(For mobile users) 00:08 Exact differentials 01:18 Condition for exactness 02:28 Combined form of First Law 02:59 Exact differntial for dU 03:54 Partial derivatives for temperature (T) and entropy (S) 04:58 Applying exactness condition 06:25 Finding dH from dU 08:18 Exact differential for dH 09:19 Partial derivatives for temperature (T) and volume (V) 10:08 Applying exactness condition
Something is weird... If you go to the time 4:28 - when the initial proofs are written - and substitute the results back into the original expression, you get T = -P. What is the significance of this result? It's written there clear as day.
What do you mean by the "original expression"? (∂U/∂S)v and (∂U/∂V)s are certainly not equal. (∂²U/∂S∂U) and (∂²U/∂U∂S) certainly ARE equal, since mixed partials are equal if they are both continuous.
We get dU = TdS - p dV from the First Law (dU = dq + dw), since entropy (S) = q / T, and the work of expansion (w) = - p dV. We get that dU = (∂U/∂S) dS + (∂U/∂V) dV, because U is a "state function". This is equivalent to being a "conservative" function - it does not depend on the size of the system. We can find the change in a state function ΔF by F(final) - F(initial). We also say that the function is "independent of path." There is a very practical and useful result of working with state functions (U, A, G, H, T, S, and V are all state functions). Let's call the function "Z" and let it be a function of two variables x and y. Then we can immediately write down the relationship dZ = (∂Z/∂x) dx + (∂Z/∂y) dy. Since U is a state function of S and V, we had: dU = (∂U/∂S) dS + (∂U/∂V) dV. (This "trick" is used in almost every derivation in thermodynamics). Since we have two different formulae for dU, we can set them equal. Since they both have "dS" and "dV" parts, the coefficients in front of "dS' in both formulae must be equal, and the two coefficients of "dV" must be equal as well. This is also a technique that is used very often. I hope that makes those steps clearer.
M dx + N dy is an exact differential if and only if : ⟨∂M/∂y⟩ = ⟨∂N/∂x⟩. This means that there exists some function f such that df = ⟨∂f/∂x⟩ dx + ⟨∂f/∂y⟩ dy. One property of the "state functions" of thermodynamics (enthalpy, entropy, Helmholtz energy, Gibbs energy) is that they are "independent of path. This means that, to calculate a change in the function, you only need to know the end points; for example, ΔG = G (final) - G (initial). However, a function that is "independent of path" is a state function, one that can be written as an exact differential. Many of the derivations in thermodynamics begin with writing a thermodynamic property ("state function") .as an exact differential
There are four (4) Maxwell relations. Two (2) were derived in this video, and the other two (2) are derived in: ua-cam.com/video/r4WW4uho16o/v-deo.html
(For mobile users)
00:08 Exact differentials
01:18 Condition for exactness
02:28 Combined form of First Law
02:59 Exact differntial for dU
03:54 Partial derivatives for temperature (T) and entropy (S)
04:58 Applying exactness condition
06:25 Finding dH from dU
08:18 Exact differential for dH
09:19 Partial derivatives for temperature (T) and volume (V)
10:08 Applying exactness condition
6 years ago and ure still helping students. thank you so much
thank you. the first 4 minutes of your video alone made everything click for me.
Sir you literally save my grade. Thank you very much
Hello
Something is weird...
If you go to the time 4:28 - when the initial proofs are written - and substitute the results back into the original expression, you get T = -P.
What is the significance of this result? It's written there clear as day.
What do you mean by the "original expression"?
(∂U/∂S)v and (∂U/∂V)s are certainly not equal.
(∂²U/∂S∂U) and (∂²U/∂U∂S) certainly ARE equal, since mixed partials are equal if they are both continuous.
Love your content, super helpful
Glad to hear it!
Very informative lecture thank you for sharing
Glad it was helpful!
Youre great! Thank you
:)
I did not understand the steps that start at 03:11 and end at 03:40. Could you please explain the mathematics behind those steps also? Thank you.
We get dU = TdS - p dV from the First Law (dU = dq + dw), since entropy (S) = q / T, and the work of expansion (w) = - p dV.
We get that dU = (∂U/∂S) dS + (∂U/∂V) dV, because U is a "state function". This is equivalent to being a "conservative" function - it does not depend on the size of the system. We can find the change in a state function ΔF by F(final) - F(initial). We also say that the function is "independent of path."
There is a very practical and useful result of working with state functions (U, A, G, H, T, S, and V are all state functions). Let's call the function "Z" and let it be a function of two variables x and y. Then we can immediately write down the relationship dZ = (∂Z/∂x) dx + (∂Z/∂y) dy.
Since U is a state function of S and V, we had: dU = (∂U/∂S) dS + (∂U/∂V) dV. (This "trick" is used in almost every derivation in thermodynamics). Since we have two different formulae for dU, we can set them equal. Since they both have "dS" and "dV" parts, the coefficients in front of "dS' in both formulae must be equal, and the two coefficients of "dV" must be equal as well. This is also a technique that is used very often.
I hope that makes those steps clearer.
Interesting
Maxwell's relations ( U&H)
Thanks professor, helpfull! God Bless you
Glad it was helpful!
This is the best and short
If mdx+ ndy = is exact or 0?? Then
M dx + N dy is an exact differential if and only if :
⟨∂M/∂y⟩ = ⟨∂N/∂x⟩. This means that there exists some function f such that
df = ⟨∂f/∂x⟩ dx + ⟨∂f/∂y⟩ dy. One property of the "state functions" of thermodynamics (enthalpy, entropy, Helmholtz energy, Gibbs energy) is that they are "independent of path. This means that, to calculate a change in the function, you only need to know the end points; for example, ΔG = G (final) - G (initial).
However, a function that is "independent of path" is a state function, one that can be written as an exact differential. Many of the derivations in thermodynamics begin with writing a thermodynamic property ("state function") .as an exact differential
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Most welcome!
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Thanks sir u solve out my doubt
Proof other maxwells quation short and easy way
There are four (4) Maxwell relations. Two (2) were derived in this video, and the other two (2) are derived in:
ua-cam.com/video/r4WW4uho16o/v-deo.html
sir thank u
Proof other eqation in short way
Thanks bro
cool kid didn't ask