Solved Examples

How to Use This Problem Bank

Ten sections of theory, and now the part that actually earns marks. What follows is 48 worked problems covering the whole chapter, from single-substitution drills up to the multi-step questions that decide ranks. They are arranged easy first, hard last, in eight parts that follow the order the chapter was taught in.

Work them with a pen. Cover the solution, try it, then compare — and read the check at the end of each one, because the check is where the marks usually leak away.

Notation for This Chapter

Key Point — THIS CHAPTER'S NOTATION:

• $n$ is the NUMBER DENSITY, molecules per cubic metre, $n = \frac{N}{V}$.
• $\mu$ is the number of MOLES. The previous chapter used $n$ for moles; here it is the reverse, and so is every kinetic-theory formula and every exam paper.
• $N$ is the number of molecules, $N_A$ the Avogadro number, and $\mu = \frac{N}{N_A} = \frac{M}{M_0}$.
• $M_0$ is the molar mass in kg/mol; $M$ is the total mass of the sample; $m = \frac{M_0}{N_A}$ is the mass of one molecule.
• $E$ is translational kinetic energy only. $U$ is the full internal energy. They coincide only for a monatomic gas.
• $C_p$ and $C_v$ are molar specific heats in J/(mol K); lowercase $c_p$, $c_v$ are per kilogram.

So every formula carried over from thermodynamics gets rewritten here: $PV = \mu RT$, not $nRT$; $\Delta Q = \mu C_v \Delta T$, not $nC_v\Delta T$. Get used to it now — it is the convention you will meet in the question paper.

The four questions to ask before you write anything

1. Is the temperature in kelvin? Every ratio $\frac{T_2}{T_1}$, every speed formula, every gas-law step needs absolute temperature. Only a bare difference $\Delta T$ is the same number in Celsius and in kelvin. This is the single commonest wrong answer in the chapter.
2. Is the molar mass in kilograms per mole? Oxygen is $0.032$, not $32$. Getting this wrong makes a speed come out $\sqrt{1000} = 31.6$ times too small, and the answer still looks like a number, which is why it slips through.
3. Which of the three speeds does the question want? Energy wants $v_{rms}$; a collision count wants $\bar{v}$; the peak of the distribution is $v_{mp}$. Name it out loud before you substitute.
4. Is the amount of gas fixed? If it is, $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ does the whole job. If gas leaks in or out, $\mu$ changes and you must go through $PV = \mu RT$ twice, once at each state.

Key Point — every formula this section uses, in one place: $$\mu = \frac{M}{M_0} = \frac{N}{N_A}, \qquad m = \frac{M_0}{N_A}, \qquad n = \frac{N}{V}$$ $$PV = \mu RT, \qquad PV = Nk_BT, \qquad P = nk_BT, \qquad P = \frac{\rho R T}{M_0}$$ $$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}, \qquad P_{\text{total}} = P_1 + P_2 + P_3 + \dots$$ $$P = \frac{1}{3}nm\overline{v^2} = \frac{1}{3}\rho\,\overline{v^2}, \qquad PV = \frac{2}{3}E, \qquad E = \frac{3}{2}Nk_BT$$ $$\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_BT, \qquad v_{rms} = \sqrt{\frac{3RT}{M_0}} = \sqrt{\frac{3k_BT}{m}}$$ $$\bar{v} = \sqrt{\frac{8RT}{\pi M_0}}, \qquad v_{mp} = \sqrt{\frac{2RT}{M_0}}, \qquad v_{mp} : \bar{v} : v_{rms} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$$ $$U = \frac{f}{2}\mu RT, \qquad C_v = \frac{f}{2}R, \qquad C_p = C_v + R, \qquad \gamma = 1 + \frac{2}{f}$$ $$l = \frac{1}{\sqrt{2}\,n\pi d^2} = \frac{k_BT}{\sqrt{2}\,\pi d^2 P}, \qquad f_{\text{coll}} = \frac{\bar{v}}{l}, \qquad \tau = \frac{1}{f_{\text{coll}}}$$ $$\frac{r_1}{r_2} = \sqrt{\frac{M_{0,2}}{M_{0,1}}} \quad \text{(equal } P \text{ and } T\text{)}$$

The Maxwell distribution, the formulas for $\bar{v}$ and $v_{mp}$, Graham's law and the pressure law all sit outside the rationalised syllabus body text, yet a large fraction of the problems below cannot be done without them, and Boards, JEE and NEET ask about them every year. They are used freely here.

