Respuesta:
12. P2 = 2.20 atm
13. V2 = 1.13 Litros
Explicación:
12.
T1 = 15 ºC + 273.15 = 288.15 K
P1 = 2 atm
P2 = ¿?
T2 = 45 ºC + 273.15 = 318.15 K
Aplicar la ley de Gay – Lussac
P1/T1 = P2/T2
P2 = P1 x T2
T1
P2 = 2 atm x 318.15 K
288.15 K
P2 = 2.20 atm
13.
P constante
V1 = 1 L
T1 = 100 ºC + 273.15 = 373.15 K
T2 = 150 ºC + 273.15 = 423.15 K
V2 = ¿?
Aplicar la ley de Charles
V1 / T1 = V2 / T2
V2 = V1 x T2
V2 = 1 L x 423.15 K
373.15 K
V2 = 1.13 Litros
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Respuesta:
12. P2 = 2.20 atm
13. V2 = 1.13 Litros
Explicación:
12.
T1 = 15 ºC + 273.15 = 288.15 K
P1 = 2 atm
P2 = ¿?
T2 = 45 ºC + 273.15 = 318.15 K
Aplicar la ley de Gay – Lussac
P1/T1 = P2/T2
P2 = P1 x T2
T1
P2 = 2 atm x 318.15 K
288.15 K
P2 = 2.20 atm
13.
P constante
V1 = 1 L
T1 = 100 ºC + 273.15 = 373.15 K
T2 = 150 ºC + 273.15 = 423.15 K
V2 = ¿?
Aplicar la ley de Charles
V1 / T1 = V2 / T2
V2 = V1 x T2
T1
V2 = 1 L x 423.15 K
373.15 K
V2 = 1.13 Litros