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Kryger [21]
3 years ago
7

.- Una esfera hueca de acero a 28°C tiene un volumen de 0.4 m³, calcular a) ¿qué volumen final tendrá a -6°C en m³ y en litros?

B) ¿Cuánto disminuyó su volumen en litros?
Physics
1 answer:
Lina20 [59]3 years ago
5 0

Answer:

Ve respuesta abajo.

Explanation:

Para hacer esto, podemos asumir que la presión es constante, pues es un proceso adiabatico, por tanto se aplica la ley de Charles a presión constante:

V₁/T₁ = V₂/T₂    (1)

De ahí podemos despejar V₂, ya que conocemos las condiciones iniciales de temperatura y volúmen, y la temperatura final:

V₂ = V₁ T₂ / T₁   (2)

Las temperaturas deben estar en grados Kelvin, y solo es cuestión de sumarle 273 al valor de la temperatura dada en °C:

T₁ = 28 + 273 = 301 K

T₂ = -6 + 273 = 267 K

El volúmen podemos pasarlo a litros ahora o al final. En este caso, podemos dejarlo en m³ como está y luego pasarlo a Litros. Resolviendo tenemos:

V₂ = 0.4 * 267 / 301

<h2>V₂ = 0.35482 m³</h2>

Pasando este volumen a Litros, sabiendo que 1 m³ son 1000 L:

V₂ = 0.35482 * 1000

<h2>V₂ = 354.82 L</h2>

Finalmente para saber cuanto disminuyó el volumen en Litros, pasemos el volumen inicial a Litros y luego se hace la resta con el volumen final:

V₁ = 0.4 * 1000 = 400 L

V₁ - V₂ = 400 - 354.82

<h2>V₁ - V₂ = 45.18 L</h2>

Esto es lo que disminuyó. Espero te ayude.

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