Considering Enthelpy Change when external pressure changes

In summary, in problem 1, we have an ideal gas in a cylinder with a piston. The initial temperature, volume, and pressure of the gas are T1, V, and P1=nRT1/V. The system is put into contact with a constant temperature bath at T2, and the piston is not allowed to move while the system re-equilibrates. The final temperature, volume, and pressure of the gas can be calculated using the ideal gas law. The amount of work done is zero since the volume is constant. The change in internal energy U can be calculated using the first law of thermodynamics, and the heat transferred can be calculated by subtracting
  • #1
sgstudent
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3
We know that ΔH=qp=ΔU+w where w=PextΔV so that means that the enthalpy change is only for constant external pressure processes. So how can we say that in a adiabatic reversible expansion, ΔH=CpΔT? Since the temperature case for that system won't be for a constant pressure system?

And is the work in the equation always based on the external pressure, PextΔV? For example if the internal pressure is 10Pa and the external 20Pa would the work we calculate be 20Pa?
 
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  • #2
sgstudent said:
We know that ΔH=qp=ΔU+w where w=PextΔV so that means that the enthalpy change is only for constant external pressure processes.
This is not correct. It is important to think of enthalpy as a physical property of a system, rather than something associated with any particular kind of process. It is defined as H = U + PV for a system (e.g., a gas) at thermodynamic equilibrium. We use constant pressure processes to measure the change in enthalpy of a system between two thermodynamic equilibrium states at constant pressure by measuring the amount of heat added in the process. For an ideal gas, the enthalpy is independent of the pressure, so once we do that experiment, the results apply at all pressures.
So how can we say that in a adiabatic reversible expansion, ΔH=CpΔT? Since the temperature case for that system won't be for a constant pressure system?
See my previous answer. For an adiabatic reversible expansion, this equation applies only for an ideal gas. For a non-ideal gas, this equation needs to be corrected for the effect of pressure on enthalpy.
And is the work in the equation always based on the external pressure, PextΔV? For example if the internal pressure is 10Pa and the external 20Pa would the work we calculate be 20Pa?
The work is always PextΔV, if Pext represents the external force per unit area applied to the gas, and where it applies at the interface between the system and the surroundings. At this interface, the internal pressure and the external pressure always match one another. For a reversible process, the gas pressure is uniform within the cylinder and is thus equal to Pext. For an irreversible process, the pressure in the cylinder is not uniform (so the internal pressure is not well defined), and, in addition, the external force per unit area Pext also includes a contribution from viscous stress, over and above the thermodynamic pressure.

For more on this, see my Physics Forums Insight article at the following link: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
 
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  • #3
Chestermiller said:
This is not correct. It is important to think of enthalpy as a physical property of a system, rather than something associated with any particular kind of process. It is defined as H = U + PV for a system (e.g., a gas) at thermodynamic equilibrium. We use constant pressure processes to measure the change in enthalpy of a system between two thermodynamic equilibrium states at constant pressure by measuring the amount of heat added in the process. For an ideal gas, the enthalpy is independent of the pressure, so once we do that experiment, the results apply at all pressures.

See my previous answer. For an adiabatic reversible expansion, this equation applies only for an ideal gas. For a non-ideal gas, this equation needs to be corrected for the effect of pressure on enthalpy.

The work is always PextΔV, if Pext represents the external force per unit area applied to the gas, and where it applies at the interface between the system and the surroundings. At this interface, the internal pressure and the external pressure always match one another. For a reversible process, the gas pressure is uniform within the cylinder and is thus equal to Pext. For an irreversible process, the pressure in the cylinder is not uniform (so the internal pressure is not well defined), and, in addition, the external force per unit area Pext also includes a contribution from viscous stress, over and above the thermodynamic pressure.

For more on this, see my Physics Forums Insight article at the following link: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
Sorry for the late reply. Thanks for the detailed reply this clarified a lot of my doubts.

