Thermodynamics: derivation for q @ constant P

1. Oct 17, 2014

iScience

general equation of q in terms of S,T

$$q=d(ST)=SdT+TdS$$

deivation of ΔS at constant pressure(in terms of heat cap C_p:

$$dq=C_{p}dT=TdS$$

$$\frac{C_{p}}{T}dT=dS$$

$$C_{p}ln(T_{f}/T_{i}=ΔS$$

why do we keep T constant on TdS side?

2. Oct 17, 2014

Philip Wood

Maybe you're not getting replies because others are as baffled as I am by your very first equation. Or maybe I'm just being stupid. Could you please justify your first equation?

3. Oct 18, 2014

iScience

oh, sorry.

well in my thermo class we used to use dq=TdS for heat. in certain equations though id seethe term "SdT", whenever i see situations like this i assume dq is actually the derivative of ST ie d(ST) and that it was due to some special case that we were able to ignore the SdT term.

but if TdS is dq and SdT does not come from q, then what is SdT?

4. Oct 19, 2014

stevendaryl

Staff Emeritus
I'm not sure what kind of answer you are looking for. S dT is an inexact differential thermodynamic quantity with no particular name, as far as I know.

The difference between an exact differential and an inexact differential is that an exact differential can be written as $dX$ for some function of state $X$. $dq$ is not an exact differential, because $q$ is not a function of state. The amount of heat you put into a system to end up at a particular state depends on how you get to that state. Similarly, $dW$, the amount of work done on a system, is not an exact differential, either. In contrast, the temperature, the entropy, the pressure, the volume, the internal energy, etc. are functions of state, and so their differentials are exact differentials.

$T dS$ is not an exact differential, and neither is $S dT$, but the sum is an exact differential: $T dS + S dT = d(ST)$

The only nonexact differentials that I know of that have names are work and heat:
$dW = -P dV$ and $dq = T dS$

Last edited: Oct 19, 2014
5. Oct 20, 2014

Staff: Mentor

dq is equal to TdS only for a reversible process.

Chet