Chapter 12 ยท Kinetic Theory

12.1 The Molecular Nature of Matter and the Behaviour of Gases

Matter is made of molecules, and in a gas they are far apart and almost free. Count them with the mole and the Avogadro constant, and every dilute gas obeys one equation: $pV=nRT=Nk_BT$.

Put a drop of ink into a glass of still water and, without any stirring, the colour slowly spreads through the whole glass. Squeeze the air in a bicycle pump with your finger over the outlet and it pushes back harder the more you squeeze; let it go and it springs back. A sealed packet of crisps carried up a mountain puffs up like a pillow. These everyday observations are hard to explain if matter is a smooth, continuous jelly. They are easy to explain if matter is made of an enormous number of tiny particles that are always moving โ€” and, in a gas, are mostly empty space with a few particles flying about in it.

This section collects the evidence for that molecular picture and puts numbers on it: how big molecules are, how far apart they sit in a solid, a liquid and a gas, and how to count them with the mole and the Avogadro constant. Then it re-reads the gas laws you met with the ideal-gas equation in Chapter 10 (Thermal Properties of Matter) โ€” Boyle's law, Charles's law, Avogadro's hypothesis and Dalton's law of partial pressures โ€” in the language of molecules, through the molecular form of the gas equation, $pV=Nk_BT$. Section 12.2 will then explain why gases obey it.

The molecular nature of matter

The idea that matter is made of indivisible particles is very old โ€” it appears in ancient Indian philosophy (Kanada's paramanu) and in Greek thought (Democritus's atomos) โ€” but for two thousand years it was a speculation. It became science around 1800, when chemists found that elements always combine in fixed, simple ratios by mass (Dalton's atomic theory), and that gases react in simple ratios by volume (Gay-Lussac). Today the evidence is overwhelming: the jittering of pollen grains in water (Brownian motion, explained by Einstein in 1905 as the grains being knocked about by water molecules), X-ray diffraction from crystals, and microscopes that image individual atoms.

Atoms are about $10^{-10}\ \text{m}$ ($1$ รฅngstrรถm, $1$ ร…) across. In solids they are packed tightly, at a spacing of about $2$ ร…, held near fixed positions by strong forces, though they vibrate about them. In liquids the spacing is about the same, but the molecules are not tied to fixed positions and can slide past one another. In gases at ordinary pressures the molecules are about ten times further apart, tens of รฅngstrรถms, so the forces between them are negligible except during brief collisions: each molecule travels freely in a straight line until it hits another molecule or a wall. That is why gases are easy to compress, fill any container and mix freely, and why their behaviour is so simple.

๐Ÿ“˜ Definition 12.1 โ€” The mole, the Avogadro constant and molar mass

One mole ($1\ \text{mol}$) of a substance is the amount that contains exactly $N_A$ elementary entities (atoms or molecules), where the Avogadro constant is

$$N_A=6.022\times10^{23}\ \text{mol}^{-1}\quad(\text{we use }6.02\times10^{23}\ \text{mol}^{-1}).$$

A sample of $N$ molecules contains $n=\dfrac{N}{N_A}$ moles. The molar mass $M$ is the mass of one mole: numerically the molecular mass in grams, so $M=0.028\ \text{kg/mol}$ for nitrogen, $\mathrm{N_2}$. The mass of one molecule is $m=\dfrac{M}{N_A}$, and a sample of mass $M_s$ contains $n=\dfrac{M_s}{M}$ moles.

โœ๏ธ Example 1 โ€” How far apart are the molecules of water and of steam?
Water has molar mass $18\ \text{g/mol}$ and density $1.0\times10^3\ \text{kg/m}^3$. (a) Find the volume available to one molecule in liquid water and hence the typical spacing between molecules. (b) Steam at $100\ ^\circ\text{C}$ and atmospheric pressure ($1.013\times10^5\ \text{Pa}$) behaves nearly as an ideal gas; the volume per molecule is then $\dfrac{k_BT}{p}$ (see Result 12.2 below). Find the typical spacing in steam and compare.
(a) Mass of one molecule $m=\dfrac{0.018}{6.02\times10^{23}}=2.99\times10^{-26}\ \text{kg}$. Volume per molecule $\dfrac{m}{\rho}=2.99\times10^{-29}\ \text{m}^3$. Imagining each molecule in a small cube of that volume, the spacing is its side: $\left(2.99\times10^{-29}\right)^{1/3}=3.1\times10^{-10}\ \text{m}=3.1$ ร….
(b) $\dfrac{k_BT}{p}=\dfrac{1.38\times10^{-23}\times373}{1.013\times10^5}=5.08\times10^{-26}\ \text{m}^3$; spacing $\left(5.08\times10^{-26}\right)^{1/3}=3.7\times10^{-9}\ \text{m}=37$ ร….
Compare The volume per molecule is about $1700$ times larger in steam, and the spacing about $12$ times larger ($\sqrt[3]{1700}\approx12$). A molecule of water is roughly $3$ ร… across, so in steam the molecules are far apart compared with their size.
Check $1\ \text{g}$ of water ($1\ \text{cm}^3$) boils to about $1.7\ \text{L}$ of steam โ€” a familiar figure for how much a kettle's water expands into vapour โœ“.
๐Ÿ“˜ Definition 12.2 โ€” Ideal gas

An ideal gas is a model gas in which (i) the molecules are so small compared with their separation that their own volume is negligible, and (ii) they exert no forces on one another except during collisions, which are elastic and take negligible time. Real gases behave almost ideally when they are dilute โ€” at low pressure and at temperatures well above the temperature at which they liquefy.

