left.dH\right.=dU+pdV+Vdp</math>
:<math>\left.dH\right.=dU+pdV+Vdp</math>
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:<math>\left.dH\right.=TdS +Vdp</math>
:<math>\left.dH\right.=TdS +Vdp</math>


For ''H(S,p)'' we have the following ''total differential''
For <math>H(S,p)</math> we have the following ''total differential''


:<math>dH=\left(\frac{\partial H}{\partial S}\right)_p dS + \left(\frac{\partial H}{\partial p}\right)_S dp</math>
:<math>dH=\left(\frac{\partial H}{\partial S}\right)_p dS + \left(\frac{\partial H}{\partial p}\right)_S dp</math>

Revision as of 15:06, 12 March 2012

Enthalpy (H) [1][2] is defined as:

H=U+pV

where U is the internal energy, p is the pressure, V is the volume. pV is a conjugate pair. The differential of this function is

dH=dU+pdV+Vdp

From the Second law of thermodynamics one obtains

dH=TdS−pdV+pdV+Vdp

thus we arrive at

dH=TdS+Vdp

For H(S,p) we have the following total differential

dH=(∂H∂S)pdS+(∂H∂p)Sdp

References