Corrected exercises: gases and kinetic theory (202-SN1-RE)

The container is rigid, so $V_{1}=V_{2}$ and the relation reduces to $\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}$: this is Gay-Lussac's law. With $T_{1}=20+273=293$ K and $T_{2}=55+273=328$ K, we get $P_{2}=P_{1}\dfrac{T_{2}}{T_{1}}=20{,}265\times\dfrac{328}{293}=22{,}686$ kPa, that is 224 atm. The volume does not appear because it cancels on both sides: only its constancy matters, not its value. A rise of 35 °C therefore pushes the pressure up by 24 atmospheres, which is why a cylinder must never be left in the sun.

Here all three quantities change, so we need the combined gas law: $V_{2}=V_{1}\dfrac{P_{1}}{P_{2}}\times\dfrac{T_{2}}{T_{1}}=25.0\times\dfrac{101.3}{5.5}\times\dfrac{223}{293}=25.0\times 18.42\times 0.761=350$ m³. The balloon therefore swells by a factor of 14. The two effects work in opposite directions: the drop in pressure multiplies the volume by 18.4 while the cooling divides it by 1.31; the pressure wins by a wide margin. This is why a weather balloon is released half empty and ends up bursting at altitude.

At ground level: $n=\dfrac{PV}{RT}=\dfrac{101.3\times 25{,}000}{8.314\times 293}=1040$ mol, after converting the 25.0 m³ to 25,000 L. At altitude: $n=\dfrac{5.5\times 350{,}450}{8.314\times 223}=1040$ mol. We find exactly the same value, which is what we should get since the balloon is closed: the amount of substance is conserved. This is in fact the proof of the combined gas law, which is nothing but the statement of $\dfrac{PV}{T}=nR$ at two instants for the same $n$. This check through the amount of substance is the best way to verify a combined gas law exercise.

With degrees Celsius, the student would get $P_{2}=20{,}265\times\dfrac{55}{20}=55{,}729$ kPa, that is 550 atm: the cylinder would explode. The error is major because the ratio $\dfrac{55}{20}=2.75$ has nothing to do with the true ratio $\dfrac{328}{293}=1.12$. The reason is that the gas laws involve a RATIO of temperatures, and a ratio only makes sense on a scale whose zero is the true zero, that is, the kelvin. On the Celsius scale the zero is arbitrary: saying that it is twice as hot at 20 °C as at 10 °C has no physical meaning. A temperature DIFFERENCE, on the other hand, is the same on both scales, since the offset of 273 cancels: this is why a heat capacity can be written equally well in J/(g·°C) or J/(g·K).

First convert: $T=25+273=298$ K. Amounts of substance: $n(He)=\dfrac{2.00}{4.003}=0.4996$ mol and $n\left(O_{2}\right)=\dfrac{8.00}{32.00}=0.2500$ mol. Each gas alone occupies the 5.00 L: $P(He)=\dfrac{0.4996\times 8.314\times 298}{5.00}=247.6$ kPa and $P\left(O_{2}\right)=\dfrac{0.2500\times 8.314\times 298}{5.00}=123.9$ kPa. The total pressure is their sum: $P_{\text{total}}=371.5$ kPa. Note that helium, which has one quarter of the mass, gives twice the pressure: it is the amount of MATTER that counts, never the mass.

$x(He)=\dfrac{0.4996}{0.4996+0.2500}=\dfrac{0.4996}{0.7496}=0.6665$. Then $P(He)=x(He)\times P_{\text{total}}=0.6665\times 371.5=247.6$ kPa: we do recover the value from a). The two routes are equivalent, and the second is faster when the composition is given in percentages.

The gas collected over water is saturated with vapour: the pressure measured is that of the mixture. So $P\left(H_{2}\right)=101.3-2.6=98.7$ kPa. With $T=22+273=295$ K and $V=0.250$ L: $n=\dfrac{PV}{RT}=\dfrac{98.7\times 0.250}{8.314\times 295}=1.006\times 10^{-2}$ mol. The mass is $m=1.006\times 10^{-2}\times 2.016=2.03\times 10^{-2}$ g, or about 20 mg.