The constants used throughout

Every solution also restates the constants it uses inside itself, so you never have to scroll back.

Quantity	Value used
Boltzmann constant $k_B$	$1.38 \times 10^{-23}$ J/K
Gas constant $R$	8.314 J/(mol K)
Avogadro number $N_A$	$6.022 \times 10^{23}$ per mol
1 atmosphere	$1.013 \times 10^{5}$ Pa
Absolute zero offset	$0°$C $= 273.15$ K
Molar volume at STP	22.4 litres
$g$	9.8 m/s$^2$
Density of water	1000 kg/m$^3$
$C_v$ monatomic / rigid diatomic / rigid non-linear polyatomic	12.47 / 20.79 / 24.94 J/(mol K)
$\gamma$ for the same three	$\frac{5}{3}$ / $\frac{7}{5}$ / $\frac{4}{3}$

[Board Important] Every solution below writes the formula on its own line before any number goes into it, converts the molar mass on a line of its own, and names which speed it is using. Do all three in the exam. A correct formula with an arithmetic slip still earns most of the marks; a gram-per-mole inside a square root loses them all.

Solved Examples

Part 1: Molecules, Sizes and Counting

Six warm-ups. No gas laws yet — just Avogadro's number used carefully, and the two length scales that make a gas a gas.

Example 1: From grams to molecules, and back

A sealed flask holds 5.6 g of nitrogen gas ($M_0 = 28$ g/mol). Find (a) the number of moles, (b) the number of molecules, (c) the number of nitrogen atoms, and (d) the mass of a single molecule.

Solution:

Constants: $N_A = 6.022 \times 10^{23}$ per mole.

1. Convert the molar mass first, as a habit, even where the answer does not need it: $$M_0 = 28 \ \text{g/mol} = 0.028 \ \text{kg/mol}$$

2. (a) Moles. $\mu = \frac{M}{M_0}$, with both masses in the same unit: $$\mu = \frac{5.6}{28} = 0.20 \ \text{mol}$$

3. (b) Molecules. $$N = \mu N_A = 0.20 \times 6.022 \times 10^{23} = 1.20 \times 10^{23}$$

4. (c) Atoms. Nitrogen gas is $N_2$ — two atoms per molecule: $$\text{atoms} = 2N = 2.41 \times 10^{23}$$

5. (d) Mass of one molecule. $$m = \frac{M_0}{N_A} = \frac{0.028}{6.022 \times 10^{23}} = 4.65 \times 10^{-26} \ \text{kg}$$ Check: $N \times m = 1.204 \times 10^{23} \times 4.65 \times 10^{-26} = 5.6 \times 10^{-3}$ kg, which is the 5.6 g we started from.

Final Answer: $0.20$ mol; $1.20 \times 10^{23}$ molecules; $2.41 \times 10^{23}$ atoms; $4.65 \times 10^{-26}$ kg per molecule.

Takeaway: Molecules and atoms are not the same count. Any diatomic gas doubles the atom count, and questions that ask for "the number of atoms in 5.6 g of nitrogen" are testing exactly that.

Example 2: Sizing a single atom from a bucket of liquid

Liquid argon has a density of about 1400 kg/m$^3$ and a molar mass of 39.9 g/mol. Estimate the diameter of one argon atom, assuming the atoms in the liquid are packed so as to fill the space.

Solution:

Constants: $N_A = 6.022 \times 10^{23}$ per mole.

