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natita [175]
2 years ago
10

Una cantidad de aire se lleva del estado a al estado b siguiendo una trayectoria recta en una grafica PV como es mostrado en la

figura. Suponga que la expansión es desde un volumen de 0.06 metros cúbicos a un volumen de 0.13 metros cúbicos y que la presión aumenta desde 125,432 pascales a 168,793 pascales.¿Cuánto trabajo efectúa el gas en este proceso?
Physics
1 answer:
allochka39001 [22]2 years ago
8 0

Answer:

El trabajo de frontera ejercido por el gas es 10,297.875 joules.

Explanation:

Supongamos que el gas se comporta idealmente y que el proceso es isotérmico y adiabático. El trabajo efectuado por el gas en el proceso equivale al área bajo la curva de la gráfica PV. Puesto que es una línea recta y que tanto la presión como el volumen son variables positivas, tenemos que esa área es la suma del rectángulo inferior (límite inferior de la presión) y el triángulo superior (entre los límites inferior y superior de la presión). Es decir:

W = P_{min}\cdot (V_{2}-V_{1})+\frac{1}{2}\cdot (P_{max}-P_{min})\cdot (V_{2}-V_{1}) (1)

Donde:

W - Trabajo de frontera ejercido por el gas, medida en joules.

V_{1}, V_{2} - Volúmenes del gas antes y después de la expansión, medidas en metros cúbicos.

P_{min} - Límite inferior de la presión del gas, medida en pascales.

P_{max} - Límite superior de la presión del gas, medida en pascales.

Si sabemos que V_{1} = 0.06\,m^{3}, V_{2} = 0.13\,m^{3}, P_{min} = 125,432\,Pa y P_{max} = 168,793\,Pa, entonces el trabajo de frontera ejercido por el gas es:

W = (125,432\,Pa)\cdot (0.13\,m^{3}-0.06\,m^{3})+\frac{1}{2}\cdot (168,793\,m-125,432\,m) \cdot (0.13\,m^{3}-0.06\,m^{3})

W = 10,297.875\,J

El trabajo de frontera ejercido por el gas es 10,297.875 joules.

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