The gas equation in molecular form

Experiments on a fixed amount of any dilute gas give three simple laws, which you met with the ideal-gas equation in Chapter 10 (Thermal Properties of Matter). They are worth restating, because each says something about molecules.

๐Ÿ“ Law 12.1 โ€” Boyle's law, Charles's law and the pressure law

For a fixed amount of a dilute gas:

  • Boyle's law: at constant temperature, $pV=\text{constant}$, so $p_1V_1=p_2V_2$;
  • Charles's law: at constant pressure, $V\propto T$, so $\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$;
  • pressure law (Gay-Lussac): at constant volume, $p\propto T$, so $\dfrac{p_1}{T_1}=\dfrac{p_2}{T_2}$,

where $T$ is the absolute (kelvin) temperature. Together they say $\dfrac{pV}{T}$ is constant: $\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$.

The molecular reading is natural. In Boyle's law, squeezing the gas into half the volume packs twice as many molecules into each cubic metre, so twice as many hit each square metre of wall every second: the pressure doubles. In the pressure law, heating the gas makes the molecules move faster, so they hit the walls harder and more often. The constant $\dfrac{pV}{T}$ is found to be proportional to the amount of gas โ€” and, remarkably, it is the same for every gas when the amount is measured in moles, or simply by counting molecules.

๐Ÿ“ Result 12.2 โ€” The ideal-gas equation in molecular form

For $n$ moles, or $N=nN_A$ molecules, of an ideal gas at absolute temperature $T$,

$$pV=nRT=Nk_BT,$$

where $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$ is the gas constant and

$$k_B=\frac{R}{N_A}=\frac{8.31}{6.02\times10^{23}}=1.38\times10^{-23}\ \text{J/K}$$

is the Boltzmann constant โ€” the gas constant per molecule. Dividing by $V$, the pressure depends only on the number density $n_V=\dfrac NV$ (molecules per cubic metre) and the temperature: $p=n_Vk_BT$. In terms of the density $\rho$ and molar mass $M$, $p=\dfrac{\rho RT}{M}$.

The equation contains no property of the particular gas. A cubic metre of hydrogen and a cubic metre of carbon dioxide at the same temperature and pressure hold the same number of molecules, even though each $\mathrm{CO_2}$ molecule is $22$ times heavier. Notice the care needed with symbols: $n$ is the number of moles, $N$ the number of molecules, and we write $n_V$ for the number density.

โœ๏ธ Example 2 โ€” Molecules in one breath
You breathe in $0.50\ \text{L}$ of air, which warms to body temperature, $37\ ^\circ\text{C}$, at atmospheric pressure $1.013\times10^5\ \text{Pa}$. (a) How many molecules is that? (b) How many moles? (c) What is the number density of air molecules in your lungs?
(a) $T=310\ \text{K}$, $V=0.50\times10^{-3}\ \text{m}^3$. $N=\dfrac{pV}{k_BT}=\dfrac{1.013\times10^5\times5.0\times10^{-4}}{1.38\times10^{-23}\times310}=1.2\times10^{22}$ molecules.
(b) $n=\dfrac{N}{N_A}=\dfrac{1.18\times10^{22}}{6.02\times10^{23}}=0.020\ \text{mol}$ (check: $\dfrac{pV}{RT}=\dfrac{50.65}{8.31\times310}=0.0197$ โœ“).
(c) $n_V=\dfrac{p}{k_BT}=\dfrac{1.013\times10^5}{1.38\times10^{-23}\times310}=2.4\times10^{25}\ \text{m}^{-3}$.
Check There are more molecules in one breath than there are breaths of air in the whole atmosphere (its mass, $5\times10^{18}\ \text{kg}$, is about $8\times10^{21}$ breaths of $0.6\ \text{g}$). So, once the atmosphere has mixed, each breath you take contains on average one or two molecules from any given breath of anyone in history.
๐Ÿงญ Problem-solving strategy โ€” gas-law problems
  1. Convert every temperature to kelvin: $T=t_C+273$ (use $273.15$ when the data justify it).
  2. Use absolute pressure. A tyre gauge reads gauge pressure: add atmospheric pressure first, and subtract it again at the end if a gauge reading is asked for.
  3. If the amount of gas is fixed, use $\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$ โ€” any consistent units for $p$ and $V$ will do, since they cancel. If the amount is asked for or changes, use $pV=nRT$ (or $Nk_BT$) in SI units.
  4. Check the direction: hotter at constant volume means higher pressure; compressed at constant temperature means higher pressure.
โœ๏ธ Example 3 โ€” A car tyre on a hot road
A car tyre is pumped to a gauge pressure of $2.0\times10^5\ \text{Pa}$ on a cool morning at $27\ ^\circ\text{C}$. After a long drive the air in it is at $47\ ^\circ\text{C}$. Assuming the volume of the tyre does not change, find the new gauge pressure. Take atmospheric pressure as $1.0\times10^5\ \text{Pa}$.
Absolute $p_1=2.0\times10^5+1.0\times10^5=3.0\times10^5\ \text{Pa}$; $T_1=300\ \text{K}$, $T_2=320\ \text{K}$.
Pressure law $p_2=p_1\dfrac{T_2}{T_1}=3.0\times10^5\times\dfrac{320}{300}=3.2\times10^5\ \text{Pa}$.
Gauge $p_2-p_a=2.2\times10^5\ \text{Pa}$ โ€” a $10\%$ rise in the gauge reading.
Check The common mistake, $2.0\times10^5\times\dfrac{320}{300}=2.13\times10^5\ \text{Pa}$, applies the law to the gauge pressure; and using Celsius, $\dfrac{47}{27}$, would predict an absurd $74\%$ rise. Tyre makers tell you to check pressures "cold" for exactly this reason.
Interactive 1 โ€” Molecular Piston Lab (Boyle and Charles)

