How to deal with units in logarithms?

Question

I'm trying to take a formula, take it's integral which yields a formula with logarithms, and then directly plug in values with units and get an answer. However if the units go into any of the logarithms I get conditional expressions. Is it possible to set up the equations so I can just plug in units?

Here's my code:

Pideal[n_, T_, V_] := (n*R*T)/V
PvdW[n_, T_, V_, a_: a, b_: b] := (n*R*T)/(V - n*b) - (n^2*a)/V^2
W[V1_, V2_, P_] := Integrate[P, {V, V1, V2}]
Wideal[V1_, V2_, n_: n, T_: T] = W[V1, V2, Pideal[n, T, V]]

Here's the problem spot:

Wideal[5 Liters, 10 Liters, 1 Mols, 298 Kelvins]

I can work around this fairly easily, just leave out the Liters because they cancel out anyway. (Liters = Quantity["Liters"], etc)

However when I get to a more complicated equation that doesn't work so easily without making sure to put back in what I leave out.

WvdW[V1_, V2_, n_: n, T_: T, a_: a, b_: b] = 
    W[V1, V2, PvdW[n, T, V, a, b]]
WvdW[5*Liters, 10*Liters, 1 Mols, 298 Kelvins, 0, 0]

This doesn't work for the same reason, even though the Liters cancel out Mathematica doesn't assume that. However if I leave them out, I still have problems with the Mols, which appear in and outside of Logarithms, even though they're multiplied by 0.

WvdW[5, 10, 1 Mols, 298 Kelvins, 0, 0]

Finally, I haven't even gotten to using b which also appears in the logarithm, and even though the volumes add up and cancel out, Mathematica doesn't assume so.

WvdW[V1*Liters, V2*Liters, 1 Mols, 298 Kelvins, 0, 0 Mols Liters^(-1)]

Answer 1

You might use Assumptions in one form or another.

Block[{$Assumptions = Liters > 0 && Mols > 0 && Kelvins > 0}, 
 Simplify[Wideal[5 Liters, 10 Liters, 1 Mols, 298 Kelvins]]
 ]
(* -> 298 Kelvins Mols R Log[2] *)

One disappointment for me is that the following doesn't work:

Block[{$Assumptions = Liters > 0 && Mols > 0 && Kelvins > 0}, 
 Wideal[5 Liters, 10 Liters, 1 Mols, 298 Kelvins]
 ]
(* -> ConditionalExpression[298 Kelvins Mols R (-Log[5 Liters] + Log[10 Liters]), Liters != 0] *)

Nor does putting the assumptions directly in the Integrate:

W[V1_, V2_, P_] := Integrate[P, {V, V1, V2}, Assumptions :> Liters > 0 && Mols > 0 && Kelvins > 0];
Wideal[5 Liters, 10 Liters, 1 Mols, 298 Kelvins]
(* -> ConditionalExpression[298 Kelvins Mols R (-Log[5 Liters] + Log[10 Liters]), Liters != 0] *)

It seems using Simplify is necessary. See this answer for more on Integrate and assumptions.

Update

I think Silvia has made the best point, that the units should be canceled out first. Simply using SetDelayed lets Integrate cancel the units in the given example. If there are cases where one might have to use Simplify or FullSimplify first, then remove the comments.

Pideal[n_, T_, V_] := (n*R*T)/V;
PvdW[n_, T_, V_, a_: a, b_: b] := (n*R*T)/(V - n*b) - (n^2*a)/V^2;
W[V1_, V2_, P_] := Integrate[(*Simplify@*)P, {V, V1, V2}];
Wideal[V1_, V2_, n_: n, T_: T] := W[V1, V2, Pideal[n, T, V]];

Wideal[5 Liters, 10 Liters, 1 Mols, 298 Kelvins]
(* -> 298 Kelvins Mols R Log[2] *)

Answer 2

Try this:

