A) \[-T\frac{{{d}^{2}}V}{d{{T}^{2}}}\]
B) \[{{T}^{2}}\frac{dV}{dT}\]
C) \[-\frac{1}{T}\frac{{{d}^{2}}V}{d{{T}^{2}}}\]
D) \[-\frac{1}{{{T}^{2}}}\frac{dV}{dT}\]
Correct Answer: A
Solution :
The Thomson coefficient is given by \[\mu =-T\frac{dS}{dt}\] Now, S = Seeback coefficient \[=\frac{dV}{dT}\] Hence, \[\mu =-\frac{T{{d}^{2}}V}{d{{T}^{2}}}\]
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