A cylinder holds $0.010\ \text{mol}$ of an ideal gas; the dots are its molecules (each stands for about $10^{20}$ real ones). The pressure gauge is not calculated from the gas law โ€” it counts the momentum the molecules actually deliver to the piston, $2mv_x$ per hit, and averages it over time. In Boyle mode the temperature is fixed and you move the piston; in Charles mode the pressure is held at $1.00\times10^5\ \text{Pa}$ and you change the temperature. Press Record point at several settings to build the graph.

0.25 L
volume $V$โ€”
gauge (from hits)โ€”
$nRT/V$ predictsโ€”
$pV$โ€”

Avogadro's hypothesis and Dalton's law of partial pressures

Since $pV=Nk_BT$ contains no property of the gas, equal values of $p$, $V$ and $T$ force equal values of $N$. Avogadro guessed this in 1811, from the simple volume ratios in which gases react, long before anyone could count molecules.

๐Ÿ“ Law 12.3 โ€” Avogadro's hypothesis and the molar volume

Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. In particular, one mole of any ideal gas at standard temperature and pressure (STP: $273\ \text{K}$ and $1\ \text{atm}=1.013\times10^5\ \text{Pa}$) occupies the molar volume

$$V_m=\frac{RT}{p}=\frac{8.31\times273}{1.013\times10^5}=2.24\times10^{-2}\ \text{m}^3=22.4\ \text{L}.$$

The number density at STP is $n_V=\dfrac{p}{k_BT}=2.69\times10^{25}\ \text{m}^{-3}$ for every gas.

โœ๏ธ Example 4 โ€” Identify the gas
A gas has a density of $1.25\ \text{kg/m}^3$ at STP. (a) Find its molar mass. (b) Which common gas could it be? (c) Find the mass of one of its molecules.
(a) From $p=\dfrac{\rho RT}{M}$: $M=\dfrac{\rho RT}{p}=\dfrac{1.25\times8.31\times273}{1.013\times10^5}=0.0280\ \text{kg/mol}=28.0\ \text{g/mol}$. Equivalently, $22.4\ \text{L}$ of it has mass $1.25\times0.0224=0.0280\ \text{kg}$.
(b) $28\ \text{g/mol}$ fits nitrogen, $\mathrm{N_2}$ (or carbon monoxide, CO โ€” density alone cannot tell them apart).
(c) $m=\dfrac{M}{N_A}=\dfrac{0.0280}{6.02\times10^{23}}=4.65\times10^{-26}\ \text{kg}$.
Check $\rho=n_Vm=2.69\times10^{25}\times4.65\times10^{-26}=1.25\ \text{kg/m}^3$ โœ“.

When several gases share a container โ€” air is mainly nitrogen, oxygen and argon โ€” each kind of molecule behaves as if the others were not there, because in an ideal gas the molecules do not interact. Each gas hits the walls and contributes its own share of the pressure.

๐Ÿ“ Law 12.4 โ€” Dalton's law of partial pressures

The partial pressure of a gas in a mixture is the pressure it would exert if it alone filled the whole volume at the same temperature. For a mixture of ideal gases the total pressure is the sum of the partial pressures:

$$p=p_1+p_2+\dots=(n_1+n_2+\dots)\frac{RT}{V}.$$

Each partial pressure is the total pressure times the gas's mole fraction: $p_1=\dfrac{n_1}{n_1+n_2+\dots}\,p$.

