I A-level differentiation/derivative dilemma

[Mark44]

However I still don't get from where that partial derivative of V appeared from?I mean this ∂V
How was the transistion made from this −ln(NΛ3)+1 to the ∂V

From an earlier post.

This gives $A = -Nk_BT[\ln(\frac V {N \Lambda^3}) + 1]$, which can be simplified to $-Nk_BT[\ln(V) - \ln(N \Lambda^3) + 1]$
Next, he takes the partial derivative of A with respect to V, getting
$\frac{\partial A}{\partial V} = -Nk_BT \cdot \frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1]$
$= -Nk_BT \frac {\partial \ln(V)} {\partial V}$

He's taking the partial with respect to V of both sides of the equation.
$\frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1] = \frac {\partial} {\partial V}[\ln(V)]$. Those two other terms drop out because their partial derivatives are zero.

 

[Martin Harris]

From an earlier post.He's taking the partial with respect to V of both sides of the equation.
$\frac {\partial} {\partial V}[\ln(V) - \ln(N \Lambda^3) + 1] = \frac {\partial} {\partial V}[\ln(V)]$. Those two other terms drop out because their partial derivatives are zero.

Riiiight, makes sense thank you very much , now understood this bit.
Also looking at picture, could you please tell me how he made transition from nkBT* dlnV/dv to NKbT/V ?

Thank you sir very much

 

[Mark44]

Also looking at picture, could you please tell me how he made transition from nkBT* dlnV/dv to NKbT/V ?

What is $\frac d {dx} [\ln(x)]$?
This is one of the first differentiation rules that you learned in calculus. You won't be able to follow a discourse in thermodynamics using partial derivatives if their are gaps in you knowledge of single-variable calculus.

 

[Martin Harris]

What is $\frac d {dx} [\ln(x)]$?
This is one of the first differentiation rules that you learned in calculus. You won't be able to follow a discourse in thermodynamics using partial derivatives if their are gaps in you knowledge of single-variable calculus.

Yes, I understood is 1/V because of the ln derivation.
In this 2nd picture could you please tell me how he recognized , how did he found out that those terms are 1st term 2nd term 3rd term ?What was the criteria?How he identified which term is 1st which is 2nd and how?
Please see below

Thank you in advance

 

[Martin Harris]

And last confusion I have is regarding this problem.
Please have a look below if possible.

In 1st picture we have the formulas for various energies.
In 2nd picture I solved the exercise plugging in numbers in the formula and I got right answer.
My problem is in the 3rd picture.
that 35335 comes from ΔU=ϑCvΔT I guess, but why?
And also when I try to compute the other 2 terms from the brackets I cannot manage to get that values?
I use same formulas as in 1st picture but instead of T=300K I put T=2000K and still don't get the values.

Thank you very much for your understanding and cooperation.

 

[Mark44]

Yes, I understood is 1/V because of the ln derivation.
In this 2nd picture could you please tell me how he recognized , how did he found out that those terms are 1st term 2nd term 3rd term ?What was the criteria?How he identified which term is 1st which is 2nd and how?
Please see below
View attachment 107257
Thank you in advance

Referring to the equation that starts with
"$\beta A^{id} = \dots$"
He's taking the partial with respect to $\beta$ of both sides of that equation.
The order "first", "second", "third" is simply the order of the terms on the right side of the equation.
First term -- is the first term on the right side of the equation.
Second term -- is the second term on the right side of the equation, but I don't see the connection between the second term in the equation above and the work he shows here. There is apparently some substitution going on, but what the image you attached doesn't show what it is.
Third term -- is -N times the first term in the last pair of brackets; namely, $-N\ln(q_{rot})$. How that is related to $\ln(\beta)$, I have no idea -- there is not enough information shown in the image.
The remaining terms are independent of T (so it says), so can be ignored.

 

[Mark44]

My problem is in the 3rd picture.
that 35335 comes from ΔU=ϑCvΔT I guess, but why?

No idea. I looked pretty closely for this formula, but I was not able to find it.
BTW, this character -- ϑ -- is "theta" in something that I think is a cursive script.
Do you mean this character -- ∂ -- ? I don't know a name for it, but it's the one used in partial derivatives. You can find it at the right end of the first row of the symbols under the Σ menu item.

And also when I try to compute the other 2 terms from the brackets I cannot manage to get that values?
I use same formulas as in 1st picture but instead of T=300K I put T=2000K and still don't get the values.

 

[Martin Harris]

No idea. I looked pretty closely for this formula, but I was not able to find it.
BTW, this character -- ϑ -- is "theta" in something that I think is a cursive script.
Do you mean this character -- ∂ -- ? I don't know a name for it, but it's the one used in partial derivatives. You can find it at the right end of the first row of the symbols under the Σ menu item.

Thank you for your answers, I was trying to refer to that symbol as Niu(Number of moles)

 

[Mark44]

I was trying to refer to that symbol as Niu(Number of moles)

Greek letter nu? That would be $\nu$ (lowercase). Uppercase Nu looks just like the Roman letter N.