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Smoke and Daggers

Age 16 to 18
Challenge Level

The moles of oxygen may be determined through the use of the ideal gas equation, PV = nRT, which relates pressure and volume of a gas. Of course this ideal modelling is full of assumptions such as the particles are point like and collisions between particles are infinitely elastic. Care needs to be taken with the units used!

The volume of Ellie's lungs is 4.7 dm$^3$ which is equivalent to 0.0047 m$^3$. Thus the corresponding volume of oxygen in her fully inflated lungs may be calculated as:
$$0.0047 \times 0.2095 = 9.8465 \times 10^{-4}\ \text{m}^3$$
Then using the ideal gas equation in the form, n= $\frac{PV}{RT}$.
$$n = \frac{PV}{RT} = \frac{760 \times 133.322 \times 9.8465 \times 10^{-4}}{8.3145 \times 298} = 0.0403\ \text{moles (3sf)}$$

A periodic table will give you the relative atomic masses of all elements considering the terrestrial abundance of various isotopes. The listed compounds therefore have the following relative molecular masses:

C=O 12.0107 + 15.9994 = 28.0101 g mol$^{-1}$
HCN 1.00794 + 12.0107 + 14.00674 = 27. 02538 g mol$^{-1}$
Nicotine [C$_{10}$H$_{14}$N$_{2}$] (1.00794 $\times$ 14) + (12.0107 $\times$ 10) + (14.00674 $\times$ 2) = 162.23164 g mol$^{-1}$
Cyclohexane [C$_{6}$H$_{12}$] $(12.0107 \times 6) + (1.00794 \times 12) = 84.15948$ g mol$^{-1}$

The moles of each can be simply calculated from $\frac{\text{mass in grams}}{\text{RMM}}$:
C=O $\frac{16 \times 10^{-3}}{28.0101}$ = 5.71 $\times$ 10$^{-4}$ moles (3sf)
HCN $\frac{300 \times 10^{-6}}{27.02538}$ = 1.11 $\times$ 10$^{-5}$ moles (3sf)
Nicotine $\frac{10 \times 10^{-3}}{162.23164}$ = 6.16 $\times$ 10$^{-5}$ moles (3sf)
Cyclohexane $\frac{300 \times 10^{-9}}{84.15948}$ = 3.56 $\times$ 10$^{-9}$ moles (3sf)

The average number of puffs per cigarette is quite simply:
$$\frac{17 + 21 + 23 + 19 + 23 + 20 + 18 + 19}{8} = 20\ \text{puffs}$$
We know that a 2 second puff has a volume of 45 ml. 20 of these 5 second puffs would give a volume of:
$$45 \times \frac{5}{2} \times 20 = 2250\ \text{ml} = 2.25\ \text{dm}^3$$
The concentrations of all the following compounds in the cigarette smoke can be easily determined:

C=O $\frac{5.71 \times 10^{-4}}{2.25} = 2.54 \times 10^{-4}\ \text{mol dm}^{-3}$ (3sf)
HCN $\frac{1.11 \times 10^{-5}}{2.25} = 4.93 \times 10^{-6}\ \text{mol dm}^{-3}$ (3sf)
Nicotine $\frac{6.16 \times 10^{-5}}{2.25} = 2.74 \times 10^{-5}\ \text{mol dm}^{-3}$ (3sf)
Cyclohexane $\frac{3.56 \times 10^{-9}}{2.25} = 1.58 \times 10^{-9}\ \text{mol dm}^{-3}$ (3sf)

If the cigarette smoke was such that there was a 1 moldm$^{-3}$ concentration of a substance initially, then there are 0.39 moles in 390 ml. The volume of her inflated lungs is 0.93 + 0.93 + 0.39 = 2.25 dm$^3$. Which means there is a new concentration of $\frac{0.39}{2.25}$ = $\frac{13}{75}$. The dilution factor is therefore $\frac{1}{\frac{13}{75}}$ = $\frac{75}{13}$ which is roughly 5.8 (1dp).

The moles of each compound present in her lungs is 0.39 times the respective concentrations.