1. Volume of one mole of the liquid. Convert the molar mass: $$M_0 = 39.9 \ \text{g/mol} = 0.0399 \ \text{kg/mol}$$ $$V_{\text{mole}} = \frac{M_0}{\rho} = \frac{0.0399}{1400} = 2.85 \times 10^{-5} \ \text{m}^3$$

2. Volume per atom. $$V_{\text{atom}} = \frac{V_{\text{mole}}}{N_A} = \frac{2.85 \times 10^{-5}}{6.022 \times 10^{23}} = 4.73 \times 10^{-29} \ \text{m}^3$$

3. Turn a volume into a diameter. Model the atom as a sphere of diameter $d$, so $V_{\text{atom}} = \frac{\pi}{6}d^3$: $$d = \left(\frac{6 V_{\text{atom}}}{\pi}\right)^{1/3} = \left(\frac{6 \times 4.73 \times 10^{-29}}{3.1416}\right)^{1/3} = \left(9.04 \times 10^{-29}\right)^{1/3}$$ $$d = 4.49 \times 10^{-10} \ \text{m} = 4.5 \ \text{Å}$$

4. Sanity check. Atoms come out a few angstroms across whatever substance you start from. A liquid is close-packed rather than perfectly space-filling, so this slightly overestimates $d$ — an estimate, not a measurement.

Final Answer: $d \approx 4.5 \times 10^{-10}$ m, about 4.5 Å.

Takeaway: A density plus a molar mass gives you the size of an atom. The whole trick is $V_{\text{atom}} = \frac{M_0}{\rho N_A}$, and then a cube root. It works for any liquid or solid.

Example 3: How far apart are the molecules of a gas?

Nitrogen gas is held at 1 atm and 27°C. Find its number density, the volume available to each molecule, and the average spacing between neighbouring molecules. Compare that spacing with a molecular diameter of 3.7 Å.

Solution:

Constants: $k_B = 1.38 \times 10^{-23}$ J/K; 1 atm $= 1.013 \times 10^{5}$ Pa.

1. Kelvin first. $$T = 27 + 273.15 = 300.15 \ \text{K} \approx 300 \ \text{K}$$

2. Number density from $P = nk_BT$. Here $n$ is molecules per cubic metre, not moles: $$n = \frac{P}{k_BT} = \frac{1.013 \times 10^{5}}{1.38 \times 10^{-23} \times 300} = 2.45 \times 10^{25} \ \text{per m}^3$$

3. Volume per molecule is just the reciprocal: $$V_1 = \frac{1}{n} = \frac{1}{2.45 \times 10^{25}} = 4.09 \times 10^{-26} \ \text{m}^3$$

4. Spacing. Give each molecule a little cube of side $\bar{r}$: $$\bar{r} = V_1^{1/3} = (4.09 \times 10^{-26})^{1/3} = 3.44 \times 10^{-9} \ \text{m} = 34.4 \ \text{Å}$$

5. Compare with the molecule itself. $$\frac{\bar{r}}{d} = \frac{34.4}{3.7} = 9.3$$

Final Answer: $n = 2.45 \times 10^{25}$ per m$^3$; $4.09 \times 10^{-26}$ m$^3$ each; spacing $34.4$ Å, about 9 molecular diameters.

Takeaway: In a gas at ordinary pressure the molecules sit roughly ten diameters apart. That single number is why the ideal gas model works: for most of its life a molecule is nowhere near another one, so the intermolecular forces have nothing to act on.

Example 4: What fraction of a compressed gas is actually molecule?

Nitrogen is compressed to 5 atm at 300 K. Taking the molecular diameter as 3.7 Å, find the fraction of the container's volume that the molecules themselves occupy.

Solution:

Constants: $k_B = 1.38 \times 10^{-23}$ J/K; 1 atm $= 1.013 \times 10^{5}$ Pa.

