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Chemistry 351

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When deriving an expression describing how a thermodynamic function (U, H, A, or G) changes with p, V or T, there are three basic starting points from which you can commonly choose:

  1. An assumed dependence: \( H(p, T) \) or \( U(V, T) \)
  2. The compbined 1st and 2nd Law: \( dU = TdS - pdV \)
  3. A definition:

Other tools that may be useful include the Maxwell Relations, especially those on A and G:

\( \left( \frac{\partial p}{\partial T} \right)_V = \left( \frac{\partial S}{\partial V} \right)_T \) and \( \left( \frac{\partial V}{\partial T} \right)_p = - \left( \frac{\partial S}{\partial p} \right)_T \)

Worked examples

Worked example: Deriving \(\left(\frac{\partial U}{\partial p}\right)_T\)

Derive an expression for \[ \left(\frac{\partial U}{\partial p}\right)_T \] in terms of measurable properties.

Start by treating internal energy as a function of \(V\) and \(T\):

\[ U = U(V,T) \]

Therefore,

\[ dU = \left( \frac{\partial U}{\partial V} \right)_T dV + \left( \frac{\partial U}{\partial T} \right)_V dT \]

Divide by \(dp\) and constrain the change to constant temperature:

\[ \left. \frac{\partial U}{\partial p} \right|_T = \left( \frac{\partial U}{\partial V} \right)_T \left. \frac{\partial V}{\partial p} \right|_T + \left( \frac{\partial U}{\partial T} \right)_V \left. \frac{\partial T}{\partial p} \right|_T \]

Since the temperature is held constant,

\[ \left. \frac{\partial T}{\partial p} \right|_T = 0 \]

so the second term vanishes. Converting to partial derivatives:

\[ \left( \frac{\partial U}{\partial p} \right)_T = \left( \frac{\partial U}{\partial V} \right)_T \left( \frac{\partial V}{\partial p} \right)_T \]

From the previous Maxwell-relation result,

\[ \left( \frac{\partial U}{\partial V} \right)_T = T \left( \frac{\partial p}{\partial T} \right)_V - p \]

Also, the isothermal compressibility is defined as

\[ \kappa_T = - \frac{1}{V} \left( \frac{\partial V}{\partial p} \right)_T \]

so

\[ \left( \frac{\partial V}{\partial p} \right)_T = -V\kappa_T \]

Finally, using

\[ \left( \frac{\partial p}{\partial T} \right)_V = \frac{\alpha}{\kappa_T} \]

gives

\[ \left( \frac{\partial U}{\partial V} \right)_T = T\frac{\alpha}{\kappa_T} - p \]

Therefore,

\[ \left( \frac{\partial U}{\partial p} \right)_T = \left( T\frac{\alpha}{\kappa_T} - p \right) (-V\kappa_T) \]


\[ \boxed{ \left( \frac{\partial U}{\partial p} \right)_T = V \left( p\kappa_T - T\alpha \right) } \]

Interpretation: This expression shows how internal energy changes with pressure at constant temperature. The result is written entirely in terms of measurable quantities: \(V\), \(p\), \(T\), \(\alpha\), and \(\kappa_T\).

Worked example: Deriving \(\left(\frac{\partial G}{\partial V}\right)_T\)

Derive an expression for \[ \left(\frac{\partial G}{\partial V}\right)_T \] starting from \[ G = U+pV-TS. \]

Differentiate \(G\):

\[ dG = dU + p\,dV + V\,dp - T\,dS - S\,dT \]

Divide by \(dV\) and constrain to constant temperature:

\[ \left(\frac{\partial G}{\partial V}\right)_T = \left(\frac{\partial U}{\partial V}\right)_T + p + V\left(\frac{\partial p}{\partial V}\right)_T - T\left(\frac{\partial S}{\partial V}\right)_T \]

Now use the Maxwell-relation result

\[ \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial p}{\partial T}\right)_V - p \]

and the Maxwell Relation on \(A\):

\[ \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial p}{\partial T}\right)_V \]

Substituting gives

\[ \left(\frac{\partial G}{\partial V}\right)_T = \left[ T\left(\frac{\partial p}{\partial T}\right)_V - p \right] + p + V\left(\frac{\partial p}{\partial V}\right)_T - T\left(\frac{\partial p}{\partial T}\right)_V \]

The entropy-related terms cancel, and the \(p\) terms cancel:

\[ \left(\frac{\partial G}{\partial V}\right)_T = V\left(\frac{\partial p}{\partial V}\right)_T \]

Using the definition of the isothermal compressibility,

\[ \kappa_T = - \frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_T \]

it follows that

\[ \left(\frac{\partial p}{\partial V}\right)_T = - \frac{1}{V\kappa_T} \]

Therefore,

\[ \boxed{ \left(\frac{\partial G}{\partial V}\right)_T = -\frac{1}{\kappa_T} } \]

Interpretation: At constant temperature, changing the volume changes the Gibbs function through the pressure response of the system. Since \(\kappa_T\) measures how compressible the system is, a less compressible substance has a larger magnitude for \(\left(\frac{\partial G}{\partial V}\right)_T\).

Practice

Key points (one glance)