diff --git a/notes/.obsidian/plugins/obsidian-activity-history/data.json b/notes/.obsidian/plugins/obsidian-activity-history/data.json index 2b6cf24..90b8e41 100644 --- a/notes/.obsidian/plugins/obsidian-activity-history/data.json +++ b/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 } ] } diff --git a/notes/.obsidian/workspace.json b/notes/.obsidian/workspace.json index 2b5d02b..11c303a 100644 --- a/notes/.obsidian/workspace.json +++ b/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, - "breadcrumbs:Breadcrumbs Visualisation": false + "obsidian-excalidraw-plugin:New Excalidraw drawing": false } }, "active": "f48f7dc6676987ee", diff --git a/notes/fyz/Mechanika tekutin/Termodynamika/Ideální plyn.md b/notes/fyz/Mechanika tekutin/Termodynamika/Ideální plyn.md index bbb38f8..0af45f5 100644 --- a/notes/fyz/Mechanika tekutin/Termodynamika/Ideální plyn.md +++ b/notes/fyz/Mechanika tekutin/Termodynamika/Ideální plyn.md @@ -34,4 +34,37 @@ 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. \ No newline at end of file +při stavové změně ideálního plynu se stálou hmotností je výraz $\frac{pV}T$ konstantní. + +--- + +$\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$ + +$p_2=4p_1$ +$V_2=\frac13V_1$ +$t_1=10\degree C$ +$T_1=283K$ +$T_2=?$ + +$\frac{p_1V_1}{283}=\frac{4p_1\frac13V_1}{T_2}$ +$T2=\frac{4p_1\frac13V_1}{\frac{p_1V_1}{283}}$ +$T_2=\frac{283*4_p1\frac13V_1}{p_1V_1}$ +$T_2=283*4*\frac13$ +$T_2=\frac432823$ +$T_2\dot=377K$ +$t_2\dot=104\degree C$ + +--- + +$p_1=0.1MPa$ +$t_1=-23\degree C$ +$p_2=?$ +$V_2=\frac1{10}V_1$ +$t_2=3\degree C$ + +$\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$ +$p_2=\frac{p_1V_1}{T_1}*\frac{T_2}{V_2}$ +$\{p_2\}=\frac{0.1*10^6*V_1}{250}*\frac{276}{\frac1{10}V_1}$ +$p_2=\frac{0.1*10^6*276}{250/10}$ +$p_2=\frac{10^5*276}{25}$ +$p_2=1.1MPa$