15g de un gas cuyo comportamiento es ideal , se halla en un recipiente de 19,5l a 145grados centigrados si la temperatura se disminuye hasta 20 grados centigrados ¿cual es el volumen final ?
Respuestas
Respuesta dada por:
43
MASA Y PRESION SE CONSIDERAN CONSTANTES
V1 = V2
```` `````
T1 T2
V1 = 19.5 L
T1 = 145 + 273 = 418 K
V2 = ?
T2 = 20 + 273 = 293 K
V2 = 19.5 L x 293 K
``````````````````````
418 K
V2 = 13.66 L
``````````````````````````````````````````````````````````````````````````````
NOTA SI LA TEMPERATURA DISMINUYE A 20 ºC ENTONCES SERÍA -20 ºC
V1 = 19.5 L
T1 = 145 + 273 = 418 K
V2 = ?
T2 = -20 + 273 = 253 K
V2 = 19.5 L x 253 K
``````````````````````
418 K
V2 = 11.80 L
V1 = V2
```` `````
T1 T2
V1 = 19.5 L
T1 = 145 + 273 = 418 K
V2 = ?
T2 = 20 + 273 = 293 K
V2 = 19.5 L x 293 K
``````````````````````
418 K
V2 = 13.66 L
``````````````````````````````````````````````````````````````````````````````
NOTA SI LA TEMPERATURA DISMINUYE A 20 ºC ENTONCES SERÍA -20 ºC
V1 = 19.5 L
T1 = 145 + 273 = 418 K
V2 = ?
T2 = -20 + 273 = 253 K
V2 = 19.5 L x 253 K
``````````````````````
418 K
V2 = 11.80 L
Respuesta dada por:
7
Respuesta:
28 litros
Explicación:
Datos
T1: 15°C+273: 288°K
V1: 19.5 litros
T2: 145°C+273: 418°K
V2: ?
V2= 19.5L*418°K/288°K= 28 litros
V2: 28 litros
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