Without the correction, the student would use 101.3 kPa and find $n=1.033\times 10^{-2}$ mol, an error of $+2.6$%. The error is hard to detect because the result remains perfectly PLAUSIBLE: right order of magnitude, right units, no absurd value. Only a careful reading of the problem, which contains the words “collected over water” and a vapour pressure given for no apparent reason, can reveal it. RULE: if a water vapour pressure is supplied in the problem, it is there to be used.

$v_{rms}=\sqrt{\dfrac{3RT}{M}}$ with $M$ in kg/mol. For helium, $M=4.003\times 10^{-3}$ kg/mol: $v_{rms}=\sqrt{\dfrac{3\times 8.314\times 298}{4.003\times 10^{-3}}}=\sqrt{1.857\times 10^{6}}=1363$ m/s. For oxygen gas, $M=32.00\times 10^{-3}$ kg/mol: $v_{rms}=\sqrt{\dfrac{7433}{0.03200}}=482$ m/s. An $O_{2}$ molecule therefore moves at nearly 1700 km/h at room temperature.

$\dfrac{v(He)}{v\left(O_{2}\right)}=\dfrac{1363}{482}=2.83$. On the other hand, $\sqrt{\dfrac{M\left(O_{2}\right)}{M(He)}}=\sqrt{\dfrac{32.00}{4.003}}=\sqrt{7.99}=2.83$: the two coincide. Indeed, in the ratio $\sqrt{3RT/M_{1}}\div\sqrt{3RT/M_{2}}$, the factor $3RT$ cancels, and only the molar masses remain. This is exactly Graham's law.

$\overline{E_{k}}=\dfrac{3}{2}RT=\dfrac{3}{2}\times 8.314\times 298=3716$ J/mol, or 3.72 kJ/mol. This value is the SAME for both gases, and for every ideal gas: the average kinetic energy depends only on the temperature. This is in fact the microscopic definition of temperature. Light molecules make up for their small mass with a higher speed, so that the product $\frac{1}{2}mv^{2}$ stays the same.

Since $v_{rms}\propto\sqrt{T}$, doubling the absolute temperature multiplies the speed by $\sqrt{2}\approx 1.41$, not by 2: it would take QUADRUPLING $T$ to double the speed. Doubling the pressure at constant temperature changes NOTHING about the speed: $v_{rms}$ depends only on $T$ and $M$. Compressing a gas increases the FREQUENCY of the collisions, not their individual violence. This is the most common misconception of the chapter.

$\dfrac{v(He)}{v\left(SF_{6}\right)}=\sqrt{\dfrac{M\left(SF_{6}\right)}{M(He)}}=\sqrt{\dfrac{146.06}{4.003}}=\sqrt{36.49}=6.04$. Helium therefore effuses about six times faster.

Two effects combine, but the main one is this: helium, with a molar mass of 4.003 g/mol, effuses $\sqrt{29.0/4.003}=2.69$ times faster than air, whose average molar mass is 29.0 g/mol. It therefore passes through the pores of the latex much more quickly. In addition, the helium atom is monatomic and very small, which makes its passage even easier. A helium balloon thus loses its lift in about a day.

“Effusing 2.92 times more slowly” means $\dfrac{v(He)}{v(X)}=2.92$, so $\sqrt{\dfrac{M(X)}{4.003}}=2.92$ and $M(X)=4.003\times 2.92^{2}=4.003\times 8.53=34.1$ g/mol. This mass matches hydrogen sulfide $H_{2}S$, whose molar mass is 34.08 g/mol. Check: $\sqrt{34.08/4.003}=2.92$. Reading pitfall: “more slowly” puts the unknown mass in the NUMERATOR under the square root; inverting it would give $M=0.47$ g/mol, an impossible value since it is smaller than that of hydrogen.