โœ๏ธ Example 5 โ€” A heliumโ€“oxygen diving mixture
A $10.0\ \text{L}$ cylinder at $300\ \text{K}$ contains $2.0\ \text{g}$ of helium ($M=4.0\ \text{g/mol}$) and $8.0\ \text{g}$ of oxygen ($M=32\ \text{g/mol}$). Find the partial pressure of each gas and the total pressure.
Moles $n_\text{He}=\dfrac{2.0}{4.0}=0.50\ \text{mol}$, $n_{\mathrm{O_2}}=\dfrac{8.0}{32}=0.25\ \text{mol}$.
Partial $p_\text{He}=\dfrac{0.50\times8.31\times300}{1.00\times10^{-2}}=1.25\times10^5\ \text{Pa}$; $p_{\mathrm{O_2}}=0.62\times10^5\ \text{Pa}$ (exactly half, $6.23\times10^4$).
Total $p=1.87\times10^5\ \text{Pa}$, about $1.8\ \text{atm}$.
Check Mole fraction of oxygen $\dfrac{0.25}{0.75}=\dfrac13$, and $\dfrac13\times1.87\times10^5=6.2\times10^4\ \text{Pa}$ โœ“. Notice that the light helium supplies two-thirds of the pressure though it is only one-fifth of the mass: pressure counts molecules, not kilograms.
โœ๏ธ Example 6 โ€” A bubble rising from a lake bed
An air bubble of volume $1.0\ \text{cm}^3$ is released by a diver $40\ \text{m}$ below the surface of a lake, where the water temperature is $12\ ^\circ\text{C}$. It rises to the surface, where the temperature is $35\ ^\circ\text{C}$. Find its volume just below the surface. Take atmospheric pressure as $1.013\times10^5\ \text{Pa}$ and ignore surface tension.
Pressures At depth (Section 9.1): $p_1=p_a+\rho gh=1.013\times10^5+1000\times9.8\times40=4.93\times10^5\ \text{Pa}$; at the surface $p_2=1.013\times10^5\ \text{Pa}$.
Gas law The bubble holds a fixed amount of air: $V_2=V_1\dfrac{p_1}{p_2}\dfrac{T_2}{T_1}=1.0\times\dfrac{4.93}{1.013}\times\dfrac{308}{285}$.
Answer $V_2=1.0\times4.87\times1.081=5.3\ \text{cm}^3$.
Check The pressure falls almost five-fold, so Boyle alone gives $4.9\ \text{cm}^3$; the warmer surface water adds another $8\%$. This is why a diver must breathe out while rising: air in the lungs expands in the same way.

Real gases and the limits of the ideal-gas model

How good is $pV=nRT$? Measure $\dfrac{pV}{nRT}$ for a real gas over a range of pressures. For an ideal gas this ratio would be exactly $1$ at every pressure. For real gases it is very close to $1$ at low pressures โ€” within a fraction of a per cent at atmospheric pressure โ€” but deviates at hundreds of atmospheres. Two molecular effects compete:

Both effects vanish as the gas is made more dilute. So real gases approach ideal behaviour at low pressures and high temperatures, and every gas law in this chapter is a low-density law. Air at room conditions obeys it to better than $0.1\%$.

Interactive 2 โ€” How Ideal Is a Real Gas?

The curves show $\dfrac{pV}{nRT}$ against pressure for four real gases at the chosen temperature, computed from a simple model of molecular attraction and size (the van der Waals model, which you do not need to learn). The dashed line at $1$ is the ideal gas. Lower the temperature and watch carbon dioxide and nitrogen dip below $1$ (attraction wins); look at very high pressures, where every curve rises above $1$ (size wins). Move the pressure marker to read off the error of $pV=nRT$.

320 K100 atm
$pV/nRT$ (model)โ€”
molar volume (model)โ€”
ideal $RT/p$โ€”
error of $pV=nRT$โ€”
โš ๏ธ Common mistakes

1. Celsius in the gas laws. $V\propto T$ and $p\propto T$ hold for kelvin temperatures only. Doubling the temperature from $20\ ^\circ\text{C}$ to $40\ ^\circ\text{C}$ raises $T$ by under $7\%$.

2. Gauge pressure in $pV=nRT$. Always use absolute pressure; convert back at the end if a gauge reading is wanted.

3. Mixing up $n$ and $N$. $pV=nRT$ uses moles; $pV=Nk_BT$ uses the number of molecules. They differ by a factor of $6.02\times10^{23}$.

4. Units of volume and molar mass. In SI, $V$ is in $\text{m}^3$ ($1\ \text{L}=10^{-3}\ \text{m}^3$, $1\ \text{cm}^3=10^{-6}\ \text{m}^3$) and $M$ in $\text{kg/mol}$ ($28\ \text{g/mol}=0.028\ \text{kg/mol}$).

5. "Equal masses, equal pressures." At the same $V$ and $T$, pressure is proportional to the number of moles, not the mass: $1\ \text{g}$ of hydrogen exerts $16$ times the pressure of $1\ \text{g}$ of oxygen.

Exercises

Take $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$, $k_B=1.38\times10^{-23}\ \text{J/K}$, $N_A=6.02\times10^{23}\ \text{mol}^{-1}$, $1\ \text{atm}=1.013\times10^5\ \text{Pa}$, STP $=273\ \text{K}$ and $1\ \text{atm}$, and $g=9.8\ \text{m/s}^2$. Treat all gases as ideal unless told otherwise.

