Physics Asked by Daniel Gusmão on May 1, 2021
Suppose that there exists an equation of state given by $PV=ST$ ( Notice that the units are correct). What would be the implication of this? Is there something that prohibits this from being true? Please let me know.
Imagine that PV is constant, then the entropy S would decrease with the increase of temperature. That's completely erroneous, due to the fact that increasing the temperature, decrease entropy. Plus, if entropy decrease with the increase of temperature, the equipartition of energy would tear apart.
Answered by Almighty on May 1, 2021
Your equation $$PV=ST$$
pass the criterion to be a state equation. But I can't say that such a system does exist or not. I have added implication in the last. In sort
The third law of thermodynamics states as follows, regarding the properties of closed systems in thermodynamic equilibrium:
The entropy of a system approaches a constant value as its temperature approaches absolute zero.
Your equation disagrees with this.
Reasons for passing as state :
An equation of state is a thermodynamic equation relating to state variables that describe the state of matter under a given set of physical conditions, such as pressure, volume, temperature ($PVT$), or internal energy.
The number of state variables required to specify the thermodynamic state depends on the system, and is not always known in advance of the experiment; it is usually found from experimental evidence. Always the number is two or more; usually it is not more than some dozen. Though the number of state variables is fixed by experiment, there remains the choice of which of them to use for a particular convenient description; a given thermodynamic system may be alternatively identified by several different choices of the set of state variables.
The relationship expressed as increments of entropy equal to the ratio of incremental heat transfer divided by temperature, which was found to vary in the thermodynamic cycle but eventually returns to the same value at the end of every cycle. Thus it was found to be a function of the state, specifically a thermodynamic state of the system.
Is entropy of all systems zero at absolute zero?
Not quite. Some systems can be in their ground state and still have a nontrivial state. But It's possible! (at least as an approximation) Per the definition, a perfect crystal at $0 K$ would have an entropy $S=0$.
Implications
Such a system seems to be impossible (or highly improbable) due to reason of the defining properties of entropy
$$S=k_BlnOmega$$
as you know that you would expect the number of states will increase if you increase the temperature and so the entropy but your formula suggest that If you kept $PV$ to be constant $$Spropto frac{1}{T}$$
which is not physically possible!
The third law of thermodynamics states as follows, regarding the properties of closed systems in thermodynamic equilibrium:
The entropy of a system approaches a constant value as its temperature approaches absolute zero.
Your equation disagrees with this.
Answered by Young Kindaichi on May 1, 2021
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