Does taking the $\log$ of a quantity change the unit of the quantity? 
\begin{align}
&= c [P - b \ln(P+b)]_1^{500} \\
&= 0.125 \frac{\text{cm}^3}{\text{g}} \times \left[\begin{aligned}&500 \text{ bar} - 2700 \text{ bar} \times \ln(500 \text{ bar} + 2700 \text{ bar}) - \\ &\big(1 \text{ bar} - 2700 \text{ bar} \times \ln(1 \text{ bar} + 2700 \text{ bar})\big)\end{aligned}\right]
\end{align}
Solve the above calculation and get,
$$W = 5.16 \frac{\text{cm}^3 \cdot \text{bar}}{\text{g}}$$

Not sure if this is the right place to post this but doing this question I thought that the $b\ln(P+b)$ portion would result in units of $\text{bar}^2$ given both $b$ and $P$ have the unit bar. I fail to see how that's not the case unless $\ln$ alters the units.
 A: In the calculation you posted, the author meant to say
$$\log\left(3200 \text{ bar}\right) - \log\left(2701 \text{ bar}\right) = \log\left(\frac{3200}{2701}\right),$$
where the right-hand side actually defines what is meant in the left-hand side. Hence, the logarithm has a dimensionless argument and also returns a dimensionless value.

Remark: I'm keeping the following paragraphs because the comment by Chemomechanics pointed out to an interesting reference (DOI: 10.1021/ed1000476) that exhibits a problem with the argument I presented using a Taylor series. John Davis also pointed out that a similar argument for the function $\frac{1}{x}$ expanded about $x=1$ would lead to an inconsistency. While my argument is wrong, I think that keeping these opposite views in here is interesting.
It doesn't really make sense to take the logarithm of a quantity with units. Both the argument and the result should be dimensionless numbers. The reason can be seen by expanding it in a Taylor series. We get
$$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + \cdots$$
Each of these terms has a different power. Hence, if $x$ has units, we'd run into trouble with attempting to compute a quantity that is given by a sum of a meter, with meter squared, with meter cubed, and so on. It is inconsistent.
Due to the same argument, any function that can be written as a Taylor series and it not just a monomial only receives dimensionless arguments.
A: Actually your expression only involves taking
the logarithm of a unit-less quantity.
To see this remember $\ln x - \ln y = \ln \frac{x}{y}$
and rewrite the expression:
$$\begin{align}
 & c\left[P - b\ln(P+b)\right]_{P_1}^{P_2} \\
=& c\left[P_2 - b\ln(P_2+b)-P_1+\ln(P_1+b)\right] \\
=& c\left[P_2 - P_1 + \ln\frac{P_1+b}{P_2+b} \right]
\end{align}$$
Now we can see that we have the logarithm of a unit-less quantity.
And therefore the unit of $P$ and $b$ doesn't matter.
