En un recipiente se tienen 16,4 litros de un gas ideal a 116ºF y una presión de 1 atm. Si el gas se expande hasta ocupar un volumen de 22000ml, y la presión se reduce a 0.8 atm. ¿Cuál será la temperatura final del sistema?
a. 273 ºK
b. 441,7 ºK
c. 220,85 ºK
d. 441,7 ºC
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ley general de los gases
V1 · P1 = V2 · P2
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T1 T2
V1 = 16.4 L
P1 = 1 atm
T1 = 116 ºF
TRANSFORMAR ºF a ºC
ºC = ºF - 32
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1.8
ºC = 116 - 32 / 1.8
ºC = 46.67
transformar ºC a K
K = ºC + 273
K =46.67 + 273
K = 319.67
V2 = 22000 mL / 1000 = 22 L
P2 = 0.8 atm
T2 = ?
T2 = 22 L · 0.8 atm · 319.67 K
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16.4 L · 1 atm
T2 = 343.06 K
V1 · P1 = V2 · P2
`````````` `````````````
T1 T2
V1 = 16.4 L
P1 = 1 atm
T1 = 116 ºF
TRANSFORMAR ºF a ºC
ºC = ºF - 32
````````````
1.8
ºC = 116 - 32 / 1.8
ºC = 46.67
transformar ºC a K
K = ºC + 273
K =46.67 + 273
K = 319.67
V2 = 22000 mL / 1000 = 22 L
P2 = 0.8 atm
T2 = ?
T2 = 22 L · 0.8 atm · 319.67 K
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16.4 L · 1 atm
T2 = 343.06 K
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