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\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.

Níckolas Alves
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2 Answers2

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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.

Níckolas Alves
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  • I skipped a few lines in the calculation, but you can rewrite the expression in terms of the combination I wrote explicitly. – Níckolas Alves Jun 26 '22 at 00:58
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    I don't quite buy this argument. For example the reciprocal of quanity with dimension is a quanitity with different dimension, but the Taylor series of $1/x$ at $1$ is $1 - (x-1) +(x-1)^2 - (x-1)^3+...$ – John Davis Jun 26 '22 at 09:24
  • @JohnDavis That's a good point... I'll have to think about it – Níckolas Alves Jun 26 '22 at 21:57
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    The problem with this Taylor-series argument is discussed in Matta et al.'s "Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions". Briefly, you forgot to include the units of the partial derivatives in your expansion (e.g., $\frac{(dx)^n}{n!}\frac{\partial^nf(x)}{\partial x^n}$. If you had, you would have found that all the terms have the same units after all. In other words, the Taylor series is not the reason that we can't take logarithms of unitfull quantities. – Chemomechanics Jun 27 '22 at 00:59
  • See my comment on Thomas Fritisch's answer. I tried to make an argument using the fact that the derivative of $ln x$ is $\frac{1}{x}$, but I think it just boils down to always being able to divide the units out of the argument of the logarithm. – John Davis Jun 27 '22 at 01:39
  • @Chemomechanics Interesting. I can't access that paper, does it make any similair points to my comment on the other answer? – John Davis Jun 27 '22 at 01:51
  • @Chemomechanics thanks for the reference! It makes a lot of sense. I modified my answer accordingly. – Níckolas Alves Jun 27 '22 at 15:40
  • @JohnDavis It is fairly similar to what you wrote. The authors essentially say that one should always add in a reference constant when taking logarithms of dimensional quantities, such that the actual argument of the logarithm is dimensionless. By exponentiating the expression, I think one will always be able to cancel out the constant – Níckolas Alves Jun 27 '22 at 15:43
  • I don't have access to the paper linked by @Chemomechanics, but from the abstract and the summary here I am deeply skeptical. The expansion of $\frac1x$ about $x=1$ is equivalent to the expansion of $\frac1{1+\epsilon}$ about $\epsilon=0$, which moves the problem of dimensionful $x$ to a problem of dimensional consistency in $1+\epsilon$. A useful exercise is to work through the Taylor series for $\sin x$ so that the appropriate powers of $\frac{2\pi\rm,rad}{360^\circ}$ appear to allow $x$ in degrees (or grads, or your favorite non-radian angle measurement). – rob Jun 27 '22 at 16:52
  • @John Davis: " The expansion of $1/x$ about 1" requires $x$ to have the same units as "1". If $x$ is not dimensionless, the series is similar to $1/(a-x) = = (1/a)(1+ x/a + (x/a)^2)+\ldots$, all terms with the same units. – mike stone Jun 27 '22 at 17:37
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    Anyone have a sense of deja vu? – Chemomechanics Jun 27 '22 at 18:51
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    Accessible version here. – Chemomechanics Jun 27 '22 at 19:37
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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.

Thomas Fritsch
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  • 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: $\log{a} = \log{a} - 0 = \log{a} - \log{\alpha} = log{\frac{a}{\alpha}}$ – John Davis Jun 27 '22 at 01:30