However, there is one part I don't really understand. We measure the enthalpy change by measuring the heat evolved at constant pressure. But even if we were to measure it at different pressure it would be the same for an ideal gas. So wouldn't ΔH=ΔU?
 
  • #4
sgstudent said:
However, there is one part I don't really understand. We measure the enthalpy change by measuring the heat evolved at constant pressure. But even if we were to measure it at different pressure it would be the same for an ideal gas. So wouldn't ΔH=ΔU?
I don't quite follow what your question is driving at. Maybe it would help if you defined an initial- and final thermodynamic equilibrium state of the gas, and we could use that as a basis of discussion. For example, initial state P1, T1 and final state P2 and T2. Would something like this work for you?

Chet
 
  • #5
I have a series of problems for you to work out that I think will help solidify your understanding.

You have an ideal gas contained in a cylinder with a piston.

PROBLEM 1: CONSTANT VOLUME HEATING
The initial temperature, volume, and pressure of the gas are T1, V, and P1=nRT1/V. The cylinder is suddenly put into contact with a constant temperature bath at T2, and the piston is not allowed to move while the system re-equilibrates. What is the final temperature, volume, and pressure of the gas? How much work was done? What is the change in internal energy U? How much heat was transferred? What is the change in enthalpy H?

After you finish this, we can move on to PROBLEM 2.

Chet
 
  • #6
Chestermiller said:
I have a series of problems for you to work out that I think will help solidify your understanding.

You have an ideal gas contained in a cylinder with a piston.

PROBLEM 1: CONSTANT VOLUME HEATING
The initial temperature, volume, and pressure of the gas are T1, V, and P1=nRT1/V. The cylinder is suddenly put into contact with a constant temperature bath at T2, and the piston is not allowed to move while the system re-equilibrates. What is the final temperature, volume, and pressure of the gas? How much work was done? What is the change in internal energy U? How much heat was transferred? What is the change in enthalpy H?

After you finish this, we can move on to PROBLEM 2.

Chet
The final temperature is T2, final volume is still V, final pressure is nRT2/V. There is no work done since the volume does not change. So ΔU=q. So the amount of heat transferred will be based on the heat capacity at constant volume Cv which would be q=Cv(T2-T1).

There is an equation relating the Cv and Cp which is Cp-Cv=R and so Cp=Cv+R multiplying the equation by ΔT gives us ΔH=ΔU+RΔT.

Would this be correct?

To rephrase my initial question, you said that "We use constant pressure processes to measure the change in enthalpy of a system between two thermodynamic equilibrium states at constant pressure by measuring the amount of heat added in the process. For an ideal gas, the enthalpy is independent of the pressure, so once we do that experiment, the results apply at all pressures."

So i was thinking say in a constant volume scenario like in the one you gave, the pressure changes from the initial and final state and you said that enthalpy is independent of the pressure. With that I thought that the amount of heat absorbed/released in the constant volume scenario would be both ΔH and ΔU.
 
  • #7
sgstudent said:
The final temperature is T2, final volume is still V, final pressure is nRT2/V. There is no work done since the volume does not change. So ΔU=q. So the amount of heat transferred will be based on the heat capacity at constant volume Cv which would be q=Cv(T2-T1).

There is an equation relating the Cv and Cp which is Cp-Cv=R and so Cp=Cv+R multiplying the equation by ΔT gives us ΔH=ΔU+RΔT.

Would this be correct?
Not quite. It should be q = nCv(T2-T1)=ΔU. Note the n for the number of moles.

I would get ΔH a slightly different (but equivalent) way: $$ΔH=ΔU+Δ(PV)=ΔU+V(P_2-P_1)=nC_v(T_2-T_1)+nR(T_2-T_1)=nC_p(T_2-T_1)$$
To rephrase my initial question, you said that "We use constant pressure processes to measure the change in enthalpy of a system between two thermodynamic equilibrium states at constant pressure by measuring the amount of heat added in the process. For an ideal gas, the enthalpy is independent of the pressure, so once we do that experiment, the results apply at all pressures."

