minor mods on note 1

big change on notes 2 and 3, changed order and changed eos to vdw

Get_involved/1_Beginner/1_Jupyter_notebook_Python_and_Applied_Thermodynamics.ipynb

@@ -26,13 +26,13 @@
     "Supose you have to solve the following problem in a thermodynamics course:\n",
     ">\"Consider a closed system containing `1 mol` of a gas. This system undergoes cyclic trasnformations consisting of four steps. The following table display variations in internal energy as well as total heat and total work input/output in each step. Determine the values of the missing quantities.\"\n",
     ">\n",
-    "> Steps | $\\Delta U$ (`J`) | $Q$ (`J`) |$W$ (`J`) \n",
+    "> Steps | $\\Delta U$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ | $Q$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ |$W$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ \n",
     "> :--: | --: | --: | --:\n",
-    "> $A\\to B$ | -200. | \\_\\_\\_ | -6,000. \n",
-    "> $B\\to C$ | \\_\\_\\_ | -3,800. | \\_\\_\\_\n",
-    "> $C\\to D$ | \\_\\_\\_ | -800. | 300.\n",
-    "> $D\\to A$ | 4,700. | \\_\\_\\_ | \\_\\_\\_\n",
-    "> $A_{\\to B \\to}^{\\leftarrow D \\leftarrow} C$ | \\_\\_\\_ | \\_\\_\\_ | -1,400.\n",
+    "> $A$$\\to$$B$ | -200. | \\_ | -6,000.\n",
+    "> $B$$\\to$$C$ | \\_ | -3,800. | \\_\n",
+    "> $C$$\\to$$D$ | \\_ | -800. | 300.\n",
+    "> $D$$\\to$$A$ | 4,700. | \\_ | \\_\n",
+    "> $A$$_{\\to B \\to}^{\\leftarrow D \\leftarrow}$$C$ | \\_ | \\_ | -1,400.\n",
     ">> Adapted from `Smith, van Ness, 0000, 00ed. Ex0.`"
    ]
   },
@@ -45,7 +45,8 @@
     "# Solution:\n",
     "## Fundamentals:\n",
     "According to the first law of thermodynamics, in any process, the variation in energy is equal to the sum of heat and work inputs/outputs\n",
-    ">- $\\Delta U_{A\\to B} = Q_{A\\to B} + W_{A\\to B}$ \n",
+    "\n",
+    "$$\\Delta U_{A\\to B} = Q_{A\\to B} + W_{A\\to B}$$ \n",
     "\n",
     "where, by convention, \n",
     "* $\\Delta U_{A\\to B}$ means difference between internal energy in state $B$ and  nternal energy in  state $A$: $U_B - U_A$\n",
@@ -54,13 +55,13 @@
     "\n",
     "Net Heat and Work in a cyclic operation is the sum of heat and work in each step process\n",
     "\n",
-    ">- $Q_{A\\to B} + Q_{B\\to C} + Q_{C\\to D} +  Q_{D\\to A} = Q_{net}$\n",
+    "$$Q_{A\\to B} + Q_{B\\to C} + Q_{C\\to D} +  Q_{D\\to A} = Q_{net}$$\n",
     "\n",
-    ">- $W_{A\\to B} + W_{B\\to C} + W_{C\\to D} +  W_{D\\to A} = W_{net}$\n",
+    "$$W_{A\\to B} + W_{B\\to C} + W_{C\\to D} +  W_{D\\to A} = W_{net}$$\n",
     "\n",
     "Also, as internal energy is a *function of state*, in a cyclic operation its net variation, when a full cycle is complete and the system is found to be in a previously visited state, is zero.\n",
     "\n",
-    ">- $\\Delta U_{A\\to B} + \\Delta U_{B\\to C} + \\Delta U_{C\\to D} +  \\Delta U_{D\\to A}= 0$\n",
+    "$$\\Delta U_{A\\to B} + \\Delta U_{B\\to C} + \\Delta U_{C\\to D} +  \\Delta U_{D\\to A}= 0$$\n",
