Thermodynamics-a small doubt

vaishakh

In thermodynamics it is said that S = dq/T. Then how can we say that S is directly proportional to T.

LeonhardEuler

That isn't exactly what is said. What is said is that
[tex]dS=\frac{dQ_{rev}}{T}[/tex]
The subscript "rev" means that the quantity dq represents the heat that would be transferred if the process were carried out reversibly.

Also, S is not proportional to T. Suppose you heat a system reversibly at constant pressure. Then we have that
[tex]dQ_{rev}=C_pdT[/tex]
The entropy change can be found by integrating the equation for dS. (You may notice that the as T approaches 0, dS seems to approach infinity. This is not actually the case since in reality the heat capacity is a function of temperature and the Debye extrapolation tells us that at very low temperatures the heat capacity varies like T^3) Suppose the heat capacity is constant over a temperature range of interest. Then
[tex]\Delta S=\int_{T_1}^{T_2}\frac{C_p}{T}dT= C_p \int_{T_1}^{T_2}\frac{1}{T}dT=C_p\ln{\frac{T_2}{_T_1}[/tex]
So in this case - as in most cases - S is not proportional to T, but does increase with incresing T.

vaishakh

by what ratio does S increase with increase in T-Sorry Im confused.

LeonhardEuler

There is no definite ratio-it depends on the situation.