Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm reading some papers for calculating the vapor pressure of alkali metals as a function of temperature, and I've come across some familiar-looking virial expansions, but when I tried to work out the unit analysis, it was very much unclear how to make everything work out.

Here's the first equation (where they give the form in terms of the physical quantities):

$$ \mathrm{ln}\left(p\right) = I_{c} - \frac{\Delta H^{\circ}_{v}}{RT} + \frac{\left(C^{g}_{p} - C_{p}^{l}\right)}{R}\mathrm{ln}\left(T\right) $$

This is equation 4 of the following paper:

(1) Alcock, C. B.; Itkin, V. P.; Horrigan, M. K. Canadian Metallurgical Quarterly 1984, 23, 309–313., (doi: 10.1179/000844384795483058

From the definitions section, $\Delta H_{v}^{\circ}$ is the molar enthalpy of vaporization in $\frac{BTU}{lb\cdot mol}$, $R$ is the gas constant. No units are given for this, but given the other values, we'll call it $\frac{BTU}{lb\cdot mol ^{\circ}R}$. $C_{p}^g$ and $C_{P}^l$ are the isobaric (constant pressure) specific heats, given in $\frac{BTU}{lb\cdot mol^{\circ}\!R}$. So working this out, I'm going to assume that this is supposed to be a unitless quantity somehow. I'm assuming that $I_{C} = \mathrm{ln}\left(p_{0}\right)$, so we get:

$$ \mathrm{ln}\left(\frac{p}{p_{0}}\right) = -\frac{\Delta H_{v}^{\circ}}{RT} + \frac{\left(C_{p}^{g}-C_{p}^{l}\right)}{R}\mathrm{ln}\left(T\right) $$ Everything there is unitless except for $\mathrm{ln}\left(T\right)$. (Alternately, maybe he is secretly assuming that $p = \frac{p}{torr}$ - the unit analysis is basically the same). What is going on there? The other form I see for this is:

$$ \mathrm{log}\left(p\right) = A_{c} - \frac{B_{c}}{T} + C_{c}\mathrm{log}\left(T\right) $$

Again, there's a logarithmic temperature in there! And, bizarrely, he defines the following quantities (A is nowhere to be found on this list):

$$B = \text{second virial coefficient, }\frac{ft^3}{mol}$$$$ C = \text{third virial coefficient, }\frac{ft^6}{mol^2} $$

According to the paper, $_{C}$ is the indicator for "empirical constant". I'm guessing $A$ is supposed to be unitless (don't know how to scale it to change the pressure units, but one thing at a time). Trying to find a second opinion, I went to this paper:

(1) Ewing, C. T.; Chang, D.; Stone, J. P.; Spann, J. R.; Miller, R. R. Journal of Chemical & Engineering Data 1969, 14, 210–214. doi: http://dx.doi.org/10.1021/worsp).

But they have listed no units, and suffer from the same problems. Here is their version of the expansion:

$$ A + \frac{B}{T} + C\cdot\mathrm{log}\left(T\right) + D\cdot T\cdot10^{-3} $$

They plug in some values for this:

$$ \mathrm{log}\left(p(atm)\right) = 7.55339-\frac{27007.7}{T} + 0.19699\mathrm{log}(T) - \frac{0.171188\cdot T}{1000} $$

Is temperature a unitless quantity now? What am I missing here?

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

No, temperature is not a unitless quantity. :-) (unless you're working in Planck units, one could argue).

As you know, you can't take the logarithm of a unit. When you see $\ln X$, where $X$ is some quantity with units, it actually means $\ln\frac{X}{X_0}$ for some fixed reference value (a.k.a. unit) $X_0$. (Even if the author doesn't realize it!)

In your case, there are some unspecified values $p_0$ and $T_0$ such that the equation is actually

$$\mathrm{ln}\left(\frac{p}{p_0}\right) = I_{0} - \frac{\Delta H^{\circ}_{v}}{RT} + \frac{\left(C^{g}_{p} - C_{p}^{l}\right)}{R}\mathrm{ln}\left(\frac{T}{T_0}\right)$$