$\dfrac{v\left(^{235}UF_{6}\right)}{v\left(^{238}UF_{6}\right)}=\sqrt{\dfrac{352.04}{349.03}}=\sqrt{1.00862}=1.00430$. The difference is only 0.43% per pass. To enrich natural uranium, which contains 0.7% of $^{235}U$, up to the 3 to 5% of nuclear fuel, THOUSANDS of effusion stages must therefore be chained in a cascade. This is what makes enrichment technically heavy, energy-hungry and easy to detect, a point with direct geopolitical consequences. Modern centrifuges exploit the same difference in mass, with a better yield per stage.

$P=\dfrac{nRT}{V}=\dfrac{1.00\times 8.314\times 300}{0.500}=4988$ kPa, nearly 49 atmospheres.

$P=\dfrac{nRT}{V-nb}-\dfrac{an^{2}}{V^{2}}=\dfrac{8.314\times 300}{0.500-0.0427}-\dfrac{364\times 1.00}{0.500^{2}}=\dfrac{2494}{0.4573}-1456=5454-1456=3998$ kPa. The relative difference is $\dfrac{3998-4988}{4988}=-19.9$%. The ideal gas law therefore OVERESTIMATES the pressure by nearly 20% under these conditions.

The term $nb$ subtracted from the volume represents the VOLUME OF THE MOLECULES THEMSELVES: the volume actually available for movement is smaller than the volume of the container, which INCREASES the pressure. The term $\dfrac{an^{2}}{V^{2}}$ represents the ATTRACTIVE FORCES between molecules: a molecule about to hit the wall is held back by its neighbours, so it hits less hard, which DECREASES the pressure. Here the two corrections are $+466$ kPa and $-1456$ kPa respectively: the attractive term clearly dominates, which makes sense for $CO_{2}$, a molecule that is easily liquefied and therefore has strong interactions.

At LOW pressure and HIGH temperature. At low pressure, the molar volume $\dfrac{V}{n}$ is large compared with $b$, so $V-nb\approx V$; and $\dfrac{an^{2}}{V^{2}}$ becomes negligible compared with $\dfrac{nRT}{V}$ since it decreases as $\dfrac{1}{V^{2}}$ while the main term only decreases as $\dfrac{1}{V}$. At high temperature, the average kinetic energy $\frac{3}{2}RT$ becomes large compared with the attraction energy, and the molecules pass each other too quickly to be held back. In practice, at ambient conditions and up to a few atmospheres, the difference is below 1%: this is why secondary school can ignore this correction entirely.

From $PV=nRT$ and $n=\dfrac{m}{M}$, we get $PV=\dfrac{m}{M}RT$, hence $M=\dfrac{mRT}{PV}$. Dividing by the volume, the density $\rho=\dfrac{m}{V}$ satisfies $\rho=\dfrac{PM}{RT}$. This last relation shows that, for a given gas, the density is proportional to the pressure and inversely proportional to the absolute temperature.

$T=100+273=373$ K. First $n=\dfrac{PV}{RT}=\dfrac{98.0\times 0.500}{8.314\times 373}=1.580\times 10^{-2}$ mol. Then $M=\dfrac{0.887}{1.580\times 10^{-2}}=56.1$ g/mol. This value matches butene $C_{4}H_{8}$, whose molar mass is 56.11 g/mol. One could also think of iron, but iron is obviously not a gas at 100 °C: the physical state is part of the identification.

$T=298$ K. For air: $\rho=\dfrac{101.3\times 29.0}{8.314\times 298}=1.186$ g/L. For helium: $\rho=\dfrac{101.3\times 4.003}{8.314\times 298}=0.164$ g/L. The ratio is 7.24, exactly that of the molar masses: at the same pressure and temperature, densities are proportional to molar masses, which is the direct translation of Avogadro's law.