C=O $2.54 \times 10^{-4} \times 0.39 = 9.90 \times 10^{-5}\ \text{moles}$ (3sf)
HCN $4.93 \times 10^{-6} \times 0.39 = 1.92 \times 10^{-6}\ \text{moles}$ (3sf)
Nicotine $2.74 \times 10^{-5} \times 0.39 = 1.07 \times 10^{-5}\ \text{moles}$ (3sf)
Cyclohexane $1.58 \times 10^{-9} \times 0.39 = 6.18 \times 10^{-10}\ \text{moles}$ (3sf)

Given that the pressure in her lungs can be assumed to still be 760mmHg, the partial pressures can be found by multiplying the molar ratios by the total pressure.The total number of moles of the components in the cigarette smoke (on the assumption that they are the only components of the smoke) is:

$9.90 \times 10^{-5} + 1.92 \times 10^{-6} + 1.07 \times 10^{-5} + 6.18 \times 10^{-10} = 1.12 \times 10^{-4}\ \text{moles}$ (3sf)

Using the ideal gas equation, this number of moles of gas would occupy the following volume:

$V = \frac{nRT}{P} = \frac{1.12 \times 10^{-4} \times 8.3145 \times 310}{760 \times 133.322} = 2.84 \times 10^{-6}\ $m$^3$. Then if we assume that the expansion of gas breathed in due to increase in temperature from 25$^o$C to body temperature is minimal (meaning that the volume of Ellie's inflated lungs is still 2.25 dm$^3$ once she has breathed in), the volume of "air" in her lungs is:

$2.25 - 2.84 \times 10^{-3} = 2.24716059\ \text{dm}^3$

The moles of components of the air (e.g. N$_2$, O$_2$, Ar...) can be given as:

$n = \frac{PV}{RT} = \frac{760 \times 133.322 \times 2.24716059 \times 10^{-3}}{8.3145 \times 310} = 0.08833883967 = 8.83 \times 10^{-2}\ $ moles (3sf)

The total number of moles in her lungs is therefore:

$8.83 \times 10^{-2} + 1.12 \times 10^{-4} = 0.08845046062 = 8.85 \times 10^{-2}\ $ moles

Now we can begin to work out the partial pressures of each of the components of cigarette smoke occupying Ellie's lungs:

$P_{C=O} = \frac{9.90 \times 10^{-5}}{8.85 \times 10^{-2}} \times 760 =$ 0.851 mmHg (3sf) = 113 Pa (3sf)
$P_{HCN} = \frac{1.92 \times 10^{-6}}{8.85 \times 10^{-2}} \times 760 =$ 0.0165 mmHg (3sf) = 2.20 Pa (3sf)
$P_{Nicotine} = \frac{1.07 \times 10^{-5}}{8.85 \times 10^{-2}} \times 760 =$ 0.0918 mmHg (3sf) = 12.2 Pa (3sf)
$P_{Cyclohexane} = \frac{6.18 \times 10^{-10}}{8.85 \times 10^{-2}} \times 760 =5.31 \times 10^{-6}$ mmHg (3sf) = 7.08 $\times$ 10$^{-4}\ $ Pa (3sf)

Assuming that the alveolar air is of given composition, an average molecular mass can be calculated.

$AMM = 0.0405 \times (12.0107 + (2 \times 15.9994)) + 0.161 \times (2 \times 15.9994) + 0.7985 \times (14.00674 \times 2) = 29.30295533\ $ g mol$^{-1}$

The number of moles of "air" has already been calculated using a rearrangement of $n = \frac{PV}{RT}$ using a valid volume. The mass of this "air" component can be calculated using the AMM.

mass$_{air} = 29.30295533 \times 8.83 \times 10^{-2} = 2.588589073 = 2.59$ g (3sf)

The number of moles of all other compounds have been calculated previously. Multplying these values with the relative molecular mass values previously obtained gives the masses of these components in Ellie's lungs.

