entropy change and temperature?

Question

change in entropy =change in enthalpy / temperature
From this formula ,entropy change seems to be inversely propostional related with temperature,wheras it is found that on increasing temperature,entropy of system increases.
Kindly explain the contradiction.Thanks

Answer 1

This is not a contradiction.

The enthalpy is the amount of heat created. So it makes sense that entropy increases as heat is created.

This formula merely says that as the temperature increases additional heat generated will cause a smaller increase in entropy.

This also makes sense if you think of entropy as a measure of disorder. As you increase the temperature you are moving towards a maximum state of disorder. Hence it require more effort/heat to produce more disorder.

Answer 2

An alternative description is to view the enthalpy change of condensation as the heat which must be released to the surroundings to compensate for the drop in entropy when a gas condenses to a liquid. As the liquid and gas are in equilibrium at the boiling point (Tb), ΔvG = 0, which leads to:

ΔvS = Sgas − Sliquid = ΔvH/Tb

As neither entropy nor enthalpy vary greatly with temperature, it is normal to use the tabulated standard values without any correction for the difference in temperature from 298 K

Answer 3

Lancenigo di Villorba (TV), Italy

Dear Mrs. Cute, please, follows mine arguments, the following themes.

Energetical studies in Europe of XIX century permit scientist stated existence of a new physic quantity, the "Inner Energy" of a system. This is the former "state's function", that is a physic quantity who is defined by state's parameters of initial and final states, butit depends not by transformation's pathway followed.
Moreover, it stated who "Inner Energy" is quantified like energetical balance by heat and (mechanical) work exchanged with surroundigs. The latter reation can be written in differential amount :
d(E) = d(Q) + d(W)
where E is "Inner Energy", Q and W are heat and work exchanged, finally d(.) is notation for differential operation. This relationship is mathematical expression of "First Principle of Thermodynamics".
A german scientist, R. J. Clausius, stated the "Second Principle of Thermodynamics" which derive by absolute impossibility of some processes who are not executed in any conditions. It is defined another "state's function" as "Entropy" which is related to "caos's level" of system considerated. Briefly, Clausius let us :
d(S) > or = d(Q) / T
where S is "Entropy" and T is temperature measured in Kelvin degrees. The overwritten is not an equation, since if process is not reversible then entropy's differential can be greater than Q, T's fraction.
The american scientist J. W. Gibbs (XIX-XX centuries) overcame to a well-known differential relation, thus the "chemical thermodynamics" can born. The relation whole I refer derive by "Principles of Thermodynamics" new elaboration, as Gibbs did it. Thinking to work exchanged by the simplest mechanical system (pure gas compressed by a plug in a tube), you can surely writes :
d(W) + P*d(V) < or = 0
where P is gas's pressure (e.g. plug's pressure) and V is gas's volume (e.g. tube volume occupied) ; you note "minus or equal" symbols, since if process is not reversible then you have not an equation. Gibbs united the following relations :
d(E) = d(Q) + d(W) (1st P. T.)
T*d(S) > or = d(Q) (2nd P. T.)
d(W) < or = 0 - P*d(V) (mechanical behaviour)
and he obtained
d(E) < or = T*d(S) - P*d(V)
Gibbs gave a mathematical relation for previding "Inner Energy" starting from state's parameters, among the latters is "Entropy" as generalizated one.
In the succeeding times, Gibbs introduced "Enthalpy" like "thermodynalic potential" as it is known. "Enthalpy" derives by "Inner Energy" as follows :
d(H) = d(E) + d(P*V)
where H is "Enthalpy".
By developing previous disequations, you overcome to :
d(H) < or = T*d(S) + V*dP
HOW ARE RELATED ENTHALPY AND ENTROPY?
Now, you have the answer.
Iso-P processes lead to :
d(H) < or = T*d(S) (iso-P)
thus reversible (and iso-P) processes respect your simple prevision, but not-reversible processes state an upper limit for enthalpy's exchanges, limit entropy's governed.
These suggestions derive by "S, P-expression of Enthalpy" ; it exists also "T, P-expression of Enthalpy". The latter is obtained by several steps, in first place an rearrangement
d(H) = (V + T*((d(S)/d(P)))*dP
hence an expansion
d(H) = Cp*dT + (V + T*((dS/d(P))(iso-T))*dP
where ((dS)/d(P))(iso-T) is an partial derivative of Entropy.
An estimate of the latter going by one among the Maxwell's realtions (J. C. Maxwell, british scientist), the one who follows :
d(H) = Cp*dT+ V*(alpha-1)*d(P)
where "alpha" is "volumic expansion" of system, one of its properties.
If you consider again iso-P processes, you can see that the greater temperature, the greater enthalpy's exchange, as you wrote. Entropy's exchanges are not inversely proportional to exchange's temperature, generally entropy and enthalpy's exchanges are proportional both to temperature.

I hope this helps you.

Answer 4

Second Law of Thermodynamics. I think, maybe third..... No contradiction, law vs equation = law. When an equation results in a law breaking we toss the equation.