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  • The entropy change for a process is given by <img alt="\mathrm{\Delta S=2 T

The entropy change \mathrm{(\Delta S)} for a process is given by \mathrm{\Delta S=2 T+0.01 T^2} (in \mathrm{\mathrm{J} / \mathrm{mol} \cdot \mathrm{K})}, where T is the temperature in Kelvin. Determine the temperature at which the process becomes spontaneous.

Option: 1

0 K


Option: 2

100 K


Option: 3

200 K


Option: 4

300 K


Answers (1)

best_answer

Solution: A process becomes spontaneous when the entropy change \mathrm{(\Delta S)} is positive \mathrm{(\Delta S>0)}. Given \mathrm{\Delta S=2 T+0.01 T^2}, we need to find the temperature at which \mathrm{\Delta S} becomes positive:

                                                 \mathrm{ \begin{aligned} 2 T+0.01 T^2 & >0 \\ T(2+0.01 T) & >0 \end{aligned} }

This is true for all temperatures \mathrm{T>0}, so the process becomes spontaneous at any temperature above absolute zero.

So, the correct answer is: D) \mathrm{ 300 \mathrm{~K}}

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