Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 15 insertions, 15 deletions

diff --git a/source/know/concept/laws-of-thermodynamics/index.md b/source/know/concept/laws-of-thermodynamics/index.md
index 7758446..3605a0e 100644
--- a/source/know/concept/laws-of-thermodynamics/index.md
+++ b/source/know/concept/laws-of-thermodynamics/index.md
@@ -19,9 +19,9 @@ is a consequence of these laws.
 
 The **first law of thermodynamics** states that energy is conserved.
 When a system goes from one equilibrium to another,
-the change $\Delta U$ of its energy $U$ is equal to
-the work $\Delta W$ done by external forces,
-plus the energy transferred by heating ($\Delta Q > 0$) or cooling ($\Delta Q < 0$):
+the change $$\Delta U$$ of its energy $$U$$ is equal to
+the work $$\Delta W$$ done by external forces,
+plus the energy transferred by heating ($$\Delta Q > 0$$) or cooling ($$\Delta Q < 0$$):
 
 $$\begin{aligned}
     \boxed{
@@ -29,9 +29,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The internal energy $U$ is a state variable,
+The internal energy $$U$$ is a state variable,
 so is independent of the path taken between equilibria.
-However, the work $\Delta W$ and heating $\Delta Q$ do depend on the path,
+However, the work $$\Delta W$$ and heating $$\Delta Q$$ do depend on the path,
 so the first law means that
 the act of transferring energy is path-dependent,
 but the result has no "memory" of that path.
@@ -44,9 +44,9 @@ the total entropy never decreases.
 An important consequence is that
 no machine can convert energy into work with 100% efficiency.
 
-It is possible for the local entropy $S_{\mathrm{loc}}$
+It is possible for the local entropy $$S_{\mathrm{loc}}$$
 of a system to decrease, but doing so requires work,
-and therefore the entropy of the surroundings $S_{\mathrm{sur}}$
+and therefore the entropy of the surroundings $$S_{\mathrm{sur}}$$
 must increase accordingly, such that:
 
 $$\begin{aligned}
@@ -57,19 +57,19 @@ $$\begin{aligned}
 
 Since the total entropy never decreases,
 the equilibrium state of a system must be a maximum
-of its entropy $S$, and therefore $S$ can be used as
+of its entropy $$S$$, and therefore $$S$$ can be used as
 a [thermodynamic "potential"](/know/concept/thermodynamic-potential/).
 
-The only situation where $\Delta S = 0$ is a reversible process,
+The only situation where $$\Delta S = 0$$ is a reversible process,
 since then it must be possible to return to
 the previous equilibrium state by doing the same work in the opposite direction.
 
 According to the first law,
 if a process is reversible, or if it is only heating/cooling,
 then (after one reversible cycle) the energy change
-is simply the heat transfer $\dd{U} = \dd{Q}$.
-An entropy change $\dd{S}$ is then expressed as follows
-(since $\ipdv{S}{U} = 1 / T$ by definition):
+is simply the heat transfer $$\dd{U} = \dd{Q}$$.
+An entropy change $$\dd{S}$$ is then expressed as follows
+(since $$\ipdv{S}{U} = 1 / T$$ by definition):
 
 $$\begin{aligned}
     \boxed{
@@ -85,7 +85,7 @@ Confusingly, this equation is sometimes also called the second law of thermodyna
 ## Third law
 
 The **third law of thermodynamics** states that
-the entropy $S$ of a system goes to zero when the temperature reaches absolute zero:
+the entropy $$S$$ of a system goes to zero when the temperature reaches absolute zero:
 
 $$\begin{aligned}
     \boxed{
@@ -93,8 +93,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-From this, the absolute quantity of $S$ is defined, otherwise we would
-only be able to speak of entropy differences $\Delta S$.
+From this, the absolute quantity of $$S$$ is defined, otherwise we would
+only be able to speak of entropy differences $$\Delta S$$.