I
Molecules and the mole.
  • 1 How many molecules are there in $1.0\ \text{g}$ of water ($M=18\ \text{g/mol}$)?
  • 2 Find the mass of one oxygen molecule, $\mathrm{O_2}$ ($M=32\ \text{g/mol}$).
  • 3 Copper has molar mass $63.5\ \text{g/mol}$ and density $8.96\times10^3\ \text{kg/m}^3$. Estimate the spacing between neighbouring copper atoms.
  • 4 How many nitrogen molecules are there in a $2.0\ \text{L}$ flask of nitrogen at STP?
  • 5 Using the molecular picture, explain why gases are easy to compress but liquids and solids are not.
  • 6 Treat each molecule of a gas at STP as a sphere of diameter $3.0\times10^{-10}\ \text{m}$. What fraction of the volume of the gas do the molecules themselves occupy?
  • 7 What is Brownian motion, and why is it evidence that the molecules of a liquid are in constant random motion?
1 $n=\dfrac{1.0}{18}=0.0556\ \text{mol}$; $N=0.0556\times6.02\times10^{23}=3.3\times10^{22}$.
2 $m=\dfrac{0.032}{6.02\times10^{23}}=5.3\times10^{-26}\ \text{kg}$.
3 Volume per atom $\dfrac{M}{\rho N_A}=\dfrac{0.0635}{8960\times6.02\times10^{23}}=1.18\times10^{-29}\ \text{m}^3$; spacing $\approx(1.18\times10^{-29})^{1/3}=2.3\times10^{-10}\ \text{m}$.
4 $n=\dfrac{2.0}{22.4}=0.0893\ \text{mol}$; $N=5.4\times10^{22}$.
5 In a gas the molecules are about ten molecular diameters apart and there is mostly empty space between them, so pushing them closer needs little force. In liquids and solids they are already touching; squeezing them closer is resisted by strong repulsive forces.
6 Volume per molecule $\dfrac{1}{n_V}=\dfrac{1}{2.69\times10^{25}}=3.72\times10^{-26}\ \text{m}^3$. Molecular volume $\tfrac{\pi}{6}d^3=1.41\times10^{-29}\ \text{m}^3$. Fraction $=3.8\times10^{-4}$, less than a twentieth of one per cent.
7 Small particles (pollen grains, smoke particles) suspended in a fluid jiggle about irregularly and never stop. Each is struck unevenly from different sides by the fluid's invisible molecules; the motion is livelier at higher temperature and for smaller particles, exactly as the molecular theory predicts.
II
The gas laws and $pV=nRT=Nk_BT$.
  • 1 A gas occupies $3.0\ \text{L}$ at $2.0\times10^5\ \text{Pa}$. It is compressed at constant temperature to $1.2\ \text{L}$. Find its new pressure.
  • 2 A balloon has a volume of $2.0\ \text{L}$ at $27\ ^\circ\text{C}$. It is cooled to $-73\ ^\circ\text{C}$ at constant pressure. Find its new volume.
  • 3 A steel gas cylinder is filled to an absolute pressure of $1.5\times10^7\ \text{Pa}$ at $15\ ^\circ\text{C}$. It is safe up to $1.8\times10^7\ \text{Pa}$. Find the highest safe temperature in $^\circ\text{C}$.
  • 4 How many moles of gas are in a $5.0\ \text{L}$ vessel at $2.0\ \text{atm}$ and $300\ \text{K}$?
  • 5 Find the density of oxygen ($M=32\ \text{g/mol}$) at $27\ ^\circ\text{C}$ and $2.0\times10^5\ \text{Pa}$.
  • 6 Show that $pV=nRT$ and $pV=Nk_BT$ are the same equation, and calculate $k_B$ from $R$ and $N_A$.
  • 7 A $50\ \text{L}$ oxygen cylinder at $300\ \text{K}$ shows a gauge pressure of $1.5\times10^6\ \text{Pa}$. After use, still at $300\ \text{K}$, it shows $5.0\times10^5\ \text{Pa}$. What mass of oxygen was used? (Atmospheric pressure $1.0\times10^5\ \text{Pa}$.)
  • 8 For a fixed amount of ideal gas, sketch (a) $p$ against $V$ at constant $T$, (b) $pV$ against $p$ at constant $T$, (c) $V$ against Celsius temperature at constant $p$. Where does the graph in (c) cut the temperature axis?
1 $p_2=\dfrac{2.0\times10^5\times3.0}{1.2}=5.0\times10^5\ \text{Pa}$.
2 $V_2=2.0\times\dfrac{200}{300}=1.3\ \text{L}$.
3 $T_2=288\times\dfrac{1.8}{1.5}=346\ \text{K}$, i.e. $73\ ^\circ\text{C}$.
4 $n=\dfrac{2.0\times1.013\times10^5\times5.0\times10^{-3}}{8.31\times300}=0.41\ \text{mol}$.
5 $\rho=\dfrac{pM}{RT}=\dfrac{2.0\times10^5\times0.032}{8.31\times300}=2.6\ \text{kg/m}^3$.
6 $N=nN_A$, so $Nk_BT=nN_Ak_BT$, which equals $nRT$ if $k_B=\dfrac{R}{N_A}=\dfrac{8.31}{6.02\times10^{23}}=1.38\times10^{-23}\ \text{J/K}$.
7 Absolute pressures $1.6\times10^6$ and $6.0\times10^5\ \text{Pa}$. $n_1=\dfrac{1.6\times10^6\times0.050}{8.31\times300}=32.1\ \text{mol}$, $n_2=12.0\ \text{mol}$. Used: $20.1\ \text{mol}\times0.032=0.64\ \text{kg}$.
8 (a) A rectangular hyperbola, $p\propto\dfrac1V$. (b) A horizontal straight line, $pV=nRT$. (c) A straight line that, extended backwards, cuts the axis at $-273\ ^\circ\text{C}$ (absolute zero).
III
Avogadro's hypothesis, Dalton's law and real gases.
  • 1 Find the volume of $4.4\ \text{g}$ of carbon dioxide ($M=44\ \text{g/mol}$) at STP.
  • 2 Two identical flasks at the same temperature and pressure contain hydrogen and oxygen respectively. Find the ratio of the numbers of molecules, and of the masses of gas, in them.
  • 3 A vessel contains $0.20\ \text{mol}$ of nitrogen and $0.30\ \text{mol}$ of oxygen at a total pressure of $2.5\times10^5\ \text{Pa}$. Find the partial pressure of each gas.
