Entropy: Why & How Does C_V Go to Zero?

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Entropy is defined by the integral S(A)=∫(0 to T_A) C_V (dT/T), where T_A is the temperature of the system. As temperature approaches absolute zero (T_A→0), the heat capacity at constant volume (C_V) must also approach zero, which raises questions about the rate of this decline. The second law of thermodynamics states that entropy cannot decrease, implying that it will always increase over time. However, in specific conditions where the system is isolated and maintained at very low temperatures, it is possible for C_V to drop rapidly to zero. This highlights the complex relationship between temperature, heat capacity, and entropy in thermodynamic systems.
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Entropy is defined by:

S(A)=\int^{T_A}_0C_V\frac{dT}{T}

where A is state of the system in which temperature is T_A. When T_A\rightarrow 0 and C_V must go to zero. Why? And how fast does it go?
 
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Your question is technically right but entropy is always increasing so it never be fall in zero.The 2nd law of thermodynamics says that (ds/dt)≥ 0. But may be in some special cases if externally keep the temperature very low depends on the state of the system its true and this is a very speedy process that is no interaction between the system and its surroundings so that the temperature is nearly= 0 and Cv falls into zero...
 
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