     "\n"
    ]
   },
@@ -86,11 +87,13 @@
     "\n",
     "Note that in lines corresponding to processes `A-B` and `C-D`, two degreees of freeedom are provided and only one value is missing, therefore we immediately solve any of these processes. Let's start with step $A\\to B$\n",
     "\n",
-    "> Steps | $\\Delta U$ (`J`) | $Q$ (`J`) |$W$ (`J`) \n",
-    "> :--: | --: | --: | --:\n",
-    "> A-B | -200. | \\_\\_\\_ | -6,000. \n",
+    " > Steps | $\\Delta U$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ | $Q$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ |$W$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$\n",
+    " > :--: | --: | --: | --:\n",
+    " > $A$ $\\to$ $B$ | -200. | \\_ | -6,000.\n",
+    " \n",
     "\n",
-    "$Q_{A\\to B}$ (`J`) = $\\Delta U_{A\\to B}$ (`J`)  - $W_{A\\to B}$ (`J`)"
+    ">> $Q_{A\\to B} \\mathrm{(J)} = \\Delta U_{A\\to B} \\mathrm{(J)}  - W_{A\\to B} \\mathrm{(J)}$\n",
+    "\n"
    ]
   },
   {
@@ -184,11 +187,11 @@
    "source": [
     "###C-D \n",
     "\n",
-    "> Steps | $\\Delta U$ (`J`) | $Q$ (`J`) |$W$ (`J`) \n",
-    "> :--: | --: | --: | --:\n",
-    "> C-D | \\_\\_\\_ | -800. | 300.\n",
+    " > Steps | $\\Delta U$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ | $Q$$\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ |$W$$\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$\n",
+    " > :--: | --: | --: | --:\n",
+    " > $C$$\\to$$D$ | \\_ | -800. | 300.\n",
     "\n",
-    "$\\Delta U_{C\\to D}$ (`J`) = $Q_{C\\to D}$ (`J`) + $W_{C\\to D}$ (`J`)\n",
+    ">> $\\Delta U_{C\\to D} \\mathrm{(J)} = Q_{C\\to D} \\mathrm{(J)} + W_{C\\to D} \\mathrm{(J)}$\n",
     "\n"
    ]
   },
@@ -245,14 +248,13 @@
    "source": [
     "Now, using the last fundamentalk exposed, concernign the fact that a cyclic process, on completion of a full cycle returns to a same equilibrium state, we can solve the last line in the table\n",
     "\n",
-    "> Steps | $\\Delta U$ (`J`) | $Q$ (`J`) |$W$ (`J`) \n",
-    "> :--: | --: | --: | --:\n",
-    "> A-B-C-D-A | \\_\\_\\_ | \\_\\_\\_ | -1,400.\n",
+    " > Steps | $\\Delta U$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ | $Q$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ |$W$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$\n",
+    " > :--: | --: | --: | --:\n",
+    " > $A$$_{\\to B \\to}^{\\leftarrow D \\leftarrow}$$C$ | \\_ | \\_ | -1,400.\n",
     "\n",
-    "> $\\Delta U_{A\\to B\\to C\\to D\\to A}=0$\n",
+    ">> $\\Delta U_{A\\to B\\to C\\to D\\to A}=0$\n",
     "\n",
-    "and, then\n",
-    "> $Q_{A\\to B\\to C\\to D\\to A}=\\Delta U_{A\\to B\\to C\\to D\\to A}-W_{A\\to B\\to C\\to D\\to A}$\n"
+    ">> $Q_{A\\to B\\to C\\to D\\to A}=\\Delta U_{A\\to B\\to C\\to D\\to A}-W_{A\\to B\\to C\\to D\\to A}$\n"
    ]
   },
   {
@@ -286,15 +288,15 @@
    "source": [
     "Now using the last fundamental again we can solve for the energy variation in step 2->3\n",
