Deriving Thermodynamics equations confusion [closed]








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Please explain:


I have taken up to Cal 3 so I understand deriving but I guess because of how it is used in context of this class I am not understanding. In the thermodynamics equation: ( as an example) what is the difference in using dU = Cvdt VS. $\Delta$U=Cv$\Delta$T. This may come across as a stupid question but I just don't understand, maybe because I have not had much practice, but would I need to be deriving something? Another example, I was asked to derive the Maxwell-Boltzmann constant, but in the book it shows the derivation and its literally just plugging in different equations;however, when I say that I don't mean its like dx/dt = v and then I substitute in v for that part but rather substituting in full equations.


EDIT: I'm sorry my question seems unclear, but I am asking it to the best of my ability. I guess all I want to know is what deriving an equation in physical chemistry means. If I derive x^2 it is 2x. Well as I have seen, if I derive the RMS equation, instead of actually (dx/dt)'ing anything, I am just plugging in different parts of different equations. Pretty ridiculous I get down voted for asking a question, unclear or not. The whole purpose of this forum is to help communicate issues with one another, not be rude.




$\mathrm{d}U = C\_V\,\mathrm{d}T$ relates an infinitesimal change in temperature to an infinitesimal change in internal energy.


$\Delta U = C\_V \Delta T$ relates a finite change in temperature to a finite change in internal energy.


To get from infinitesimal to finite you have to add up, or sum, a huge number of infinitesimal changes. This isn't done with a sum, but rather an integral (which acts like a sum for infinitesimal quantities) from an initial state to a final state. Since both $U$ and $T$ are state functions you can do this quite safely without worrying about intricacies of path functions.


$$\begin{align}
\mathrm{d}U &= C\_V\,\mathrm{d}T \tag{1} \\
\int\_\text{initial}^\text{final}\mathrm{d}U &= \int\_\text{initial}^\text{final}C\_V\,\mathrm{d}T \tag{2}
\end{align}$$


The limits of integration on the left are simply the initial internal energy and the final internal energy. You can call it $U\_1$ and $U\_2$ respectively. Likewise the limits of integration on the right are $T\_1$ and $T\_2$


$$\begin{align}
\int\_{U\_1}^{U\_2}\mathrm{d}U &= \int\_{T\_1}^{T\_2} C\_V\,\mathrm{d}T \tag{3}
\end{align}$$


In general, $C\_V$ is a function of $T$. However, to get to your desired equation, you have to assume that $C\_V$ is independent of $T$ over the temperature range you are interested in, i.e $C\_V$ is a constant. That allows you to take it out of the integral much like you would any other constant.


$$\begin{align}
\int\_{U\_1}^{U\_2}\mathrm{d}U &= C\_V\int\_{T\_1}^{T\_2} \mathrm{d}T \tag{4} \\
U\_2 - U\_1 &= C\_V(T\_2 - T\_1) \tag{5} \\
\Delta U &= C\_V\Delta T \tag{6}
\end{align}$$


So, **yes**, there is a difference. For the finite change you made an extra assumption that $C\_V$ doesn't vary over the temperature range. For the infinitesimal change the temperature range is infinitesimally small - hence $C\_V$ cannot vary and the assumption is not necessary.


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