Is the differential, dp, exact or not? [closed]








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> 
> Consider the differential $$\mathrm dp=\frac RV\,\mathrm dT+\left(\frac{2a}{V^2}-\frac{RT}{V^2}\right)\,\mathrm dV$$ (where $a$ is a constant value)
> 
> 
> (a) Determine whether the above differential, i.e. $\mathrm dp$, is exact or not. Show all your steps and evaluation of the appropriate partial differentials!
> 
> 
> 


I have no idea on how to even start this. As far as I can tell the differential is exact, but I don't know how to prove or show it. I'm really struggling to show the steps involved. I would appreciate any advice and thank you very much in advance.




Your function $p$ is a function of the independent variables $V$ and $T$ i.e., $p(V,T)$. The other variables, $a$, and $R$ are constants.


Let me rewrite the differential as


$$\mathrm{d}p = A(V)\,\mathrm{d}T + B(T,V)\,\mathrm{d}V$$


where


$$A(V) = \frac{R}{V}$$


and


$$ B(T,V) = \left( \frac{2a}{V^3} - \frac{RT}{V^2} \right) $$


A differential is exact if


$$\left( \frac{\partial A}{\partial V} \right)\_T = \left( \frac{\partial B}{\partial T} \right)\_V $$


Note that subscripts denote that the variable is being held constant during the partial differentiation. The problem already told us what our $A$ and $B$ terms are, and I wrote them out explicitly above. Now, we just need to take the appropriate partial differentials and compare. If they are equal, then $\mathrm{d}p$ is exact!


Writing out the formulas took awhile, it seems a shame to stop now... A little bit of differentiating, remembering to hold the appropriate variables constant during differentiation leads to


$$\left( \frac{\partial A}{\partial V} \right)\_T = -\frac{R}{V^2}$$


$$\left( \frac{\partial B}{\partial T} \right)\_V = -\frac{R}{V^2}$$


If the two partial derivatives are the same, the differential is exact. I will let you be the judge.


 13

For a given function, $F(x,y,z,...)$, it's differential $\text{d}F$ is given by:
$$
\text{d}F = \left(\frac{\partial F}{\partial x}\right)\_{y,z} \text d x +\left(\frac{\partial F}{\partial y}\right)\_{x,z} \text d y + \left(\frac{\partial F}{\partial z}\right)\_{x,y} \text d z +\; ...
$$


To say that a differential is an exact differential is to say that is if the differential of a function and hence is of the form given about.


For the case given:


$$
\mathrm dp(T,V)=\frac RV\,\mathrm dT+\left(\frac{2a}{V^2}-\frac{RT}{V^2}\right)\,\mathrm dV
$$


If $\mathrm dp$ is an exact differential, that would mean that:


$$
\left(\frac{\partial p}{\partial T}\right)\_{V} = \frac RV\ \text{and } \left(\frac{\partial p}{\partial V}\right)\_{T} = \left(\frac{2a}{V^2}-\frac{RT}{V^2}\right)
$$


There are two equivalent way to determine whether this is true, you can integrate the partial derivatives of $p$ to recover a form for $p$, or you can differentiate each term once more to so that both give identical values for the mixed second derivative.


### Integration


Taking indefinite integrals of the suspected derivatives:


$$\int \frac RV\ \mathrm d T = \frac{RT}{V} + g(V) \\
\int \left(\frac{2a}{V^2}-\frac{RT}{V^2}\right) \mathrm d V = \frac{-2a+RT}{V} + h(T) \\
\to p(T,V) = \frac{RT-2a}{V} + c
$$


### Differentiation


It is typically easier to compare the suspected derivative by differentiation. If $p$ is a true function of $T$ and $V$, by the symmetry of mixed derivatives:


$$
\frac{\partial }{\partial T }\_V \left(\frac{\partial p }{\partial V }\right)\_T = \frac{\partial }{\partial V }\_T \left( \frac{\partial p }{\partial T } \right)\_V
$$


Assuming:
$$ \left(\frac{\partial p}{\partial T}\right)\_{V} = \frac RV\ \\
\to \frac{\partial }{\partial V }\_T \left( \frac{\partial p }{\partial T } \right)\_V = -\frac{R}{V^2}
$$


Assuming


$$ \left(\frac{\partial p}{\partial V}\right)\_{T} = \left(\frac{2a}{V^2}-\frac{RT}{V^2}\right) \\
\to \frac{\partial }{\partial T }\_V \left( \frac{\partial p }{\partial V } \right)\_T = -\frac{R}{V^2}
$$


Both terms of the differential $\mathrm dp$ imply the same mixed second derivative, hence $\mathrm dp$ is an exact differential.


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