The buoyant force corresponds to the mass of displaced air, that is $1.186\times 10.0=11.86$ g. The balloon already contains $0.164\times 10.0=1.64$ g of helium. The maximum payload is the difference: $11.86-1.64=10.2$ g. That is little: lifting one kilogram would take nearly 1000 L of helium, a balloon about 1.2 m in diameter.

Because what matters for buoyancy is not the composition but the DENSITY. According to $\rho=\dfrac{PM}{RT}$, at constant pressure the density is inversely proportional to the absolute temperature. By heating the air from 20 °C to 100 °C, that is from 293 K to 373 K, we divide $\rho$ by $\dfrac{373}{293}=1.27$: the air inside becomes 27% less dense than the air outside, and the buoyant force wins. It is the same principle as with helium, obtained not by changing $M$ but by changing $T$.

The bag contains nitrogen gas: $n\left(N_{2}\right)=\dfrac{PV}{RT}=\dfrac{101.3\times 65.0}{8.314\times 298}=2.658$ mol. The equation gives 3 mol of $N_{2}$ for 2 mol of $NaN_{3}$, so $n\left(NaN_{3}\right)=\dfrac{2}{3}\times 2.658=1.772$ mol. With $M\left(NaN_{3}\right)=22.99+3(14.01)=65.02$ g/mol, the mass is $1.772\times 65.02=115$ g. The order of magnitude is right: a real airbag contains about a hundred grams of azide.

At constant temperature and pressure, gas volumes are proportional to amounts of substance: the coefficients of the equation can be read directly as volumes. For 10.0 L of methane we therefore need 20.0 L of oxygen gas, and 10.0 L of carbon dioxide and 20.0 L of water vapour form, since at 150 °C water remains a gas. The total volume goes from $10.0+20.0=30.0$ L (methane and oxygen) to $10.0+20.0=30.0$ L (carbon dioxide and water vapour): it does not change. This is no coincidence: the sum of the gaseous coefficients is 3 on both sides of the equation.

Before the reaction, $n_{\text{total}}=0.100+0.300=0.400$ mol and $T=150+273=423$ K, so $P=\dfrac{nRT}{V}=\dfrac{0.400\times 8.314\times 423}{2.00}=703$ kPa. The limiting reactant is methane: $\dfrac{0.100}{1}=0.100$ against $\dfrac{0.300}{2}=0.150$ for oxygen gas. The reaction consumes 0.100 mol of $CH_{4}$ and 0.200 mol of $O_{2}$, and produces 0.100 mol of $CO_{2}$ and 0.200 mol of $H_{2}O$. After the reaction, $0.300-0.200=0.100$ mol of oxygen gas remains, plus 0.100 mol of $CO_{2}$ and 0.200 mol of water vapour, that is $n_{\text{total}}=0.400$ mol: exactly the same total as before. The pressure is therefore UNCHANGED, 703 kPa. This is the counterintuitive result of the exercise: a violent combustion can take place in a rigid container without changing the pressure once the temperature is back to its initial value, simply because the number of moles of gas is conserved, as b) already showed.

The partial pressures are calculated from the mole fractions: $P\left(O_{2}\right)=\dfrac{0.100}{0.400}\times 703=176$ kPa, $P\left(CO_{2}\right)=176$ kPa and $P\left(H_{2}O\right)=\dfrac{0.200}{0.400}\times 703=352$ kPa. At 25 °C, water vapour condenses almost entirely: only 3.2 kPa of vapour remains in the gas phase, the saturation value. The permanent gases, for their part, remain at $0.100+0.100=0.200$ mol and give $P=\dfrac{0.200\times 8.314\times 298}{2.00}=248$ kPa. The total pressure is therefore $248+3.2=251$ kPa, against 703 kPa before cooling. The drop by a factor of 2.8 combines two causes that must be told apart: the cooling from 423 K to 298 K accounts for a factor of $\dfrac{423}{298}=1.42$, and the disappearance of half of the gas moles by condensation accounts for a factor of 2. The product $1.42\times 2=2.84$ does give back the observed ratio. It is the same principle as the tin can that crumples as it cools, and as the vapour pressure correction of exercise 2.