C=O $9.90 \times 10^{-5} \times 28.0101 \text{gmol}^{-1}= 2.773333333 \times 10^{-3}\ $g (3sf)
HCN $1.92 \times 10^{-6} \times 27. 02538 \text{gmol}^{-1}= 5.2 \times 10^{-5}\ $g (3sf)
Nicotine $1.07 \times 10^{-5} \times 162.23164 \text{gmol}^{-1}= 1.733333333 \times 10^{-3}\ $g (3sf)
Cyclohexane $6.18 \times 10^{-10} \times 84.15948\ \text{g mol}^{-1}=5.2 \times 10^{-8}\ $g (3sf)

Total mass of these components $= 4.558718666 \times 10^{-3}\ $g

Adding this to the mass of "air" in Ellie's lungs, we obtain a value of:

mass$_{total} = 2.588589073 + 4.558718666 \times 10^{-3} = 2.593147792\ $ g

The density of the smoke filled air in Ellie's lungs is therefore:

density $= \frac{2.593147792}{2.25 \times 10^{3}} = 1.15251013 \times 10^{-3} = 1.15 \times 10^{-3}\ \text{g cm}^{-3}$ (3sf)

The following unlikely situation would of course require Ellie to smoke several cigarettes voraciously such that the air in her lungs was actually a uniform smoke in composition. Nevertheless, performing the calculation gives an insight into the different molar values obtained depending on the amount of smoke inhaled and what these mean in real terms.

The total volume occupied by smoke is equivalent to the total lung capacity, 0.93 + 0.93 + 0.39 + 2.45 = 4.7 dm$^3$. Thus the moles of each compound present is simply this value multiplied by the concentration of each compound in the smoke, which can then be multiplied by Avogadro's number ($6.022 \times 10^{23}$) to give the number of actual molecules this corresponds to.

C=O $2.54 \times 10^{-4} \times 4.7 = 1.193220382 \times 10^{-3}$ moles
No. of molecules: $1.193220382 \times 10^{-3} \times 6.022 \times 10^{23} = 7.18557314 \times 10^{20} = 7.19 \times 10^{20}\ $ molecules (3sf)

HCN $4.93 \times 10^{-6} \times 4.7 = 2.318807975 \times 10^{-5} $ moles (3sf)
No. of molecules: $2.318807975 \times 10^{-5} \times 6.022 \times 10^{23} = 1.396386162 \times 10^{19} = 1.40 \times 10^{19}\ $ molecules (3sf)

Nicotine $2.74 \times 10^{-5} \times 4.7 = 1.287596482 \times 10^{-4}$ moles (3sf)
No. of molecules: $1.287596482 \times 10^{-4} \times 6.022 \times 10^{23} = 7.753906013 \times 10^{19} = 7.75 \times 10^{19}\ $ molecules (3sf)

Cyclohexane $1.58 \times 10^{-9} \times 4.7 = 7.446180355 \times 10^{-9}$ moles (3sf)
No. of molecules: $7.446180355 \times 10^{-9} \times 6.022 \times 10^{23} = 4.48408981 \times 10^{15} = 4.48 \times 10^{15}\ $ molecules (3sf)

The answer to the next part of the question assumes that Ellie stops breathing for a period and that the diffusion of nicotine into her blood is constant over time, despite a decreasing concentration of nicotine over time in the alveolar air.

We know that Ellie's lungs contain $1.287596482 \times 10^{-4}$ moles of nicotine, which corresponds to a mass of $1.287596482 \times 10^{-4} \times 162.23164 \times 10^{6} =$ 20888.88889 $\mu$g.

This gives a time of $\frac{20888.88889}{400}$ = 52.22222223 = 52 seconds (to the nearest second)

The only way we know how to work out a concentration from a given pressure and number of moles is using the ideal gas equation. Of course this requires modelling Ellie's blood as an ideal gas which clearly has some drawbacks considering the physical nature of blood. We will come back to this point.

The blood pressure values given are those when there is a point of maximal aortic contraction and when the heart is in a relaxed diastolic state and the atria are filling with blood. Assuming that these two phases are of the same length in the human body, an average figure of 100 mmHg can be generated. This pressure is far lower than the atmospheric value of 760 mmHg and it can be considered that the blood pressure is the pressure measured over and above the atmospheric pressure. Thus we can take the absolute pressure of Ellie's blood as 860 mmHg. Rearranging the ideal gas equation:

$V_{Nicotine} = \frac{nRT}{P} = \frac{1.287596482 \times 10^{-4} \times 8.3145 \times 310}{860 \times 133.322} = 2.894525245 \times 10^{-6}\ \text{m}^{3}$

This can give a volume ratio which can determine a partial pressure for nicotine in Ellie's "blood" as if pressure and temperature remain constant the $PV = nRT$ relationship means that volumes are proportional to molar quantities.

