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Para CHCl 3 (g) C p = 7,05 + 35,60 x 10 -3 T - 216,9 x 10 -7 T 2 cal mol -1 grado

Para CHCl 3 (g) C p = 7,05 + 35,60 x 10 -3 T - 216,9 x 10 -7 T 2 cal mol -1 grado -1. Suponiendo que este gas sea ideal, calcule el cambio de entropa que implica calentar 2 moles del gas de un volumen de 100 litros a 500 K a un volumen de 70 litros a 700 K.

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