  • 4 Dry air is $78\%$ nitrogen, $21\%$ oxygen and $1\%$ argon by number of molecules. Find the partial pressure of oxygen at sea level ($1.013\times10^5\ \text{Pa}$) and at the summit of Everest ($3.3\times10^4\ \text{Pa}$).
  • 5 A $2.0\ \text{L}$ vessel of gas at $3.0\times10^5\ \text{Pa}$ is connected by a thin tube to a $3.0\ \text{L}$ vessel at $1.0\times10^5\ \text{Pa}$, at the same temperature. Find the final common pressure.
  • 6 Under what conditions does a real gas deviate most from ideal behaviour? Explain why $\dfrac{pV}{nRT}$ can be either less than or greater than $1$.
  • 7 Using the composition in part 4, find the average molar mass of dry air and its density at STP.
1 $n=0.10\ \text{mol}$; $V=0.10\times22.4=2.24\ \text{L}$.
2 Same $p$, $V$, $T$: equal numbers of molecules (Avogadro), ratio $1:1$. Masses in the ratio of molar masses, $2:32=1:16$.
3 Mole fractions $0.40$ and $0.60$: $p_{\mathrm{N_2}}=1.0\times10^5\ \text{Pa}$, $p_{\mathrm{O_2}}=1.5\times10^5\ \text{Pa}$.
4 $p_{\mathrm{O_2}}=0.21\times1.013\times10^5=2.1\times10^4\ \text{Pa}$ at sea level; $0.21\times3.3\times10^4=6.9\times10^3\ \text{Pa}$ on Everest โ€” a third as much, which is why climbers carry oxygen.
5 The total amount ($\propto pV$) is conserved: $p=\dfrac{3.0\times10^5\times2.0+1.0\times10^5\times3.0}{5.0}=1.8\times10^5\ \text{Pa}$.
6 At high pressure and low temperature (near liquefaction). Intermolecular attraction reduces the momentum delivered to the walls, making $\dfrac{pV}{nRT}\lt1$; at very high density the finite volume of the molecules reduces the free space, making $\dfrac{pV}{nRT}\gt1$.
7 $M=0.78(28)+0.21(32)+0.01(40)=29.0\ \text{g/mol}$; $\rho=\dfrac{M}{V_m}=\dfrac{0.0290}{0.0224}=1.29\ \text{kg/m}^3$.
IV
Applications and harder problems.
  • 1 (Modelling) A hot-air balloon has an envelope of volume $2000\ \text{m}^3$. The outside air is at $27\ ^\circ\text{C}$ and $1.013\times10^5\ \text{Pa}$ (molar mass $29\ \text{g/mol}$); the air inside is heated to $100\ ^\circ\text{C}$ at the same pressure. Find the densities of the outside and inside air and the greatest total mass (envelope, basket and passengers) the balloon can lift.
  • 2 A vessel of gas at $27\ ^\circ\text{C}$ is fitted with a valve that keeps its pressure constant by letting gas escape. The vessel is heated to $127\ ^\circ\text{C}$. What fraction of the gas escapes?
  • 3 A narrow tube closed at one end contains a $20\ \text{cm}$ column of air trapped by a $10\ \text{cm}$ thread of mercury. With the tube horizontal the air column is $20\ \text{cm}$ long. Find its length when the tube is vertical (a) open end up, (b) open end down. Atmospheric pressure is $76\ \text{cm}$ of mercury; the temperature is constant.
  • 4 The outlet of a syringe is sealed with $60\ \text{mL}$ of air inside at $1.0\times10^5\ \text{Pa}$. The plunger, of area $4.0\ \text{cm}^2$, is pushed in slowly until the volume is $20\ \text{mL}$. Find the pressure of the air and the extra force needed to hold the plunger there.
  • 5 A sealed packet of crisps contains $1.0\ \text{L}$ of air at $1.0\times10^5\ \text{Pa}$ and $300\ \text{K}$. It is carried up a mountain where the pressure is $7.0\times10^4\ \text{Pa}$ and the temperature $280\ \text{K}$. Assuming the packet is slack enough not to resist, find its new volume.
  • 6 Estimate the number of air molecules in a classroom $10\ \text{m}\times8.0\ \text{m}\times3.0\ \text{m}$ at $300\ \text{K}$ and $1.013\times10^5\ \text{Pa}$, and the mass of the air.
1 $\rho_\text{out}=\dfrac{pM}{RT}=\dfrac{1.013\times10^5\times0.029}{8.31\times300}=1.18\ \text{kg/m}^3$; $\rho_\text{in}=1.18\times\dfrac{300}{373}=0.95\ \text{kg/m}^3$. The upthrust minus the weight of hot air supports $(\rho_\text{out}-\rho_\text{in})V=0.231\times2000=4.6\times10^2\ \text{kg}$.
2 At constant $p$ and $V$, $n\propto\dfrac1T$: $\dfrac{n_2}{n_1}=\dfrac{300}{400}=0.75$, so one-quarter of the gas escapes.
3 Horizontal: $p=76\ \text{cm Hg}$, $L=20\ \text{cm}$. (a) $p=86$: $L=\dfrac{20\times76}{86}=17.7\ \text{cm}$. (b) $p=66$: $L=\dfrac{20\times76}{66}=23.0\ \text{cm}$.
4 $p=1.0\times10^5\times\dfrac{60}{20}=3.0\times10^5\ \text{Pa}$. The atmosphere already pushes on the outside of the plunger, so the extra force is $(p-p_a)A=2.0\times10^5\times4.0\times10^{-4}=80\ \text{N}$.
5 $V_2=1.0\times\dfrac{1.0\times10^5}{7.0\times10^4}\times\dfrac{280}{300}=1.3\ \text{L}$ ($1.33$).
6 $V=240\ \text{m}^3$; $N=\dfrac{pV}{k_BT}=\dfrac{1.013\times10^5\times240}{1.38\times10^{-23}\times300}=5.9\times10^{27}$. $n=\dfrac{N}{N_A}=9.75\times10^3\ \text{mol}$, so the mass is $9.75\times10^3\times0.029\approx2.8\times10^2\ \text{kg}$ โ€” more than you might guess.
๐ŸŽฏ Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board ยท 2 marks State Avogadro's hypothesis. Use the equation $pV=Nk_BT$ to show that it holds for ideal gases, and find the volume of one mole of an ideal gas at STP.

