Enthelpy and isothermal, where is my logic wrong

In summary, in an isothermal process with an ideal gas, there is no change in temperature or internal energy. However, work can still be done, causing a change in enthalpy. This is because the internal energy of an ideal gas is inversely proportional to volume. The differential of enthalpy is also dependent on temperature and pressure, and can only be used at constant pressure. Therefore, in an isothermal process, heat must be removed as work is done in order to maintain a constant temperature. Valves do not necessarily work as an isothermal process for an ideal gas, as there are other factors at play.
  • #1
EvilKermit
23
0
Assume ideal gas and isothermal :

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

[tex]\Delta H = \Delta U + W, U = 0[/tex]

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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  • #2
EvilKermit said:
There is no change in internal energy ...

What makes you think that?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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 [tex]dH = TdS + V dP[/tex].
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 [tex]TdS = -VdP[/tex] 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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  • #6
Alright so

[tex]dU = TdS - pdV[/tex]

[tex]dH = TdS + VdP[/tex]

Set dU and dH to 0. Then

[tex]-pdV = Vdp[/tex] 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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  • #7
Yes, which means PV is constant, which you could've gotten out of the gas-law, as PV = nRT = constant
 
  • #8
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

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

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.
 

Related to Enthelpy and isothermal, where is my logic wrong

1. What is the difference between enthalpy and isothermal?

Enthalpy is a measure of the total energy of a thermodynamic system, including both its internal energy and the work it can do on its surroundings. Isothermal refers to a process where the temperature remains constant throughout. So, while enthalpy is a property of a system, isothermal is a process.

2. Why is enthalpy important in thermodynamics?

Enthalpy is important in thermodynamics because it helps us understand and predict how energy is exchanged between a system and its surroundings during a process. It also allows us to calculate the work done and heat transferred in a system.

3. Can enthalpy and isothermal coexist in a system?

No, enthalpy and isothermal cannot coexist in a system. Enthalpy is a property of a system, while isothermal is a process. A system can have a constant temperature (isothermal), but its enthalpy will change if work is done on or by the system.

4. How do enthalpy and isothermal affect chemical reactions?

Enthalpy and isothermal can affect chemical reactions in different ways. Enthalpy change (ΔH) is a measure of the heat released or absorbed during a chemical reaction, while isothermal conditions can influence the rate of the reaction. In an isothermal process, the temperature remains constant, so the energy input is only used to drive the reaction forward, rather than increasing the temperature of the system.

5. What is an example of a real-life application of enthalpy and isothermal?

One example of a real-life application of enthalpy and isothermal is in refrigeration systems. These systems use the enthalpy of a refrigerant to absorb heat from inside a space and release it outside, resulting in a cooling effect. Isothermal conditions are also important in refrigeration to ensure that the cooling process is consistent and efficient.

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