Enthelpy and isothermal, where is my logic wrong

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    Isothermal Logic
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Discussion Overview

The discussion revolves around the concepts of enthalpy and internal energy in the context of isothermal processes for an ideal gas. Participants explore the relationships between these thermodynamic quantities and the implications of work done during such processes.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that for an ideal gas undergoing an isothermal process, the change in enthalpy is zero due to no change in temperature.
  • Another participant challenges the assumption of no change in internal energy, questioning the logic behind it.
  • There is a discussion about the validity of using the integral for enthalpy change, with concerns raised about the conditions under which it applies, specifically regarding constant pressure versus constant temperature.
  • A participant provides definitions and relationships for enthalpy and internal energy, noting that for a monoatomic ideal gas, enthalpy remains unchanged at constant temperature.
  • Participants discuss the implications of work done on the system and the need for heat removal to maintain isothermal conditions, suggesting that without heat removal, the process would not remain isothermal.
  • There is a mathematical exploration of the relationships between differentials of internal energy and enthalpy, with one participant proposing an equation involving pressure and volume changes.
  • Questions are raised about the behavior of valves in relation to enthalpy changes and isothermal processes, with differing views on whether the absence of enthalpy change implies isothermal behavior.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between internal energy and enthalpy in isothermal processes, with no consensus reached on the implications of these relationships or the validity of certain assumptions.

Contextual Notes

Some participants highlight limitations in the assumptions made about ideal gases and the conditions required for applying certain thermodynamic equations, indicating that the discussion may depend on specific system details.

EvilKermit
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Assume ideal gas and isothermal :

\Delta H = \int_{T_{i}}^{T_{f}}\! C_{p} dT = 0
(no change in temperature, no change in enthalpy)

\Delta H = \Delta U + W, U = 0

There is no change in internal energy but there is change in work done. How do these two contradicting statements work.
 
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EvilKermit said:
There is no change in internal energy ...

What makes you think that?
 
Well there is no change in internal energy because you're talking about an ideal gas.

But I don't get how "isothermal => no change in enthalpy" follows... Your integral makes no sense. There is no dT. And you can only use Cp if you're dealing with constant pressure.
 
mr. vodka said:
Well there is no change in internal energy because you're talking about an ideal gas.

What do you mean? The internal energy of an ideal gas is inversely proportional to the volume. So, if there is an expansion (which I am guessing there is since he is talking about "work done"), then the internal energy must be changing, right?
 
H = U + PV is the definition of enthalpy.

A monoatomic ideal gas has U = 3/2 NkT and PV = NkT, so H = 5/2 NkT and you are right that the enthalpy does not change when pressure is applied at constant temperature.

The differential of H is dH = TdS + V dP.
To get your integral equation for delta H you need to divide by dT at constant P, but you are talking about a situation where T is held constant, not P. So your integral is wrong, as it can only be used at constant pressure.

Since we know that dH is zero for an ideal gas at constant temperature, then we must have the relation TdS = -VdP where the left hand side is the differential heat added to the system. If dP is positive so pressure is increasing, then the differential heat is negative, which means that as work is done on the system, heat must be removed in order for it to remain at constant temperature. If heat does not get removed, then you will not have an isothermal process (instead it will be adiabatic).
 
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Alright so

dU = TdS - pdV

dH = TdS + VdP

Set dU and dH to 0. Then

-pdV = Vdp This correct?

Also, if there is no change in enthalpy for a valve, does that mean that valves also work as an isothermal process for an ideal gas?
 
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Yes, which means PV is constant, which you could've gotten out of the gas-law, as PV = nRT = constant
 
EvilKermit said:
Also, if there is no change in enthalpy for a valve, does that mean that valves also work as an isothermal process for an ideal gas?

I'm not really sure what you mean but the answer is probably no. Simply having no change in enthalpy means there is some specific balance between the heat and work. At constant enthalpy, you can say

T \left( \frac{\partial S}{\partial P} \right)_H = -V

which will depend on the details of the system being studied, and will definitely not be ideal gas like unless you really have something which is an ideal gas.
 

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