Statement Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. [ยฝ]
Proof $N=\dfrac{pV}{k_BT}$, and $k_B$ is the same universal constant for every gas, so equal $p$, $V$, $T$ give equal $N$. [ยฝ]
Molar volume $V=\dfrac{RT}{p}=\dfrac{8.31\times273}{1.013\times10^5}=2.24\times10^{-2}\ \text{m}^3=22.4\ \text{L}$. [1]
Exam tip Write the units at each step: $\text{J mol}^{-1}\text{K}^{-1}\times\text{K}/\text{Pa}=\text{m}^3\text{/mol}$. Answers in litres without the conversion lose the mark.

E2. NEET A closed vessel contains $8.0\ \text{g}$ of oxygen and $7.0\ \text{g}$ of nitrogen at $27\ ^\circ\text{C}$. The total pressure is $2.0\ \text{atm}$. The partial pressure of nitrogen is (a) $0.93\ \text{atm}$ (b) $1.0\ \text{atm}$ (c) $1.07\ \text{atm}$ (d) $2.0\ \text{atm}$

Idea Partial pressure $=$ mole fraction $\times$ total. $n_{\mathrm{O_2}}=\dfrac{8.0}{32}=0.25\ \text{mol}$, $n_{\mathrm{N_2}}=\dfrac{7.0}{28}=0.25\ \text{mol}$.
Answer Mole fraction $\tfrac12$: $p_{\mathrm{N_2}}=1.0\ \text{atm}$, (b).
Exam tip (a) and (c) use mass fractions, $\tfrac{7}{15}$ and $\tfrac{8}{15}$ โ€” pressure counts molecules, not grams. The temperature is not needed.

E3. JEE Main (Numerical value) A $10.0\ \text{L}$ flask at $300\ \text{K}$ contains a mixture of helium ($M=4\ \text{g/mol}$) and nitrogen ($M=28\ \text{g/mol}$) of total mass $20.0\ \text{g}$, at a total pressure of $6.0\ \text{atm}$. Find the mass of helium in the flask, in grams, to the nearest integer. ($R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$, $1\ \text{atm}=1.013\times10^5\ \text{Pa}$.)

Total moles $n=\dfrac{pV}{RT}=\dfrac{6.0\times1.013\times10^5\times1.00\times10^{-2}}{8.31\times300}=2.44\ \text{mol}$.
Masses With $x$ mol of helium: $4x+28(2.44-x)=20.0$, so $24x=68.3-20.0$ and $x=2.01\ \text{mol}$.
Answer Mass of helium $=4\times2.01=8.0\ \text{g}$: $8$.
Exam tip Check: nitrogen $0.43\ \text{mol}=12.0\ \text{g}$, and $8.0+12.0=20.0\ \text{g}$ โœ“. Two unknowns need two equations โ€” one from the gas law (moles), one from the masses.

E4. JEE Advanced (One or more correct) For ideal gases, which statements are correct? (A) At the same temperature and pressure, $1\ \text{L}$ of hydrogen and $1\ \text{L}$ of carbon dioxide contain the same number of molecules. (B) At the same temperature and pressure, the density of a gas is proportional to its molar mass. (C) If the Celsius temperature of a gas in a rigid container is doubled, its pressure doubles. (D) For a fixed mass of a given gas, $\dfrac{p}{\rho T}$ is constant, whatever the volume.

(A) $N=\dfrac{pV}{k_BT}$ is the same: Avogadro โœ“.
(B) $\rho=\dfrac{pM}{RT}$, so $\rho\propto M$ at fixed $p$ and $T$ โœ“.
(C) $p\propto T$ in kelvin. From $20\ ^\circ\text{C}$ to $40\ ^\circ\text{C}$, $\dfrac{p_2}{p_1}=\dfrac{313}{293}=1.07$, not $2$ โœ—.
(D) $\dfrac{p}{\rho T}=\dfrac{R}{M}$, a constant for the gas โœ“ (it holds even if the mass changes).
Answer (A), (B), (D).
Exam tip $p=\dfrac{\rho RT}{M}$ is the most useful form when a problem gives a density instead of a volume.