Clear[Wideal]
Wideal[V1_, V2_, n_: n, T_: T] := Evaluate[W[V1, V2, Pideal[n, T, V]]] // FullSimplify

Wideal[5 Liters, 10 Liters, 1 Mols, 298 Kelvins]

298 Kelvins Mols R Log[2]

Clear[WvdW]
WvdW[V1_, V2_, n_ : n, T_ : T, a_ : a, b_ : b] := 
 Evaluate[(Integrate[PvdW[n, T, V, a, b], V] /. {{V -> V1}, {V -> V2}}).{-1, 1}] // FullSimplify

WvdW[5 Liters, 10 Liters, 1 Mols, 298 Kelvins, 0, 0]

298 Kelvins Mols R Log[2]

(Don't know why, Integrate[-((a n^2)/V^2) + (n R T)/(-b n + V), {V, V1, V2}] feels taking forever on my laptop..)

Like Michael pointed out, additional assumptions may be needed for more complex expressions.

comment

I think a better way is to cancel the units before the quantity is feed to logarithm, because it's ill-defined of a logarithm with a parameter whose unit is other than $1$. (Think of the Taylor expansion of logarithm!) So the built-in Units functions might be a better choice than custom definitions. (Or simply do not involve units in calculations.)

Answer 3

Your problem seems to be the Integrate result: as usual it assumes all variables could be complex which results in a complicated ConditionalExpression which will not evaluate correctly when fed with variables that have units. The trick is, as Michael has shown to use assumptions, but I think it is much better to use the assumptions for the formal variables than for the units, it's not only belisarius students who probably shouldn't do that :-). To see the problem evaluate this:

Integrate[Pideal[n, T, V], {V, V1, V2}

Now if you are not working in very unusual conditions it might be safe to assume that V1 and V2 both are real valued and larger than zero:

Integrate[Pideal[n, T, V], {V, V1, V2}, Assumptions -> {V1 > 0, V2 > 0}]

which makes the result already a lot simpler, but still is a ConditionalExpression which should also evaluate for variables with units. For some reasons out of my understanding Mathematica seems to be too restrictive here, as it only returns a result for V1 < V2, but of course we know that the result is exactly the same for V2 < V1, and Mathematica also knows that:

Integrate[Pideal[n, T, V], {V, V1, V2},Assumptions -> {V1 > 0, 0 < V2 < V1}]
Integrate[Pideal[n, T, V], {V, V1, V2},Assumptions -> {V1 > 0, 0 < V1 < V2}]

so for this case you wouldn't actually need a ConditionalExpression at all (I think that Log will even give the correct results for most "academic corner cases", that is when one or two of V1,V2 will be zero). Putting all this together you could use something like the following:

Liters = Quantity["Liters"];
Kelvins = Quantity["Kelvins"];
Mols = Quantity["Moles"];
Pideal[n_, T_, V_] := (n*R*T)/V
PvdW[n_, T_, V_, a_: a, b_: b] := (n*R*T)/(V - n*b) - (n^2*a)/V^2
W[V1_, V2_, P_] := Piecewise[{
   {Integrate[P, {V, V1, V2}, Assumptions -> V1 > V2 > 0], V1 > V2},
   {Integrate[P, {V, V1, V2}, Assumptions -> V2 > V1 > 0], V2 > V1}
   }, Indeterminate]
Wideal[V1_, V2_, n_: n, T_: T] = W[V1, V2, Pideal[n, T, V]];

Wideal[5 Liters, 10 Liters, 1 Mols, 298 Kelvins]

Given the above definitions you'll receive an answer which does evaluate correctly and I don't think it is a problem to feed quantities (with units) to those expressions: if the units don't match and the argument to the Log isn't dimensionless, you'll get a partially unevaluated result and some messages, which might even help you to find problems related to wrong units... There remains one problem: if you need to use another definition for the equation of state you might need to check whether and which assumptions you have to give to Integrate...