Question

to 2.000 kJ at 350. K. Assuming both the enthalpy and the entropy changes for the process are independent of temperature between 200.K and 400.K, determine the temperature in kelvin where this process occurs reversibly. Hint: You will find it easier to solve this problem by first calculating AH, AS for the process.

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Answer #1

\Delta G = \Delta H - T \Delta S

\\ \: \\ \: \Delta G =4000 \: J \\ \: \\ \: T = 400. K \\ \: \\ \: \Delta G = \Delta H - T \Delta S \\ \: \\ \: 4000 \: J = \Delta H - (400. K) \times \Delta S......(1)

\\ \: \\ \: \Delta G =2000 \: J \\ \: \\ \: T = 350. K \\ \: \\ \: \Delta G = \Delta H - T \Delta S \\ \: \\ \: 2000 \: J = \Delta H - (350. K) \times \Delta S......(2)

Substract equation (2) from equation (1)

\\ \: \\ \: 4000 \: J-2000 \: J = \Delta H - (400. K) \times \Delta S - \left [ \Delta H - (350. K) \times \Delta S \right ]

\\ \: \\ \: 2000 \: J = - (400. K) \times \Delta S+ \left [ (350. K) \times \Delta S \right ]

\\ \: \\ \: 2000 \: J = - (50. K) \times \Delta S

\\ \: \\ \: \Delta S=-40 \: J/K....(3)

Substitute equation (3) in equation (1)

\\ \: \\ \: 4000 \: J = \Delta H - (400. K) \times \Delta S \\ \: \\ \: 4000 \: J = \Delta H - (400. K) \times (-40 \: J/K) \\ \: \\ \: 4000 \: J = \Delta H + 16000. J \\ \: \\ \: \Delta H =- 12000. J

\\ \: \\ \: T = \dfrac {\Delta H}{ \Delta S} =\dfrac {- 12000. J}{-40 \: J/K} =300 K

Hence, the temperature at which process occurs reversibly is 300 K.

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