With $M=28.02\times 10^{-3}$ kg/mol and $T=298$ K: $v_{p}=\sqrt{\dfrac{2\times 8.314\times 298}{0.02802}}=421$ m/s; $v_{avg}=\sqrt{\dfrac{8\times 8.314\times 298}{\pi\times 0.02802}}=475$ m/s; $v_{rms}=\sqrt{\dfrac{3\times 8.314\times 298}{0.02802}}=515$ m/s. We always have $v_{p}<v_{avg}<v_{rms}$, and the ratios are $\dfrac{v_{avg}}{v_{p}}=\sqrt{\dfrac{8}{2\pi}}=1.128$ and $\dfrac{v_{rms}}{v_{p}}=\sqrt{\dfrac{3}{2}}=1.225$. These ratios are numerical constants: the factor $\sqrt{RT/M}$ is common to the three expressions and disappears in the ratios. The order is explained by the asymmetry of the curve, whose long tail toward high speeds pulls the average above the peak, and pulls the root mean square even harder since it weights the speeds by their square.

Heating from 298 K to 1273 K: $v_{rms}$ goes from 515 to $\sqrt{\dfrac{3\times 8.314\times 1273}{0.02802}}=1065$ m/s, a factor of $\sqrt{\dfrac{1273}{298}}=2.07$. The peak of the curve moves to the right, the curve flattens and widens considerably, and the tail toward high speeds thickens a great deal. Replacing with hydrogen gas at 298 K: $v_{rms}=\sqrt{\dfrac{3\times 8.314\times 298}{0.002016}}=1920$ m/s, almost four times as fast, in agreement with $\sqrt{\dfrac{28.02}{2.016}}=3.73$. Here again the maximum moves to the right and the curve spreads out. What stays invariant in both cases is the area under the curve, which is always 1: it represents all of the molecules. A curve that widens must therefore necessarily come down, which is the key to reading all these figures.

At 1000 K: $v_{rms}\left(H_{2}\right)=\sqrt{\dfrac{3\times 8.314\times 1000}{0.002016}}=3517$ m/s and $v_{rms}\left(N_{2}\right)=\sqrt{\dfrac{3\times 8.314\times 1000}{0.02802}}=943$ m/s. The ratios are $\dfrac{11{,}200}{3517}=3.18$ for hydrogen gas and $\dfrac{11{,}200}{943}=11.9$ for nitrogen gas. Nitrogen gas is well above the threshold of 6: the Earth retains it without difficulty. Hydrogen gas is at 3.2, well below: a non-negligible fraction of its molecules is constantly exceeding the escape speed, and it escapes. This is why hydrogen gas, although the most abundant element in the universe, is almost absent from our atmosphere, as is helium. For the Moon, $\dfrac{2380}{943}=2.52$ for nitrogen gas itself: even heavy gases are below the threshold, which explains why it has no atmosphere. The retention criterion therefore combines the gravity of the body and the molar mass of the gas.

In a glass at 20 °C, the average speed of the water molecules is far below the one that would correspond to boiling, but the distribution is not a single value: its tail contains, at every instant, a small fraction of molecules fast enough to overcome the cohesive forces of the liquid and escape from the surface. These molecules leave, the slower ones stay, and the liquid would cool down if it were not warmed by the table and the air, which immediately regenerates a new tail. The glass therefore ends up emptying completely, just more slowly than at 100 °C, where the whole mass of the liquid can vaporize. The lesson shared with c) is that these two phenomena are governed neither by $v_{p}$, nor by $v_{avg}$, nor by $v_{rms}$, but by the fraction of molecules lying beyond a threshold. This fraction varies extremely sensitively with temperature: a slight warming thickens the tail far more than it moves the peak. It is the same reasoning that will explain, in chemical kinetics, why a rise of 10 °C can double a reaction rate.