Thus the volume ratio for nicotine is $\frac{2.894525245 \times 10^{-6}}{4.7 \times 10^{-3}} = 6.158564351 \times 10^{-4}$.

The partial pressure of nicotine in the blood is therefore $6.158564351 \times 10^{-4} \times 860 \times 133.322 = 70.61220201\ $ Pa.

Rearranging the ideal gas equation, $\frac{n}{V}$ = $\frac{P}{RT}$.

So $\frac{n}{V}$ = $\frac{70.61220201}{8.3145 \times 310} = 0.02739566983 \text{mol m}^{3} = 2.74 \times 10^{-5} \text{mol dm}^{3}$ (3sf)

The assumptions of the ideal gas equation include the fact that molecules can be modelled as small, hard spheres. All collisions between molecules are elastic and the motion of molecules in the medium is frictionless. The molecules follow Newtonian dynamics. The distance between molecules is much larger than the size of the molecules on average. The direction of movement of molecules is random and the speeds follow a distribution. There are no attractive or repulsive forces between molecules and their environment.

In reality, an ideal gas does not exist as assumptions of frictionless movement and lack of forces between molecules and elastic collisions simply do not happen in real life. The assumptions are less valid for blood as this is a liquid so motion is more readily affected by friction, intramolecular forces are clearly greater and collisions less likely to be elastic. It is also true to say that the distance between molecules being larger than the size of the molecules may be less valid for a liquid.

From the information given it takes 20 $\times$ 5 = 100 seconds of continuous smoking to finish a cigarette. Each cigarette is known to deliver 10 mg of nicotine, and as uptake of nicotine was quoted as 400 $\mu$gs$^{-1}$ a rough calculation would mean that uptake could take 25 seconds if this dose was delivered all at once. These times are very small in comparison to the 2 hour half life of nicotine stated. It may be assumed that on smoking the cigarette at every half hour, there is almost instantaneous delivery of nicotine to the blood. We also know that the volume of blood in Ellie's body is 4.7 litres and that nicotine uptake makes a minimal change to the volume, so we can consider each cigarette smoked as dissolving the appropriate amount of nicotine in Ellie's blood.

At t = 0, Ellie smokes a cigarette which corresponds to an initial blood nicotine concentration of $\frac{10^7}{4700} = \frac{100000}{47}$ ng ml$^{-1}$.

Just before t = 30 minutes, the blood nicotine concentration has reduced to $\frac{100000}{47} \times (\frac{1}{2})^{(\frac{1}{4})} = 1789.141309$ ng ml$^{-1}$

At t = 30, another cigarette dose is given and over the course of the next half hour, the concentration of nicotine is again subjected to a quarter of the half life:

$(1789.141309 + \frac{100000}{47}) \times (\frac{1}{2})^{(\frac{1}{4})} = 3293.623822$ ng ml$^{-1}$

At t = 60, this process is repeated:

$(3293.623822 + \frac{100000}{47}) \times (\frac{1}{2})^{(\frac{1}{4})} = 4558.737774$ ng ml$^{-1}$

At t = 90,

$(4558.737774 + \frac{100000}{47}) \times (\frac{1}{2})^{(\frac{1}{4})} = 5622.567562$ ng ml$^{-1}$

At t = 120,

$(5622.567562 + \frac{100000}{47}) \times (\frac{1}{2})^{(\frac{1}{4})} = 6517.138216$ ng ml$^{-1}$

At t = 150,
$(6517.138216 + \frac{100000}{47}) \times (\frac{1}{2})^{(\frac{1}{4})} = 7269.379473$ ng ml$^{-1}$

At t = 180, Ellie smokes her final cigarette:
$(7269.379473 + \frac{100000}{47}) = 9397.039047$ ng ml$^{-1}$

This level of nicotine then must drop to below 200 ng ml$^{-1}$.

