Thermodynamics: derivation for q @ constant P

In summary, the general equation of q in terms of S,T states that q=d(ST)+SdT. 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. S dT is not an exact differential, because q is not a function of state. However, the sum of S dT and T dS is an exact differential: T dS + S dT = d(ST).
  • #1
iScience
466
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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?
 
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  • #2
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
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.
bad assumption i guess.

but if TdS is dq and SdT does not come from q, then what is SdT?
 
  • #4
iScience said:
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.
bad assumption i guess.

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

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 [itex]dX[/itex] for some function of state [itex]X[/itex]. [itex]dq[/itex] is not an exact differential, because [itex]q[/itex] 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, [itex]dW[/itex], 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.

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

The only nonexact differentials that I know of that have names are work and heat:
[itex]dW = -P dV[/itex] and [itex]dq = T dS[/itex]
 
Last edited:
  • #5
dq is equal to TdS only for a reversible process.

Chet
 

FAQ: Thermodynamics: derivation for q @ constant P

1. What is the derivation for q at constant pressure in thermodynamics?

The derivation for q at constant pressure is based on the First Law of Thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (q) added to the system minus the work (w) done by the system. At constant pressure, the work done is equal to the pressure (P) multiplied by the change in volume (ΔV), so the equation becomes ΔU = q - PΔV. Rearranging this equation gives us the derivation for q at constant pressure: q = ΔU + PΔV.

2. How is the derivation for q at constant pressure different from the derivation at constant volume?

The derivation for q at constant pressure and the derivation for q at constant volume are based on the same equation: ΔU = q - PΔV. However, at constant volume, there is no change in volume (ΔV = 0), so the equation simplifies to ΔU = q. This means that at constant volume, all of the heat added to the system goes towards increasing the internal energy, while at constant pressure, some of the heat is used to do work by expanding the system.

3. How does the derivation for q at constant pressure relate to enthalpy?

The derivation for q at constant pressure is closely related to the concept of enthalpy (H), which is defined as H = U + PV. At constant pressure, the change in enthalpy (ΔH) is equal to the heat added to the system, as the work term (PΔV) cancels out. This means that the derivation for q at constant pressure can also be written as q = ΔH.

4. Can the derivation for q at constant pressure be applied to all thermodynamic processes?

No, the derivation for q at constant pressure can only be applied to processes where the pressure remains constant. This is because the derivation is based on the assumption that the work done by the system is equal to the pressure times the change in volume. If the pressure is not constant, this assumption is no longer valid and the derivation cannot be used.

5. How is the derivation for q at constant pressure used in practical applications?

The derivation for q at constant pressure is used in various practical applications, such as in the design of heat exchangers and in the calculation of the enthalpy of chemical reactions. It is also used in the analysis of thermodynamic cycles, such as the Carnot cycle and the Brayton cycle. By understanding the derivation, scientists and engineers can accurately calculate and predict the heat transfer and energy changes in a system at constant pressure.

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