What is the problem with the units in this state equation of a gas?

[zenterix]

Homework Statement

3.4 (a) Calculate the work done upon expansion of 1 mol of gas quasi-statically and isothermally from volume $v_i$ to a volume $v_f$, when the equation of state is

$$\left ( P+\frac{a}{v^2}\right )(v-b)=RT$$

where $a$ and $b$ are the van der Waals constants.

(b) If $a=1.4\times 10^9\mathrm{N\cdot m^4/mol}$ and $b=3.2\times 10^{-5}\mathrm{m^3/mol}$, how much work is done when the gas expands from a volume of 10 liters to a volume of 22.4 liters at $20^{\circ} \text{C}$?

Relevant Equations

$W=-\int PdV$

My question is about units.

For the first part, we can solve the state equation for $P$ as a function of $V$ and $T$. We obtain

$$P(V,T)=\frac{RT}{V-b}-\frac{a}{V^2}\tag{1}$$

The units question already comes up here because as far as I can tell the right-hand side doesn't have the correct units.

$$\mathrm{\frac{J}{mol\cdot K}\cdot K\cdot \frac{1}{m^3-\frac{m^3}{mol}}-\frac{N\cdot m^4}{mol}\cdot \frac{1}{(m^3)^2}}\tag{2}$$

$$\mathrm{\frac{N\cdot m}{mol}\cdot \frac{mol}{m^3\cdot mol -m^3}-\frac{N\cdot m}{mol\cdot m^3}}\tag{3}$$

$$\mathrm{\frac{N\cdot m}{m^3\cdot mol-m^3}-\frac{N}{mol\cdot m^2}}\tag{4}$$

However, if we move past this for now, work in our quasi-static isothermal process is

$$W=-\int_{V_i}^{V_f} PdV$$

$$=(...)$$

$$=RT\ln{\frac{V_i-b}{V_f-b}}-\frac{a}{V_f}+\frac{a}{V_i}\tag{5}$$

Then, if we plug in the values given in (b) we get

$$W=\mathrm{8.31\frac{J}{mol\cdot K}\cdot 293 K\cdot \ln{\left ( \frac{10L-3.2\cdot 10^{-5} \frac{m^3}{mol}}{22.4L - 1.4\cdot 10^9 \frac{N\cdot m^4}{L\cdot mol}} \right )}-\frac{1.4\times 10^9 \frac{N\cdot m^4}{mol}}{22.4L}+\frac{1.4\times 10^9 \frac{N\cdot m^4}{mol}}{10L}}$$

The numerical result of this calculation is $7.75\times 10^{10}$ which matches the answer at the end of the book.

My question is again about the units (which, as expected are probably not working given that they weren't working to begin with).

Specifically, the units of what is in the logarithm.

In the logarithm numerator and denominator individually, we have liters minus $\mathrm{m^3/mol}$.

Does this subtraction make sense?

Now, for all the units to work out, we need the logarithm to have no units at all. Then, the other units all work out to be $\frac{J}{mol}$. Now, these units are not correct, because we expect just $J$.

After having written this all out, I think the problem is in the original state equation.

I think there is one or more hidden $n$'s that were replaced with $1\text{mol}$ by the problem statement.

That being said, when I look back at the main text of the book I am reading, the van der Waals state equation does indeed seem to be correct in the problem statement. Which brings me back to my main question: how do the units work in this state equation?

 

[BvU]

A logarithm needs a dimensionless argument ($\log ab = \log a + \log b$ and the log of a dimension is non-existent ). So does an exponentiation.

The ideal gas law $PV = NRT$ has the dimension of energy: N/m² times m³ is Nm

What dimensions do you attribute to the $a$ and $b$ in the vdW equation ? and to R ?

$\$

 

[zenterix]

A logarithm needs a dimensionless argument ($\log ab = \log a + \log b$ and the log of a dimension is non-existent ).

The ideal gas law $PV = NRT$ has the dimension of energy: N/m² times m³ is Nm

What dimensions do you attribute to the $a$ and $b$ in the vdW equation ? and to R ?

$\$

It seems, reading Wikipedia, that perhaps the equation given in this problem is the intensive version. That is, $v=\frac{V}{n}$.

Indeed, I have not been paying attention to whether $v$ or $V$ is being used.

 

[BvU]

perhaps the equation given in this problem is the intensive version.

Not if it says $PV = RT$ for 1 mol (when a and b values are zero )

 

[zenterix]

After redoing some calculations, it still seems like the units for $a$ given in the problem are missing a unit of mol in the denominator.

That is, it seems the units for $a$ should be $\frac{N\cdot m^4}{mol^2}$.

A quick google search seems to indicate this as well.

 

[zenterix]

A logarithm needs a dimensionless argument ($\log ab = \log a + \log b$ and the log of a dimension is non-existent ). So does an exponentiation.

The ideal gas law $PV = NRT$ has the dimension of energy: N/m² times m³ is Nm

What dimensions do you attribute to the $a$ and $b$ in the vdW equation ? and to R ?

$\$

As for the issue of units in a logarithm, consider Newton's second law

$$F=ma$$

$$L=\ln{F}=\ln{(m)}+\ln{(a)}$$

What are the units of L?

The arguments of the logarithms are not dimensionless.

What happens here exactly?

 

[Chestermiller]

In the original equation, v is the molar volume m^3/mole. So the initial molar volume is $10\ \frac{l}{mole}=0.01\ \frac{m^3}{mole}$ and the final molar volume is $0.0224\ \frac{m^3}{mole}$.

The differential work done on the gas is $$dW=-Pdv$$and the units are $\frac{Pa.m^3}{mole}=\frac{N.m}{m^2}\frac{m^3}{mole}=\frac{N.m}{mole} =\frac{J}{mole}$

The logarithmic term is $$\ln{\frac{v_i-b}{v_f-b}}=\ln{\frac{(0.01-3.2\times 10^{-5})}{(0.0224-3.2\times 10^{-5})}}$$ with the argument being dimensionless.

 

[BvU]

$$L=\ln{F}=\ln{(m)}+\ln{(a)}$$

where does this come from ?

$\$

 

[zenterix]

where does this come from ?

$\$

This is just a generic question about taking logarithms of an expression where we have units involved.

I just thought of a simple example and came up with Newton's second law.

 

[BvU]

Nothing wrong with $F=ma$, but everything with your $L$ ! No such thing in science...

I suppose closest you can get is a logarithm of a ratio.

(tried to dig up something illustrative and uncomplicated; not so easy... e.g. logarithmic spiral but even there $k\phi = \log {r\over r_0}$ )

$\$

 

[haruspex]

As for the issue of units in a logarithm, consider Newton's second law

$$F=ma$$

$$L=\ln{F}=\ln{(m)}+\ln{(a)}$$

What are the units of L?

The arguments of the logarithms are not dimensionless.

What happens here exactly?

That's quite interesting. It shows that you can take logarithms of dimensioned values, thereby creating 'log dimensions'. E.g. $\ln(m)$ has dimension $\ln(M)$. It all works out when these are combined algebraically: $\ln(MLT^{-2})=\ln(M)+\ln(LT^{-2})$.

 

[haruspex]

(b) If $a=1.4\times 10^9\mathrm{N\cdot m^4/mol}$

I cannot make that work. Should it be $\mathrm{N\cdot m^4/mol^2}$ ?

 

[Chestermiller]

I cannot make that work. Should it be $\mathrm{N\cdot m^4/mol^2}$ ?

Yes, those are the correct units for "a".