So i was thinking say in a constant volume scenario like in the one you gave, the pressure changes from the initial and final state and you said that enthalpy is independent of the pressure. With that I thought that the amount of heat absorbed/released in the constant volume scenario would be both ΔH and ΔU.
Well, you can see from our example that it's not. The enthalpy change is independent of the pressure variation (and depends only on the temperature change), but not the heat q.

Would you like to try another problem where we add heat holding the external force per unit area constant at P and see how that scenario plays out?

Chet
 
  • #8
Chestermiller said:
Not quite. It should be q = nCv(T2-T1)=ΔU. Note the n for the number of moles.

I would get ΔH a slightly different (but equivalent) way: $$ΔH=ΔU+Δ(PV)=ΔU+V(P_2-P_1)=nC_v(T_2-T_1)+nR(T_2-T_1)=nC_p(T_2-T_1)$$

Well, you can see from our example that it's not. The enthalpy change is independent of the pressure variation (and depends only on the temperature change), but not the heat q.

Would you like to try another problem where we add heat holding the external force per unit area constant at P and see how that scenario plays out?

Chet
Hmm what do you mean by independent of the pressure variation? Do you mean that when the pressure is held constant and measuring the amount of heat produced, it would be the same no matter what pressure we use - but still the pressure remains constant?

I would like to try the other problem thanks
 
  • #9
sgstudent said:
Hmm what do you mean by independent of the pressure variation?
Actually, what I mean is that, for an ideal gas, both internal energy U and enthalpy H are functions of temperature only, and are independent of pressure all together. U and H are physical properties of the gas. The changes in these equilibrium physical properties are independent of the path between the two end equilibrium states, and depend only on the temperatures of the two end states. On the other hand, q and w depend on the process path between the two states, and are not physical properties of the material.
Do you mean that when the pressure is held constant and measuring the amount of heat produced, it would be the same no matter what pressure we use - but still the pressure remains constant?
This is certainly correct for an ideal gas.
I would like to try the other problem thanks
PROBLEM 2: CONSTANT EXTERNAL FORCE (PRESSURE) HEATING
The initial temperature, pressure, and volume of the gas are T1, P, and V1=nRT1/P. The cylinder is suddenly put into contact with a constant temperature bath at T2, and the piston is controlled to move in such a way that the external force per unit area Pext stays constant at the value P throughout the gas expansion. The expansion continues until the system equilibrates. What is the final temperature, volume, and pressure of the gas? How much work was done? What is the change in internal energy U? How much heat was transferred? What is the change in enthalpy H?
 

Related to Considering Enthelpy Change when external pressure changes

1. What is enthalpy change?

Enthalpy change refers to the amount of energy transferred in a chemical reaction or physical process. It is often measured in joules (J) or kilojoules (kJ).

2. How does external pressure affect enthalpy change?

External pressure can affect enthalpy change by altering the volume of a system. For gases, an increase in external pressure will decrease the volume and therefore increase the enthalpy change. For liquids and solids, the effect of external pressure on enthalpy change is usually negligible.

3. Can enthalpy change be negative?

Yes, enthalpy change can be negative. This means that the chemical reaction or physical process releases energy rather than absorbs it. Negative enthalpy change is often associated with exothermic reactions.

4. How is enthalpy change calculated?

Enthalpy change is calculated by taking the difference between the enthalpies of the products and reactants in a chemical reaction or physical process. It can also be measured experimentally using a calorimeter.

5. What are some real-world applications of considering enthalpy change when external pressure changes?

Considering enthalpy change when external pressure changes is important in industries such as chemical manufacturing, where precise control of temperature and pressure is necessary for efficient and safe production processes. It is also relevant in fields such as thermodynamics, where understanding the relationship between energy and pressure is crucial for designing and improving energy systems.

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