$p_0$ and $T_0$ are the units of pressure and temperature. If you want to change to some other unit system, with units $p_1$ and $T_1$, you can rewrite the equation like this (glossing over some details):

$$\mathrm{ln}\left(\frac{p}{p_1}\right) + \mathrm{ln}\left(\frac{p_1}{p_0}\right) = I_{0} - \frac{\Delta H^{\circ}_{v}}{RT} + \frac{\left(C^{g}_{p} - C_{p}^{l}\right)}{R}\left[\mathrm{ln}\left(\frac{T}{T_1}\right) + \mathrm{ln}\left(\frac{T_1}{T_0}\right)\right]$$

which you can then rearrange to

$$\mathrm{ln}\left(\frac{p}{p_1}\right) = \underbrace{I_{0} - \mathrm{ln}\left(\frac{p_1}{p_0}\right) + \frac{\left(C^{g}_{p} - C_{p}^{l}\right)}{R}\mathrm{ln}\left(\frac{T_1}{T_0}\right)}_{\displaystyle I_1} - \frac{\Delta H^{\circ}_{v}}{RT} + \frac{\left(C^{g}_{p} - C_{p}^{l}\right)}{R}\mathrm{ln}\left(\frac{T}{T_1}\right)$$

This lets you define a new constant $I_1$ which makes the equation work with the new units. So all you have to do is find the value of $I_c$ corresponding to one set of pressure/temperature units (hopefully this is given in the paper), and then you can adjust the equation to work for other unit systems as needed.

share|improve this answer
    
Yes, I like this a lot. The only problem is that no units are ever given for the T in ln(T). I think based on the rest of the documents, I can assume it's Rankine. I'm going to work out the values and see if they work for what I'm expecting (the two documents give answers in two different sets of units for the same thing, so I can check the answer), and if so I'll accept this one. –  Paul Apr 12 '13 at 7:09
add comment

I don't think that you can assume that $I_c=\ln (p_0)$

A lot of equations have logs of dimensioned quantities in them. It's usually not hard to get rid of the log:

$$ \mathrm{ln}\left(\frac{p_1}{p_2}\right) = I_{c,1}-I_{c,2} - \left(\frac{\Delta H^{\circ}_{v,1}}{RT_1}-\frac{\Delta H^{\circ}_{v,2}}{RT_2}\right) + \mathrm{ln}\left(\frac{T_1^{\frac{\left(C^{g}_{p,1} - C_{p,1}^{l}\right)}{R}}}{T_2^{\frac{\left(C^{g}_{p,2} - C_{p,2}^{l}\right)}{R}}}\right) $$

It may also be that $I_c$ is just a gas-specific constant, of the form $$I_c=\ln p_0-\frac{\Delta H^{\circ}_{v,0}}{R} -\frac{\left(C^{g}_{p,0} - C_{p}^{l,0}\right)}{RT_0}\mathrm{ln}\left(T_0\right) $$

for some $p_0,T_0$.

You can do the same for the other equations as well. When there's a log of a dimensioned quantity, one of two things is possible:

  • The equation has the units implicitly assumed (this is generally not done in physics, but you see it a lot in chemistry)
  • The equation is a change equation, and you need to subtract one copy of the equation at a different point from it.

The latter form arises when one takes an indefinite integral in the last step while solving a differential equation (Which is why subtracting is the right way to fix this, since that corresponds to tacking limits onto the integral)

share|improve this answer
    
Yes, good point. In this case it basically has to be the first one, since this is a gas law - absolute. That said, I'm not sure exactly how to convert the units so that it gives me a value in... I dunno, ln(Pa) or ln(Torr) instead of ln(atm) –  Paul Apr 12 '13 at 4:46
    
@Paul: Not necessarily. A gas law can be relative, and a set of $(p_0,T_0,V_0)$ values could be considered as a gas-specific parameter. I strongly feel that it is the latter :) –  Manishearth Apr 12 '13 at 4:49
    
What would be the relevant and "natural" point to compare this to? Neither paper mentions a reference point, but as far as I know they are the fundamental papers in the literature on this. –  Paul Apr 12 '13 at 4:52
    
Sorry, to be clear, I think you're probably right, but that the "unitless" quantity I_c is where that information seems to be contained. It's also not clear how to make T dimensionless here without having A be a function of T. The natural way is to reference it to 0, but you can't divide by 0K or 0R, so what, you divide by 1K? How much sense does that really make? –  Paul Apr 12 '13 at 4:54
1  
@Paul: That too, $I_c$ could basically be$$\ln p_0-\frac{\Delta H^{\circ}_{v,0}}{R} -\frac{\left(C^{g}_{p,0} - C_{p}^{l,0}\right)}{RT_0}\mathrm{ln}\left(T_0\right) $$ –  Manishearth Apr 12 '13 at 4:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.