We apply $P=\dfrac{nRT}{V-nb}-\dfrac{an^{2}}{V^{2}}$ then $Z=\dfrac{PV}{nRT}$ with $nRT=8.314\times 300=2494$ kPa·L. For $V=5.00$ L: $P=\dfrac{2494}{4.957}-\dfrac{364}{25.0}=503.1-14.6=488.6$ kPa, so $Z=\dfrac{488.6\times 5.00}{2494}=0.979$. For $V=0.500$ L: $P=\dfrac{2494}{0.4573}-\dfrac{364}{0.250}=5454-1456=3998$ kPa, so $Z=\dfrac{3998\times 0.500}{2494}=0.801$. For $V=0.100$ L: $P=\dfrac{2494}{0.0573}-\dfrac{364}{0.0100}=43{,}530-36{,}400=7129$ kPa, so $Z=\dfrac{7129\times 0.100}{2494}=0.286$. For $V=0.0600$ L: $P=\dfrac{2494}{0.0173}-\dfrac{364}{0.00360}=144{,}173-101{,}111=43{,}062$ kPa, so $Z=\dfrac{43{,}062\times 0.0600}{2494}=1.04$.

Starting from the largest volume, $Z$ is almost 1, then it decreases clearly below 1 as the gas is compressed, goes through a minimum, then rises sharply and ends up exceeding 1. In the first regime, the attractive correction $\dfrac{an^{2}}{V^{2}}$, which varies as $\dfrac{1}{V^{2}}$, grows faster than the molecular volume correction, which only enters through $\dfrac{nb}{V}$: the attractions dominate. A $Z<1$ then means that the gas is MORE compressible than an ideal gas, because the molecules attract each other and take up less room than expected. In the second regime, the available volume $V-nb$ tends to zero and the repulsive term blows up: the volume of the molecules themselves dominates, the gas becomes LESS compressible than an ideal gas, and $Z>1$. Condensed matter cannot be compressed indefinitely. The minimum marks the point where the two effects exactly compensate.

We write $Z=\dfrac{PV}{RT}$ for one mole with $P=\dfrac{RT}{V-b}-\dfrac{a}{V^{2}}$, that is $Z=\dfrac{V}{V-b}-\dfrac{a}{RTV}=\dfrac{1}{1-\frac{b}{V}}-\dfrac{a}{RTV}$. At large volumes, $\dfrac{1}{1-\frac{b}{V}}\approx 1+\dfrac{b}{V}$, hence $Z\approx 1+\dfrac{1}{V}\left(b-\dfrac{a}{RT}\right)$. Since $\dfrac{1}{V}\approx\dfrac{P}{RT}$ to this order, we do get $Z\approx 1+\dfrac{1}{RT}\left(b-\dfrac{a}{RT}\right)P$. The coefficient of $P$ vanishes when $b=\dfrac{a}{RT}$, that is at the Boyle temperature $T_{B}=\dfrac{a}{Rb}$. For carbon dioxide: $T_{B}=\dfrac{364}{8.314\times 0.0427}=1025$ K. Below this temperature, the coefficient is negative and $Z$ starts by dropping below 1: this is just what we observed in a) at 300 K. Above it, the gas is already in the regime where the molecular volume prevails and $Z$ increases from low pressures on.

The disagreement is real: the order of magnitude is right but the model overestimates by about 40%. This does not invalidate van der Waals, it delimits its scope. The equation only corrects the ideal gas law with two fitted constants: $b$ assumes spherical, incompressible molecules, and $a$ sums up all the attractions in a single average term, independent of distance and temperature, which is very crude for a molecule like carbon dioxide. A two-parameter equation cannot reproduce quantitatively a property as fine as a compensation temperature. What the model does predict correctly, on the other hand, is essential: the existence of a minimum of $Z$, the fact that the attractive regime comes before the repulsive regime, the very existence of a Boyle temperature, liquefaction and the critical point, and the return to the ideal gas at low pressure and high temperature. It is the mark of a good physical model to give the right qualitative behaviours with two parameters, while leaving the precise numerical values to empirical equations of state.