Practica No 1 Diagrama PT de Una Sustancia Pura
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Transcript of Practica No 1 Diagrama PT de Una Sustancia Pura
Practica No 1 diagrama PT de una sustancia puraObjetivos: A travs de un experimento el estudiante obtendr valores de presin y temperatura del agua en equilibrio con su vapor, para trazar el diagrama presin-temperatura y compararlos con los de las tablas de vapor saturado agua Tabla de datos experimentales hT= 1.27mT amb= 28CHbarom=0.585 m Hg
Temperatura (t)(en C)Altura al sistema (hs)(en cm)Altura a la atmosfera (ha)(en cm)
850.510.47
800.5540.43
750.5900.38
700.6350.348
650.6550.324
600.6740.304
Calculos 1.-Calcula la densidad del mercurio (Hg) a la temperatura Tamb.Hg=13595.08-2.466(t) + 0.0003 (t2)Hg=13595.08-2.466(28C) + 0.0003 (28C) 2=13526.2672 Kg/m3 2.-Calcula la presin baromtrica (Pbarom)Pbarom=( Hg)(g loc) (hbarom)Pbarom=( 13526.26.72 Kg/m3)(9.78m/s2) (0.585 m)=77387.791Pa3.- Calcula los valores de la presin hidrosttica del mercurio (PHg) en PaPHg= (Hg) )(g loc) (hs- ha)PHg= (13526.2672Kg/m3) )( 9.78m/s2) (0.0051m 0.0047m)=52.914PHg= (13526.2672 Kg/m3) )( 9.78m/s2) (0.00554m 0.0043m)=164.035PHg= (13526.2672 Kg/m3) )( 9.78m/s2) (0.00590m 0.0038m)=277.802PHg= (13526.2672 Kg/m3) )( 9.78m/s2) (0.00635m 0.00348m)=411.412PHg= (13526.2672 Kg/m3) )( 9.78m/s2) (0.00655m 0.00324m)=437.869PHg= (13526.2672 Kg/m3) )( 9.78m/s2) (0.00674m 0.00304m)=489.4614.- Calcula la densidad del agua (H2O) a la temperatura ambiente y transfrmala a Kg/m3.
PH2O=0.99998 + 3.5 x 10-5 t amb 6.0 x 10-6 t amb2
PH2O=0.99998 + 3.5 x 10-5 (28 C) 6.0 x 10-6 (28C)2= 0.9962 g/mL0.9962 g/mL()3 = 996.2Kg/m3 5.- Calcula los valores de la presin hidrosttica del agua (PH2O) en Pa.H2O= PH2o g loc (ht- hS)PH2O= (996.2Kg/m3)( 9.78m/s2) (0.0051m 0.0047m)=3.897PH2O= (996.2Kg/m3) 9.78m/s2) (0.00554m 0.0043m)= 12.081PH2O= (996.2Kg/m3 )( 9.78m/s2) (0.00590m 0.0038m)= 20.459PH2O= (996.2Kg/m3) ( 9.78m/s2) (0.00635m 0.00348m)= 27.961PH2O= (996.2Kg/m3) ( 9.78m/s2) (0.00655m 0.00324m)= 32.248PH2O= (996.2Kg/m3)( 9.78m/s2) (0.00674m 0.00304m)= 36.0486.-Calcula los valores de la presin de vaco (Pvac) dentro del matraz en PaPvac= PH2o+ PhgPvac= 3.897+ 41.9856=45.8826PaPvac= 12.081+ 164.035= 176.116PaPvac= 20.459 + 277.802= 298.261PaPvac= 27.916 + 411.412= 439.328PaPvac= 32.248 + 437.869= 470.117PaPvac= 36.048 + 489.461=525.509Pa
7.-Calcula los valores de la presin absoluta ( Pabs) dentro del matraz en PaPabs= Pbarom PvacPabs= 77387.791Pa 45.8826Pa=77341.9084 PaPabs= 77387.791Pa 176.116Pa=77211.675PaPabs= 77387.791Pa 298.261Pa=77089.53PaPabs= 77387.791Pa 439.328Pa =77000.463 PaPabs= 77387.791Pa 470.117Pa=76917.674 PaPabs= 77387.791Pa 525.509Pa=76862.282 Pa8.- obtn en las tablas de vapor de agua saturada la presin (PTabla) para cada valor experimental de temperatura.Temperatura(t) CPTabla (bar)PTabla (Pa)
850.5783057830
800.4739047390
750.3858038580
700.3119031190
650.2503025030
600.1994119941
9.- calcula el porcentaje de erro de la presin absoluta (Pabs)%E= x 100%E= x 100=3.3740%E= x 100=6.2928%E= x 100=9.981%E= x 100=14.6875%E= x 100=20.730%E= x 100=28.544
Tabla de resultados P Hg(Pa)Ph2o(Pa)P vac(Pa)Pabs(Pa)%E de p abs
52.9143.89745.8826Pa77341.93.374%
164.03512.081176.116Pa77211.6756.2928%
277.80220.459298.261Pa77089.539.981%
411.41227.961439.328Pa77000.414.6875%
437.86932.248470.117Pa76917.620.73%
489.46136.048525.509Pa76862.28228.544%