Hi Steven, great video as always! A few questions: 1) At 1:55 I think it's ok to take just the derivative of V and not of the product pV because p is a natural variable of H. If I understand natural variables correctly, I should think of H as function of S, p and ni like this: H = H(S, p, ni) = U(S, V(p, ni), ni) + p*V(p, ni) Here in U I cannot consider V as an independent variable anymore, so the partial derivative of U with respect to ni should be dU/dV * dV/dni + dU/dni which is different from the partial derivative of U wrt ni when I think U in terms of its natural variables: U=U(S, V, ni). This is tricky. 2) In a similar way, at 5:31, because (p,T) are natural variables for G I don't need their derivative wrt ni. 3) Is the p at 5:59 the partial pressure rather than the total pressure of the system? 4) I suppose the fundamental relations in terms of the partial molar quantities work because the partial derivative of a differential is the differential of the partial derivative. Ex: f(x, y) = (x^3)(y^2) derivative of the differential: df(x, y, dx, dy) = 3(x^2)(y^2)dx + 2(x^3)(y)dy [the differential is a function of 4 arguments] (d/dx)df(x, y, dx, dy) = 6x(y^2)dx + 6(x^2)(y)dy differential of the derivative: df/dx(x, y) = 3(x^2)(y^2) d(df/dx)(x, y, dx, dy) =6x(y^2)dx + 6(x^2)(y)dy.
Thinking about this one more time, I'm not sure the pressure at 5:59 is the partial pressure. For ideal gases V(p, T, n1, n2,...) = ∑_i niRT/p = nRT/p so ∂V/∂ni = RT/p and integrating: μi = μi0 + RTln(p/p0) So it looks like p is the total pressure. Where is my mistake?
Your ∂V/∂nᵢ equation is fine. But that's partial molar volume, not partial molar Gibbs energy. Here's a thought experiment that may help make it seem reasonable that partial pressure is correct: Imagine two containers of gas, separated by a stopcock. Both are at 3 atm pressure. On the left side, the partial pressure of gas A is 2 atm and the partial pressure of gas B is 1 atm. On the right side, it's the opposite: A has partial pressure of 1 atm and B is 2 atm. Now, open the valve. Gas A will move from the left to the right, because its chemical potential is higher on the left. This is due to its higher *partial* pressure; the total pressure is the same on both sides. Now imagine these two containers in a different scenario: Both have 1 atm of A. But the left side has 2 atm of B, for a total pressure of 3 atm. The right has only 1 atm of B, for a total pressure of 2 atm. When the stopcock is opened, even though gas flows from left to right, this will be gas B, and there will be no (net) flow of gas A. This is because its chemical potential is the same on both sides. This is because its chemical potential is the same on both sides, even though the total pressure is different.
@@PhysicalChemistry Thanks, now I understand it qualitatively, but I can't figure out the mistake. I have integrated the partial molar volume as we know that ∂μᵢ/∂p = Vᵢ (partial molar volume). Probably my mistake is in defining μᵢ⁰. For the ideal gas mixture I think it's true that G(T,p,𝐧) - G(T,p₀,𝐧) = ∑ᵢnᵢRTln(p/p₀) where 𝐧 = n₁, n₂, ... but deriving wrt to a particular nᵢ gives μᵢ (T,p,𝐧) - μᵢ (T,p₀,𝐧) = RTln(p/p₀) Now I cannot define μᵢ⁰ = μᵢ(T,p₀,𝐧) because this will change with 𝐧. μᵢ⁰ should be μᵢ⁰(T₀,p₀,𝐧₀) where all variables are taken at a standard arbitrary condition. I'm doing some research and it looks like a formal proof involves the Helmholtz energy and the Gibbs-Duhem equation.
At 7:12 why did you write μ, not μi? If there are multiple components in the system then we will need more than one value of μ?
You're absolutely right, I should have written μᵢ instead of μ there.
Hi Steven,
great video as always! A few questions:
1)
At 1:55 I think it's ok to take just the derivative of V and not of the product pV because p is a natural variable of H. If I understand natural variables correctly, I should think of H as function of S, p and ni like this:
H = H(S, p, ni) = U(S, V(p, ni), ni) + p*V(p, ni)
Here in U I cannot consider V as an independent variable anymore, so the partial derivative of U with respect to ni should be dU/dV * dV/dni + dU/dni which is different from the partial derivative of U wrt ni when I think U in terms of its natural variables: U=U(S, V, ni). This is tricky.
2)
In a similar way, at 5:31, because (p,T) are natural variables for G I don't need their derivative wrt ni.
3)
Is the p at 5:59 the partial pressure rather than the total pressure of the system?
4)
I suppose the fundamental relations in terms of the partial molar quantities work because the partial derivative of a differential is the differential of the partial derivative.
Ex:
f(x, y) = (x^3)(y^2)
derivative of the differential:
df(x, y, dx, dy) = 3(x^2)(y^2)dx + 2(x^3)(y)dy [the differential is a function of 4 arguments]
(d/dx)df(x, y, dx, dy) = 6x(y^2)dx + 6(x^2)(y)dy
differential of the derivative:
df/dx(x, y) = 3(x^2)(y^2)
d(df/dx)(x, y, dx, dy) =6x(y^2)dx + 6(x^2)(y)dy.
Thinking about this one more time, I'm not sure the pressure at 5:59 is the partial pressure.
For ideal gases V(p, T, n1, n2,...) = ∑_i niRT/p = nRT/p
so ∂V/∂ni = RT/p and integrating:
μi = μi0 + RTln(p/p0)
So it looks like p is the total pressure. Where is my mistake?
Your ∂V/∂nᵢ equation is fine. But that's partial molar volume, not partial molar Gibbs energy.
Here's a thought experiment that may help make it seem reasonable that partial pressure is correct:
Imagine two containers of gas, separated by a stopcock. Both are at 3 atm pressure. On the left side, the partial pressure of gas A is 2 atm and the partial pressure of gas B is 1 atm. On the right side, it's the opposite: A has partial pressure of 1 atm and B is 2 atm. Now, open the valve. Gas A will move from the left to the right, because its chemical potential is higher on the left. This is due to its higher *partial* pressure; the total pressure is the same on both sides.
Now imagine these two containers in a different scenario: Both have 1 atm of A. But the left side has 2 atm of B, for a total pressure of 3 atm. The right has only 1 atm of B, for a total pressure of 2 atm. When the stopcock is opened, even though gas flows from left to right, this will be gas B, and there will be no (net) flow of gas A. This is because its chemical potential is the same on both sides. This is because its chemical potential is the same on both sides, even though the total pressure is different.
@@PhysicalChemistry Thanks, now I understand it qualitatively, but I can't figure out the mistake. I have integrated the partial molar volume as we know that ∂μᵢ/∂p = Vᵢ (partial molar volume). Probably my mistake is in defining μᵢ⁰. For the ideal gas mixture I think it's true that
G(T,p,𝐧) - G(T,p₀,𝐧) = ∑ᵢnᵢRTln(p/p₀)
where 𝐧 = n₁, n₂, ...
but deriving wrt to a particular nᵢ gives
μᵢ (T,p,𝐧) - μᵢ (T,p₀,𝐧) = RTln(p/p₀)
Now I cannot define μᵢ⁰ = μᵢ(T,p₀,𝐧) because this will change with 𝐧.
μᵢ⁰ should be μᵢ⁰(T₀,p₀,𝐧₀) where all variables are taken at a standard arbitrary condition.
I'm doing some research and it looks like a formal proof involves the Helmholtz energy and the Gibbs-Duhem equation.