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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.

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.