$9397.039047 \times (0.5)^n$ < 200
0.5$^n$ < 0.02128329988
n > $\frac{\log 0.02128329988}{\log 0.5}$
n > 5.554134338

This corresponds to a time greater than $5.554134338 \times 2= 11.10826868$ hours. We can reasonably say that after 14 hours and 10 minutes after smoking her first cigarette, Ellie will be able to take the nicotine detection test and have a negligible level of nicotine in her blood stream (ie. below the detectable level).

Let us assume that there is a dilution of dangerous compounds to a $\frac{1}{3}$ of the level coming out of the burning end of a cigarette. Thus the percentage of dangerous compounds in the air breathed in by Pete is now $\frac{86}{3}$ = 28.7 % (3sf).

Consider that initially the volume of air in Ellie's lungs is 1.86 dm$^{3}$ and an intake of smoke through a cigarette provides 0.39 dm$^{3}$ of air, 11.2% of which is composed of dangerous compounds. Therefore the fraction of dangerous compounds in Ellie's lungs as a decimal after 1 drag is:

$\frac{(0.39)(0.112)}{2.25} = 0.01941333333$

For the next breath out and breath in, it is easy to see that 1.86 dm$^3$ of the alveolar air will have the same composition of dangerous compounds as the decimal calculated. The additional 0.39 dm$^{3}$ of air is that taken through a cigarette meaning it again contains an 11.2% concentration of dangerous compounds. Therefore the fraction of dangerous compounds in Ellie's lungs as a decimal after 2 drags is:

$\frac{0.39 \times 0.112}{2.25} \times \frac{1.86}{2.25} + \frac{(0.39)(0.112)}{2.25} = 0.03546168889$

By the 15th inward breath, the fraction of dangerous compounds is

$^{n-1}_{k=0}\Sigma = \frac{(0.39)(0.112)}{2.25}\times (\frac{1.86}{2.25})^k$

$\frac{(0.39)(0.112)}{2.25} \times ^{n-1}_{k=0}\Sigma = (\frac{1.86}{2.25})^k$

$\frac{(0.39)(0.112)}{2.25} \times ^{14}_{k=0}\Sigma = (\frac{1.86}{2.25})^k$

$\frac{(0.39)(0.112)}{2.25} \times (1 + (\frac{1.86}{2.25}) + (\frac{1.86}{2.25})^2 + ... + (\frac{1.86}{2.25})^{14}) = 0.10555572 ...= 0.106$ (3sf)

On the last outward breath, this fractional value remains the same as a certain volume of air with this fraction of dangerous particles is merely being expired. So the percentage of dangerous compounds in Ellie's lungs after this period is 10.6% (3sf).

Similarly, the percentage of dangerous compounds in Pete's lungs can be calculated. Consider that after 1 breath of 0.125dm$^{3}$, the fraction of dangerous compounds in Pete's lungs (considering he has a functional residual capacity of 0.6 dm$^{3}$) is:

$\frac{(0.125)(\frac{0.86}{3})}{0.725} = 0.04942528736$

For the next breath out and in this becomes:

$\frac{(0.125)(\frac{0.86}{3})}{0.725} \times \frac{0.6}{0.725} + \frac{(0.125)(\frac{0.86}{3})}{0.725} = 0.09032897344$

By the 15th inward breath, the fraction of dangerous compounds is

$^{n-1}_{k=0}\Sigma = \frac{(0.125)(\frac{0.86}{3})}{0.725} \times (\frac{0.6}{0.725})^k$

$\frac{(0.125)(\frac{0.86}{3})}{0.725} \times ^{n-1}_{k=0}\Sigma = (\frac{0.6}{0.725})^k$

$\frac{(0.125)(\frac{0.86}{3})}{0.725} \times ^{14}_{k=0}\Sigma = (\frac{0.6}{0.725})^k$

$\frac{(0.125)(\frac{0.86}{3})}{0.725} \times (1 + (\frac{0.6}{0.725}) + (\frac{0.6}{0.725})^2 + ... + (\frac{0.6}{0.725})^{14}) = 0.269895... = 0.270$ (3sf)

This yields a percentage of 27.0% (3sf) for this dilution factor.

It may be easier to complete the extension exercise using a spreadsheet as the dilution value can easily be varied and the calculated output obtained. After some experimentation you might obtain a value for the dilution that makes the percentage of dangerous compounds in the lungs of Ellie and Pete to be something like $\frac{34175}{262144}$.