\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.


2 Answers 2


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.


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.

  • $\begingroup$ In fact I suppose you can always divide out the units inside the logarithm. Given quanitity $a>0$ which is in units where $\alpha$ is the unit quantity: $\log{a} = \log{a} - 0 = \log{a} - \log{\alpha} = log{\frac{a}{\alpha}}$ $\endgroup$
    – John Davis
    Jun 27 at 1:30

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