Sir you are wonderful ❤. You are one in million. My biggest wish is health for you and you make more videos for us. I really appreciate you from the bottom of my heart. I'm watching your awesome videos one by one and you beyond incredible. I'm not able to describe by words how professional you are . Thank you for thousands times 🌹
That would depend on how exactly V(T) and α(T) depend on the temperature. You would need to insert those functions into the integral, if you knew them, and perform the integral.
I have a concern about the example from the definition of the thermal expansion coefficient, As we are measuring the volume w.r.t the temperature V=V(T). So isn't it better to take the V to the dV side when we are differentiating. Then a logarithmic term appears in the relationship also and we do not have to assume that the change in V is negligible w.r.t T.
Yes, that's correct on all counts. With the approach you describe, the result is: V₂ = V₁ exp(α ΔT). That is more accurate that the result obtained here, V₂ = V₁ + α ΔT. The more accurate equation is sometimes needed, but can often be ignored. I didn't derive the more accurate equation here because the main point was how to manipulate the dA/dB at constant C relationships, and didn't want to get too bogged down in discussion of math or physical conditions.
I like how integrals and partial derivatives can look kinda scary but in practicee they are simple, but as a mathematical concept they are really useful anyways
Sir you are wonderful ❤. You are one in million. My biggest wish is health for you and you make more videos for us. I really appreciate you from the bottom of my heart. I'm watching your awesome videos one by one and you beyond incredible. I'm not able to describe by words how professional you are . Thank you for thousands times 🌹
I'm really happy that you're enjoying them so much
What would the integral be if we were extra careful and didn't assume V and α were constant
That would depend on how exactly V(T) and α(T) depend on the temperature. You would need to insert those functions into the integral, if you knew them, and perform the integral.
I have a concern about the example from the definition of the thermal expansion coefficient,
As we are measuring the volume w.r.t the temperature V=V(T). So isn't it better to take the V to the dV side
when we are differentiating. Then a logarithmic term appears in the relationship also and we do not have to assume that the change in V is negligible w.r.t T.
Yes, that's correct on all counts.
With the approach you describe, the result is: V₂ = V₁ exp(α ΔT). That is more accurate that the result obtained here, V₂ = V₁ + α ΔT.
The more accurate equation is sometimes needed, but can often be ignored. I didn't derive the more accurate equation here because the main point was how to manipulate the dA/dB at constant C relationships, and didn't want to get too bogged down in discussion of math or physical conditions.
@@PhysicalChemistry I was thinking the same thing. So we can separate the variables and get an answer?
@@Mohobofo Yes, correct
I like how integrals and partial derivatives can look kinda scary but in practicee they are simple, but as a mathematical concept they are really useful anyways