Enthalpy
In constant pressure, enthalpy, \(H\), is defined as
$$H = U + PV$$
At constant pressure, \(PV\) is the work done on the system. Thus, enthalpy is the energy required for creating a system with internal energy \(U\) and volume \(V\) at constant pressure \(P\).
By thermodynamic identity, \(dU = TdS - PdV + \mu dN\), so
$$dH = dU + PdV + VdP = TdS + VdP + \mu dN$$
Helmholtz Free Energy
In constant temperature, the Helmholtz free energy, \(F\), is defined as
$$F = U - TS$$
When there is no work done on or by the system, \(dU = dQ\), so \(Q = TS\). Thus, Helmholtz free energy is the energy required for creating a system with internal energy \(U\) at constant temperature \(T\).
By thermodynamic identity, \(dU = TdS - PdV + \mu dN\), so
$$dF = dU - TdS - SdT = -SdT - PdV + \mu dN$$
Gibbs Free Energy
In constant temperature and pressure, the Gibbs free energy, \(G\), is defined as
$$G = U - TS + PV$$
By thermodynamic first law and temperature definition, \(SdT = dQ + PdV\), so \(Q = TS - PV\). Thus, Gibbs free energy is the energy required for creating a system with internal energy \(U\) at constant temperature \(T\) and pressure \(P\).
By thermodynamic identity, \(dU = TdS - PdV + \mu dN\), so
$$dG = -SdT + VdP + \mu dN$$
Maxwell Relations
Consider system with constant number of particles, i.e. \(dN = 0\)
$$dU = TdS - PdV$$
So,
$$T = \left(\frac{\partial U}{\partial S}\right)_V$$
and
$$P = -\left(\frac{\partial U}{\partial V}\right)_S$$
Since
$$\frac{\partial}{\partial V}\left(\frac{\partial U}{\partial S}\right) = \frac{\partial}{\partial S}\left(\frac{\partial U}{\partial V}\right)$$
we have
$$ \bbox[5px,border:2px solid #666]
{
\left(\frac{\partial T}{\partial V}\right)_S = - \left(\frac{\partial P}{\partial S}\right)_V
}$$
Similarly, by \(dF = -SdT -PdV\),
$$ \bbox[5px,border:2px solid #666]
{
\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V
}$$
By \(dH = TdS + VdP\),
$$ \bbox[5px,border:2px solid #666]
{
\left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P
}$$
By \(dG = -SdT + VdP\),
$$ \bbox[5px,border:2px solid #666]
{
\left(\frac{\partial S}{\partial P}\right)_T = - \left(\frac{\partial V}{\partial T}\right)_P
}$$
These are called Maxwell relations.
Clausius-Clapeyron Relation
When the number of particles is a constant, i.e. \(dN = 0\), entropy depends on \(V\) and \(U\) hence \(T\) only, so
$$dS = \left(\frac{\partial S}{\partial V}\right) _T dV + \left(\frac{\partial S}{\partial T}\right)_V dT $$
During phase change, temperature is constant
$$dS = \left(\frac{\partial S}{\partial V}\right) _T dV$$
By Maxwell relations,
$$dS = \left(\frac{\partial P}{\partial T}\right)_V dV$$
As pressure and temperature should be constant, the partial derivative should be the same as total derivative
$$dS = \frac{dP}{dT} dV$$
So,
$$\frac{dP}{dT} = \frac{S_{p1} - S_{p2}}{V_{p1} - V_{p2}}$$
Enthalpy at constant pressure is
$$dH = dU + PdV$$
Substitute first law of thermodynamics, at constant pressure
$$dH = Q - PdV + PdV = Q$$
So the latent heat \(L\) is equal to
$$L = \Delta H$$
As \(dN = 0\),
$$dH = TdS + VdP = TdS$$
So,
$$\Delta S = \frac{\Delta H}{T} = \frac{L}{T}$$
Then,
$$\bbox[5px,border:2px solid #666]
{
\frac{dP}{dT} = \frac{L}{T\Delta V}
}$$
This is called Clausius-Clapeyron relation.
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