     "\n",
-    "> Steps | $\\Delta U$ (`J`) \n",
+    "> Steps | $\\Delta U$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$\n",
     "> :--: | --: \n",
-    "> A-B | -200. \n",
-    "> B-C | \\_\\_\\_ \n",
-    "> C-D | **-500.**\n",
-    "> D-A | 4,700. \n",
-    "> A-B-C-D-A | 0.\n",
+    "> ${A\\to B}$ | -200.$\\;\\;\\;\\;$ \n",
+    "> ${B\\to C}$ | \\_$\\;\\;\\;\\;$ \n",
+    "> ${C\\to D}$ | -500.$\\;\\;\\;\\;$ \n",
+    "> ${D\\to A}$ | 4,700. $\\;\\;\\;\\;$ \n",
+    "> $A_{\\to B \\to}^{\\leftarrow D \\leftarrow} C$ | 0.$\\;\\;\\;\\;$ \n",
     "\n",
-    "$\\Delta U_{A\\to B} + \\Delta U_{B\\to C} + \\Delta U_{C\\to D} +  \\Delta U_{D\\to A}= 0$"
+    ">> $\\Delta U_{A\\to B} + \\Delta U_{B\\to C} + \\Delta U_{C\\to D} +  \\Delta U_{D\\to A}= 0$"
    ]
   },
   {
@@ -337,11 +339,11 @@
     "according to 1st fundamental\n",
     "\n",
     "\n",
-    "> Steps | $\\Delta U$ (`J`) | $Q$ (`J`) |$W$ (`J`) \n",
+    "> Steps | $\\Delta U$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ | $Q$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ |$W$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$\n",
     "> :--: | --: | --: | --:\n",
-    "> B-C | -4,000. | -3,800. | \\_\\_\\_\n",
+    "> $B$$\\to$$C$ | -4,000. | -3,800. | \\_\n",
     "\n",
-    "$W_{B\\to C} = \\Delta U_{B\\to C} - Q_{B\\to C}$\n",
+    ">> $W_{B\\to C} = \\Delta U_{B\\to C} - Q_{B\\to C}$\n",
     "\n",
     "\n"
    ]
@@ -373,17 +375,17 @@
    "source": [
     "Now using the 2nd and 3rd fundamentals we can solve for the heat and for work in step 4->1\n",
     "\n",
-    "> Steps | $Q$ (`J`) |$W$ (`J`) \n",
+    "> Steps | $Q$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ |$W$ $\\mathrm{(}$$\\mathrm{J}$$\\mathrm{)}$ \n",
     "> :--: | --: | --:\n",
-    "> A-B | 5,800. | -6,000. \n",
-    "> B-C | -3,800. | -200.\n",
-    "> C-D | -800. | 300.\n",
-    "> D-A |  \\_\\_\\_ | \\_\\_\\_\n",
-    "> A-B-C-D-A |  1400. | -1,400.\n",
+    "> $A$$\\to$$B$ | 5,800. | -6,000. \n",
+    "> $B$$\\to$$C$ | -3,800. | -200.\n",
+    "> $C$$\\to$$D$ | -800. | 300.\n",
+    "> $D$$\\to$$A$ |  \\_ | \\_\n",
+    "> $A$$_{\\to B \\to}^{\\leftarrow D \\leftarrow}$$C$ |  1400. | -1,400.\n",
     "\n",
-    ">- $Q_{D\\to A} = Q_{net} - (Q_{A\\to B} + Q_{B\\to C} + Q_{C\\to D})$\n",
+    ">> $Q_{D\\to A} = Q_{net} - (Q_{A\\to B} + Q_{B\\to C} + Q_{C\\to D})$\n",
     "\n",
-    ">- $W_{D\\to A} = W_{net} - (W_{A\\to B} + W_{B\\to C} + W_{C\\to D})$"
+    ">> $W_{D\\to A} = W_{net} - (W_{A\\to B} + W_{B\\to C} + W_{C\\to D})$"
    ]
   },
   {
@@ -878,7 +880,8 @@
    "cell_type": "code",
    "execution_count": 20,
    "metadata": {
-    "collapsed": false
+    "collapsed": false,
+    "scrolled": true
    },
    "outputs": [
     {
@@ -945,8 +948,7 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "#References\n",
-    "`Smith, van Ness, 0000, 00ed. Ex0.`"
+    "# Conclusion\n"
    ]
   }
  ],