E5. NEET JEE Main A gas at $27\ ^\circ\text{C}$ occupies a volume $V$. At constant pressure it is heated to $327\ ^\circ\text{C}$. Its new volume is (a) $1.1V$ (b) $1.5V$ (c) $2V$ (d) $12.1V$

Idea Charles's law with kelvin temperatures: $T_1=300\ \text{K}$, $T_2=600\ \text{K}$.
Answer $V_2=V\times\dfrac{600}{300}=2V$: (c).
Exam tip (d) is $\dfrac{327}{27}$, the Celsius trap. Examiners like $27\ ^\circ\text{C}=300\ \text{K}$ because doubling or halving in kelvin then gives round numbers.

E6. Board ยท Case study A weather balloon is filled with helium at ground level, where the pressure is $1.0\times10^5\ \text{Pa}$ and the temperature $300\ \text{K}$; its volume there is $5.0\ \text{m}^3$. It rises to a height where the pressure is $2.5\times10^4\ \text{Pa}$ and the temperature $225\ \text{K}$. (i) Which gas law (or laws) must you use to find its new volume, and why? [1] (ii) Find the new volume, assuming the envelope stretches freely. [2] (iii) How many moles of helium does the balloon contain? [1]

(i) Both $p$ and $T$ change while the amount of gas is fixed, so use the combined law $\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$ (Boyle's and Charles's laws together). [1]
(ii) $V_2=V_1\dfrac{p_1}{p_2}\dfrac{T_2}{T_1}=5.0\times4.0\times0.75=15\ \text{m}^3$. [2]
(iii) $n=\dfrac{p_1V_1}{RT_1}=\dfrac{1.0\times10^5\times5.0}{8.31\times300}=2.0\times10^2\ \text{mol}$ (about $0.80\ \text{kg}$ of helium). [1]
Exam tip The pressure falls four-fold but the volume rises only three-fold, because the cold reduces it. Real weather balloons are launched only partly filled so that they can expand like this without bursting too early.
๐Ÿ”ฌ Researchโ€“Teaching NexusResearch-tutored ยท optional, about 45 minutes

Research question. Can a sealed plastic syringe and a set of kitchen weights verify Boyle's law โ€” and measure the pressure of the atmosphere at the same time?

Why it matters. Boyle's law is the working principle of the spirometers that doctors use to test lung capacity, of the pressure sensors in car tyres and weather stations, and of every scuba regulator. The idea of finding an unknown pressure from the intercept of a straight-line graph is the same one that laboratories use to locate absolute zero and to calibrate gauges.

What to do. Use a $20\ \text{mL}$ or $50\ \text{mL}$ plastic syringe. Measure the internal diameter of its barrel (or divide a marked volume by the length it occupies) to find the plunger area $A$.

  1. Draw in air to a mid-scale mark and seal the nozzle firmly (a cap or a blob of putty). Note the volume $V_0$.
  2. Clamp the syringe vertically, nozzle down, and hang or stand known masses $m$ on the plunger: $0.5$, $1.0$, $1.5$, $2.0\ \text{kg}$. Twist the plunger gently before each reading to reduce friction, wait, and record the volume $V$.
  3. The gas pressure is $p=p_a+\dfrac{mg}{A}$. Boyle's law says $p\propto\dfrac1V$, so plot $\dfrac{mg}{A}$ (the extra pressure) on the $y$-axis against $\dfrac1V$ on the $x$-axis.
  4. The graph should be a straight line whose $y$-intercept is $-p_a$. Read off the atmospheric pressure and compare with a weather report for your town.
  5. Repeat the readings while removing the masses. Do the two sets agree? What does any difference tell you about friction?

Tools: Syringe and cap or putty, a clamp stand, kitchen weights or water bottles of known mass, a ruler, a spreadsheet for the graph and a straight-line fit.

Hand in: Your table of $m$, $V$, $\dfrac1V$ and $\dfrac{mg}{A}$; the graph with its best-fit line; your value of $p_a$ with an estimate of its uncertainty; and a paragraph on the main sources of error (friction, leaks, the dead volume in the nozzle, temperature).

Going further. Put the sealed syringe in water baths at different temperatures with a fixed load and test Charles's law. Extrapolate your $V$โ€“$T$ line: how close to $-273\ ^\circ\text{C}$ does it cross the axis?

Instructor note: for a barrel of internal diameter $2.0\ \text{cm}$, $A=3.14\times10^{-4}\ \text{m}^2$, so $1.0\ \text{kg}$ adds $3.1\times10^4\ \text{Pa}$ (about $0.3\ \text{atm}$) and $2.0\ \text{kg}$ shrinks $20.0\ \text{mL}$ of air to about $12\ \text{mL}$. The line of extra pressure against $1/V$ has slope $p_aV_0$ and intercept $-p_a$. Friction typically spreads the "loading" and "unloading" readings by $1$โ€“$2\ \text{mL}$; averaging them gives $p_a$ within about $10\%$. The nozzle's dead volume (about $0.1$โ€“$0.2\ \text{mL}$) should be added to every reading.

๐Ÿงช Self-check quiz

0/5 correct

Q1. The Boltzmann constant $k_B$ is equal to

Q2. When a gas is compressed to half its volume at constant temperature, its pressure doubles because

Q3. At STP, one mole of hydrogen and one mole of oxygen occupy

Q4. A mixture of $2\ \text{mol}$ of helium and $3\ \text{mol}$ of argon has a total pressure of $5.0\ \text{atm}$. The partial pressure of helium is

Q5. A real gas behaves most nearly like an ideal gas at