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I'm currently preparing for a physical chemistry exam tomorrow, and was given the following question for practice:

Derive an expression for $\mathrm{d}U$ as a function of $\mathrm{d}V$ and $\mathrm{d}T$.

I'm confused as to whether they want me to derive the equation below:

$$ \mathrm{d}U=\left(\dfrac{\partial U}{\partial V}\right)_T \mathrm{d}V + \left(\dfrac{\partial U}{\partial T}\right)_V \mathrm{d}T $$

or this

$$ U = q + w $$

given an infinitesimal change

$$ \mathrm{d}U = \mathrm{d}q + \mathrm{d}w $$

where

$$ \mathrm{d}q = C\mathrm{d}T $$

and

$$ \mathrm{d}w = -p\mathrm{d}V $$

hence

$$ \mathrm{d}U = C\mathrm{d}T + -p\mathrm{d}V $$

Essentially, I need help interpreting the direction I should proceed, and advice regarding the correctness of my methods.

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  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. While it may seem strange at first we do not use salutations in our Questions and Answers. Please use the upvote button to say thanks, or leave a comment how the post can be improved. $\endgroup$ – Martin - マーチン Jan 28 '16 at 5:57
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I realize now what you're looking for: Consider $\mathrm{d}S(V,T)$ as

$$\mathrm{d}S(V,T) = \left(\frac{\partial S}{\partial V}\right)_{T}\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{V}\mathrm{d}T$$ and place into the first law:

$$ \mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V = T\left[\left(\frac{\partial S}{\partial V}\right)_{T}\mathrm{d}V + \left(\frac{\partial S}{\partial T}\right)_{V}\mathrm{d}T\right] - p\mathrm{d}V $$

Combine like terms of $\mathrm{d}V$:

$$ \mathrm{d}U = T \left(\frac{\partial S}{\partial T}\right)_{V} \mathrm{d}T + \left[T \left(\frac{\partial S}{\partial V}\right)_{T} - p\right]\mathrm{d}V $$

and you have $\mathrm{d}U$ a function of $V$ and $T$.

It turns out that $\displaystyle T \left(\frac{\partial S}{\partial T}\right)_{V} = \left(\frac{\partial U}{\partial T}\right)_{V} = \mathrm{C}_V$ and $\displaystyle \left(\frac{\partial S}{\partial V}\right)_{T} = \left(\frac{\partial p}{\partial T}\right)_{V}$ by a second-derivative Maxwell relations of $A$ (below) with respect to $T$ and $V$, and a "prettier" version is:

$$ \mathrm{d}U = \mathrm{C}_V\mathrm{d}T + \left[T \left(\frac{\partial p}{\partial T}\right)_{V} - p\right]\mathrm{d}V $$

(In the special case of an ideal gas, $pV = nRT$ and the term in brackets is zero - just write $p = nRT/V$ and evaluate.)


To derive $\mathrm{d}A$: What you're looking for is a Legendre transform. You know

$$ \mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V $$ from the First Law of thermodynamics.

Hence the natural variables for $U$ are $S$ and $V$, not $T$ and $V$ (which is desired). So instead of $U(S,V)$ you want $?(T,V)$ where $?$ happens to be the Helmholtz free energy $A$, sometimes denoted $F$ by physicists.

To do the transform, you take your original function - the exchange variable times its differential form, i.e.

$$ A(T,V) = U(S,V) - S\left(\frac{\partial U}{\partial S}\right)_{V}$$

You know also that $$\left(\frac{\partial U}{\partial S}\right)_{V} = T$$ because

$$\mathrm{d}U(S,V) = \left(\frac{\partial U}{\partial S}\right)_{V}\mathrm{d}S + \left(\frac{\partial U}{\partial V}\right)_{S}\mathrm{d}V = T\mathrm{d}S - p\mathrm{d}V$$ by total derivative formula for a function, in this case $U$.

So you have $A = U - ST$ or as commonly written, $A = U - TS$ so the differential form of $A$ is $\mathrm{d}A = \mathrm{d}U - T\mathrm{d}S - S\mathrm{d}T = T\mathrm{d}S - p\mathrm{d}V -T\mathrm{d}S - S\mathrm{d}T$, thus $\mathrm{d}A = -p\mathrm{d}V - S\mathrm{d}T.$ That is a function of $V$ and $T$.

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    $\begingroup$ I really have no clue how this would answer the question "Derive an expression for dU as a function dV and dT." $\endgroup$ – Martin - マーチン Jan 28 '16 at 6:39
  • $\begingroup$ Thanks for the comment -- it became more clear to me; edits made $\endgroup$ – khaverim Jan 28 '16 at 11:23

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