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Hi, is the molar Gibbs energy in this formula equivalent to chemical potential (μ)? As my lecturer showed us the same formula but with μ. I know that chemical potential is equal to the partial molar Gibbs free energy, so does that mean partial molar Gibbs free energy and molar Gibbs free energy is the same? Thank you
Yes, the same formula works with μ instead of G. The chemical potential is *partial* molar Gibbs energy, as you point out. But, for a system with only a single component, ∂G/∂n is the same as G/n. So you're right: partial molar Gibbs energy is the same as molar Gibbs energy. (As long as there is only a single component. Things get more complicated in mixtures.)
I think the standard molar Gibbs free energy needs to be at the same temperature T used here, and not at a standard, fixed, T₀: ∆G = G(T, p) - G(T, p₀) = RTln(p/p₀) Basically I need to be careful that ∆ means "variation only in pressure" here. Now G(T, p₀) will vary with T. It is not G(T₀, p₀) that I can read from a table where, for example, T₀ = 298.15 K and p₀ = 1 bar.
Yes, that's all correct. This result is for changing P at constant T. The G(P) and G° must be at the same temperature. "Standard" state can be confusing. Usually (as here), the "standard state" only implies 1 atm pressure. It does not imply anything in particular about the temperature. It is common, but incorrect, to assume that "standard state" also means 298.15 K or 273.15 K or some other particular temperature. I think this confusion probably arises from the fact that general chemistry students are often encouraged to work at "standard temperature and pressure", or STP. And, as you point out, it is common to tabulate G° values at 298.15 K.
Here are a few: (1) Which has the *larger* Gibbs energy: a gas at a pressure of 1.0 atm, or the same amount of the gas at the same temperature but a pressure of 2.0 atm? (2) By how much does the molar Gibbs energy of air (treated as an ideal gas) change when you compress it isothermally from 1.0 atm to 2.0 atm? (3) Using a standard state of 1 bar, and defining the standard-state Gibbs energy (G°) to be zero, what is the Gibbs energy of 2.0 moles of an ideal gas at a pressure of 1.0 Torr? 1.0 MPa?
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Sir I'm watching your awesome videos one by one, the only thing I can say is your are super incredible and amazing. You are wonderful. Great job, you are definitely one in million. Interesting videos. You are a very well educated professor 👏. I really appreciate you from the bottom of my heart and thank you so much for your very useful lessons. I study chemical engineering and thermodynamics, unit operations, fluid mechanics, mass/heat transfer are my main lessons. I really appreciate you for what you have done to teach us and I wish you health. I love you from the bottom of my heart and love your magical videos. You are fabulous. Please please please make more videos. Thank you soooo much 🌹🌹🌹🌹🌹🌹
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I'm glad you found it. Thanks for the comment
Hi, is the molar Gibbs energy in this formula equivalent to chemical potential (μ)? As my lecturer showed us the same formula but with μ.
I know that chemical potential is equal to the partial molar Gibbs free energy, so does that mean partial molar Gibbs free energy and molar Gibbs free energy is the same? Thank you
Yes, the same formula works with μ instead of G.
The chemical potential is *partial* molar Gibbs energy, as you point out. But, for a system with only a single component, ∂G/∂n is the same as G/n. So you're right: partial molar Gibbs energy is the same as molar Gibbs energy. (As long as there is only a single component. Things get more complicated in mixtures.)
@@PhysicalChemistry Thanks! I understand now.
I think the standard molar Gibbs free energy needs to be at the same temperature T used here, and not at a standard, fixed, T₀:
∆G = G(T, p) - G(T, p₀) = RTln(p/p₀)
Basically I need to be careful that ∆ means "variation only in pressure" here.
Now G(T, p₀) will vary with T. It is not G(T₀, p₀) that I can read from a table where, for example, T₀ = 298.15 K and p₀ = 1 bar.
Yes, that's all correct. This result is for changing P at constant T. The G(P) and G° must be at the same temperature.
"Standard" state can be confusing. Usually (as here), the "standard state" only implies 1 atm pressure. It does not imply anything in particular about the temperature.
It is common, but incorrect, to assume that "standard state" also means 298.15 K or 273.15 K or some other particular temperature. I think this confusion probably arises from the fact that general chemistry students are often encouraged to work at "standard temperature and pressure", or STP. And, as you point out, it is common to tabulate G° values at 298.15 K.
@@PhysicalChemistry Thanks
do you have any sample problem for this?
Here are a few:
(1) Which has the *larger* Gibbs energy: a gas at a pressure of 1.0 atm, or the same amount of the gas at the same temperature but a pressure of 2.0 atm?
(2) By how much does the molar Gibbs energy of air (treated as an ideal gas) change when you compress it isothermally from 1.0 atm to 2.0 atm?
(3) Using a standard state of 1 bar, and defining the standard-state Gibbs energy (G°) to be zero, what is the Gibbs energy of 2.0 moles of an ideal gas at a pressure of 1.0 Torr? 1.0 MPa?