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Lecture 13
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Clapeyron equation
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The Clausius-Clapeyron relation
was introduced for one-component systems. It can be used to calculate the shift in transition temperature.
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<math> \frac
= \frac{- \Delta X_j}
</math>
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Below this is written in terms of the enthalpy difference
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<math> \frac
= \frac{-T \Delta X_j}
</math>
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The above equation could be generalized in terms of any two conjugate pairs, with a phase curve relating their intensive variables.
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<math> \frac
= \frac{- \Delta X_j}
</math>
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Super-elasticity
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Imagine two phases, a high temperature phase, <math>P</math>, and a low temperature, matensite phase, <math>M</math>. The martensite phase is larger in one direction, and this fact can be used to control the free energy difference. Look at the equations of energy and find a differential relating length, force, and free energy.
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<math> dU = ... + F dl </math>
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<math> G = U -TS + PV -Fl </math>
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<math>l = \left (\frac{-\partial G}
\right )_
</math>
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The derivative of free energy with respect to force is equal to the length. A larger length corresponds to a larger decrease in free energy with the application of force. This fact can be used to influence phase transition. When the states of a sample are close in free energy, elongation could be induced by a phase transition. The martensite phase would decrease more in free energy, and there could be a transition from the <math>P</math> phase to the <math>M</math> phase.
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Imagine a sample at temperature <math>T</math> between <math>T_o</math> and <math>T(F)</math>. With the application of a force, there can be a phase transition. We would want a fairly reversible transition in a useful temperature range.
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Unable to render embedded object: File (Phase_transition_shift.PNG) not found.
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This example is based on a one-dimensional change. In three dimensions, there would be a change in crystal symmetry, and all lattice vectors would change. Calculations would involve more than one crystal and would be more complicated.
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Example calculation
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Calculate how much stress is needed to be applied to induce a phase transition.
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<math> T_o = 300 K</math>
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<math> \Delta H = 300 \frac
</math>
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<math> \Delta \epsilon = -0.075 </math>
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Below is a Clausius-Clapeyron equation, and the extensive variables are changed to intensive quantities.
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<math> \frac
= \frac{- \Delta S}
</math>
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<math> \Delta l = \Delta \epsilon l </math>
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<math> \frac
\left ( \frac
\right ) = \frac
\left ( \frac{- \Delta S}
\right )</math>
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<math> \frac
= \frac{- \Delta S}
</math>
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<math> \frac
= 1.66 \frac
</math>
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It is important to stay in the elastic regime. Plastic deformation is an irreversible process. Plastic deformation in metals usually occurs at a few hundred MPa, so in the example above, there could be a phase transition before plastic deformation. It's important to push the temperature from <math>T_o</math> far enough to avoid random transitions due to temperature fluctuations. With a <math>\Delta T</math> value of <math>20^\circ </math>, <math>33 Mpa</math> would need to be applied.
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The amount of stress needed to be added to the induce a phase transition is very temperature dependent. An issue is that there is a disontinuity in the length during the phase transition. We want to spread out the transformation. A way to do this is to compositionally grade the material. The composition affects <math>T_o</math>, and this shift can be calculated. Write the Clausius-Clapeyron equation and the differential form of Gibbs free energy for this system. The temperature shift can be calculated from a known chemical potential.
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<math> \frac
= \pm \frac
</math>
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The transformation first occurs where there is a higher <math>T_o</math>.
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Unable to render embedded object: File (Composition_gradient.PNG) not found.
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There is real hysterisis when this is done adiabatically. Boundary movement results in pure dissipation.
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Generalize
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What is the heat effect?
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<math>H=U+PV</math>
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<math>dH=dU+pdV+VdP</math>
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<math>dU = \delta Q - pdV + Fdl</math>
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<math>(dH)_p = (\delta Q)_p + Fdl</math>
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There is an extra work term in the free energy differential. The change in enthalpy at constant pressure is not just due to heat flow. Perform a Legendre transform.
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<math>H=U+PV-Fl</math>
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<math>dH=\delta Q + VdP -ldF</math>
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<math>(dH)_
= (\delta Q)_
</math>
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Below is the generalized enthalpy. It is the free energy minus all the work terms.
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<math>H=U-\sum_
Y_i X_i</math>
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<math>(dH)_
= (\delta Q)_
</math>
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An error in calculations can result from using the enthalpy for heat reaction when subjected to non-PdV work. The full heat release is not the standard enthalpy. The error can be large when doing electrochemistry.
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Magnetically induced transition
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Consider an example of transition between a ferromagnetic state and a paramagnetic state. Can there be a shift in transition with the application of a magnetic field? Yes, anytime there are phases that differ in extensive quantities, essentially implying a first order phase transition, there can be a shift by applying intensive variable. Look at the differentials and Clausius-Clapeyron equation.
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<math>dU=TdS+HdM</math>
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<math>dG=-SdT+...-MdH</math>
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<math>\frac
= -\frac
</math>
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How large or small is the change in transition per atom due to magnetization? How large is <math>\Delta M</math>? Consider a few aligned electron spins. A large estimate of <math>\Delta M</math> is a few Bohr magnetons per atom. Yet <math>\Delta M</math> is still small in energetic terms. Use the Boltzmann constant, <math>k_B</math>, for the <math>\Delta S</math> term. This is the atomic version of <math>R</math>. The result of the calculation is that there is a change in the transition temperature of 0.66 K per applied Tesla. A giant field is required to shift temperature. In this case the effect of magnetism on the transition temperature is small, and the only hope for getting in a reasonable region is if there is a small heat effect in the transition, which corresponds to a small value of <math>\Delta S</math>.
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Example: <math>(La, Ca)MnO_3</math>
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Consider an example of an <math>ABO_3</math> material. In the case of <math>(La, Ca)MnO_3</math>, the amount of <math>Mn^
</math> and <math>Mn^
</math> ions can be controlled by varying the amount of <math>La</math> and <math>Ca</math>, which are of charge <math>3+</math> and <math>2+</math>, respectively. Look at the electrical and magnetic phase diagram.
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Unable to render embedded object: File (Magnetic_phase_diagram.PNG) not found.
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Ferromagnetic materials are better conductors than anti-ferromagnetic. Electrons are able to hop between atoms and preserve spin. Paramagnetic materials are not as bad conductors as anti-ferromagnetic. This is due to Hund's Rule.
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Imagine one is at a point above the transition temperature in the phase diagram. The material is in the paramagnetic field, and it is possible to be in the magnetized phase by applying a field and shifting the transition temperature. Conductivity increases by four orders of magnitude, and this effect is termed the Giant Magnetoresistive Effect (GMR). There is not much changing intrinsically and is complicated in reality; there is a percolation issue. IBM discovered a bigger change and termed it the Colossal Magnetoresistance Effect (CMR).
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Remember that there are different extensive properties that result from applying a field