vault backup: 2023-01-17 12:35:02

notes/.obsidian/plugins/obsidian-activity-history/data.json

@@ -13,7 +13,7 @@
     {
       "path": "/",
       "date": "2023-01-17",
-      "size": 955685
+      "size": 956132
     }
   ],
   "activityHistory": [
@@ -1038,7 +1038,7 @@
         },
         {
           "date": "2023-01-17",
-          "value": 1160
+          "value": 1607
         }
       ]
     }

notes/.obsidian/workspace.json

@@ -225,8 +225,7 @@
       "command-palette:Open command palette": false,
       "markdown-importer:Open format converter": false,
       "table-editor-obsidian:Advanced Tables Toolbar": false,
-      "obsidian-excalidraw-plugin:New Excalidraw drawing": false,
+      "obsidian-excalidraw-plugin:New Excalidraw drawing": false
-      "breadcrumbs:Breadcrumbs Visualisation": false
     }
   },
   "active": "f48f7dc6676987ee",

notes/fyz/Mechanika tekutin/Termodynamika/Ideální plyn.md

@@ -35,3 +35,36 @@ porovnávají dva stavy téhož plynu.
 $\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}=konst.$
 
 při stavové změně ideálního plynu se stálou hmotností je výraz $\frac{pV}T$ konstantní.
+
+---
+
+$\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$
+
+$p_2=4p_1$
+$V_2=\frac13V_1$
+$t_1=10\degree C$
+$T_1=283K$
+$T_2=?$
+
+$\frac{p_1V_1}{283}=\frac{4p_1\frac13V_1}{T_2}$
+$T2=\frac{4p_1\frac13V_1}{\frac{p_1V_1}{283}}$
+$T_2=\frac{283*4_p1\frac13V_1}{p_1V_1}$
+$T_2=283*4*\frac13$
+$T_2=\frac432823$
+$T_2\dot=377K$
+$t_2\dot=104\degree C$
+
+---
+
+$p_1=0.1MPa$
+$t_1=-23\degree C$
+$p_2=?$
+$V_2=\frac1{10}V_1$
+$t_2=3\degree C$
+
+$\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$
+$p_2=\frac{p_1V_1}{T_1}*\frac{T_2}{V_2}$
+$\{p_2\}=\frac{0.1*10^6*V_1}{250}*\frac{276}{\frac1{10}V_1}$
+$p_2=\frac{0.1*10^6*276}{250/10}$
+$p_2=\frac{10^5*276}{25}$
+$p_2=1.1MPa$