State the first law of thermodynamics
Internal energy is a exact differential give three consequences of this fact
State the second law of thermodynamics
When does the equality sign in the second law hold and when is it an inequality
Write an expression that combines the first and second laws of thermodynamics
How are the temperature, pressure and chemical potential related to derivatives of the internal energy
How are the Maxwell relations derived.
The first law of thermodynamics states:
<aside> 💡 $dE = dq + dw$
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In other words, we measure the change in internal energy by adding together the work done one the system and the heat absorbed by the system. We assume that the internal energy is an exact differential. This has three important consequences:
The second law of thermodynamics states:
<aside> 💡 $\Delta_{A\rightarrow B} S \ge \int_A^B \frac{\textrm{d}q}{T}$
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The equality sign holds when the transition is reversible, the inequality holds when the transition is irreversible. The entropy is once again an exact differential so:
Importantly, however, because of the inequality sign in the second law of thermodynamics, the heat absorbed during a transition will depend on the path taken and during moves around closed paths heat will be input/output from the system. The consequence of this is that it is impossible to build an engine that converts 100 % all of the heat it absorbs into work.
Combining the first and second laws gives:
<aside> 💡 $\textrm{d}E = T\textrm{d}S - P\textrm{d}V + \mu \textrm{d}N$
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This expression is derived by considering the differentials that it is possible to derive for reversible transitions and by remembering that the internal energy is an exact differential so the change internal energy does not depend on the path taken between equilibrium states. In other words, the above expression holds for both reversible and irreversible transitions. The theory of exact differentials tells us that:
$\textrm{d}E =\left( \frac{ \partial E }{\partial S} \right){V,N} \textrm{d}S + \left( \frac{ \partial E }{\partial V} \right){S,N} \textrm{d}V + \left( \frac{ \partial E }{\partial N} \right)_{V,S} \textrm{d}N$
Equating coefficients in these two equations gives:
$T = \left( \frac{ \partial E }{\partial S} \right){V,N} \qquad P = - \left( \frac{ \partial E }{\partial V} \right){S,N} \qquad \mu = \left( \frac{ \partial E }{\partial N} \right)_{V,S}$
Furthermore, using the equality of the second, crossed derivatives we can arrive at the following Maxwell relation:
$\left(\frac{\partial T}{\partial V} \right){S,N} = - \left(\frac{\partial P}{\partial S} \right){V,N}$
<aside> 📌 SUMMARY: The first law of thermodynamics defines an exact differential knows as the internal energy. The change in internal energy that occurs during a transition is equal to the sum of the work done and heat transferred. The second law of thermodynamics defines an exact differential known as the entropy and tells us that it is impossible to build an engine that is 100 % efficient. By combining the first and second laws of thermodynamics you can arrive at an expression for the differential of the internal energy that is given entirely in terms of the thermodynamic variables. You can then connect intensive variables to derivatives of the internal energy with respect to intensive variables and derive Maxwell relations.
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