1. Number density at 5 atm. $$n = \frac{P}{k_BT} = \frac{5 \times 1.013 \times 10^{5}}{1.38 \times 10^{-23} \times 300} = 1.22 \times 10^{26} \ \text{per m}^3$$

2. Volume of one molecule, treated as a sphere of diameter $d = 3.7 \times 10^{-10}$ m: $$v_{\text{mol}} = \frac{\pi}{6}d^3 = \frac{3.1416}{6}\left(3.7 \times 10^{-10}\right)^3 = 2.65 \times 10^{-29} \ \text{m}^3$$

3. Fraction. Molecular volume per cubic metre of gas, divided by that cubic metre: $$\text{fraction} = n\,v_{\text{mol}} = 1.22 \times 10^{26} \times 2.65 \times 10^{-29} = 3.24 \times 10^{-3}$$ $$= 0.32\%$$

4. What it means. Even at five atmospheres, 99.7% of the vessel is empty space. But notice the fraction is proportional to $P$ at fixed $T$ — squeeze the same gas to 100 atm and it rises to about 6.5%, and at that point the finite size of the molecules is no longer negligible and the gas stops being ideal.

Final Answer: about $3.2 \times 10^{-3}$, i.e. 0.32% of the volume.

Takeaway: The "molecules are point-like" assumption is a statement about pressure, not about molecules. It holds because $n v_{\text{mol}} \ll 1$, and that fails the moment you compress hard enough.

Example 5: The molecules in one breath

A quiet breath draws in about 500 cm$^3$ of air at 1 atm and body temperature, 37°C. Find the number of moles, the number of molecules, and the mass of that air. Take the mean molar mass of air to be 28.9 g/mol.

Solution:

Constants: $R = 8.314$ J/(mol K); $k_B = 1.38 \times 10^{-23}$ J/K; $N_A = 6.022 \times 10^{23}$ per mole; 1 atm $= 1.013 \times 10^{5}$ Pa.

1. Everything into SI. $$T = 37 + 273.15 = 310.15 \ \text{K}, \qquad V = 500 \ \text{cm}^3 = 5.00 \times 10^{-4} \ \text{m}^3$$

2. Moles, from $PV = \mu RT$: $$\mu = \frac{PV}{RT} = \frac{1.013 \times 10^{5} \times 5.00 \times 10^{-4}}{8.314 \times 310.15} = \frac{50.65}{2578.6} = 1.96 \times 10^{-2} \ \text{mol}$$

3. Molecules. $$N = \mu N_A = 1.964 \times 10^{-2} \times 6.022 \times 10^{23} = 1.18 \times 10^{22}$$

4. Cross-check by the other road, using $P = nk_BT$: $$n = \frac{P}{k_BT} = \frac{1.013 \times 10^{5}}{1.38 \times 10^{-23} \times 310.15} = 2.37 \times 10^{25} \ \text{per m}^3$$ $$N = nV = 2.37 \times 10^{25} \times 5.00 \times 10^{-4} = 1.18 \times 10^{22} \quad \checkmark$$

5. Mass. $$M = \mu M_0 = 1.964 \times 10^{-2} \times 28.9 = 0.568 \ \text{g}$$

Final Answer: $0.0196$ mol, $1.18 \times 10^{22}$ molecules, mass about $0.57$ g.

Takeaway: $PV = \mu RT$ and $P = nk_BT$ are the same equation divided by $N_A$. Solving a counting problem both ways costs thirty seconds and catches a stray factor of $6 \times 10^{23}$ instantly.

Example 6: A drop of water, counted three ways

A drop of water has a mass of 1.8 g. Find the number of water molecules in it, the total number of atoms, and the average spacing between neighbouring molecules. Molar mass of water is 18 g/mol and its density is 1000 kg/m$^3$.

Solution:

Constants: $N_A = 6.022 \times 10^{23}$ per mole.

1. Moles and molecules. $$\mu = \frac{1.8}{18} = 0.10 \ \text{mol}, \qquad N = 0.10 \times 6.022 \times 10^{23} = 6.02 \times 10^{22}$$

2. Atoms. Each $H_2O$ carries 3 nuclei: $$\text{atoms} = 3N = 1.81 \times 10^{23}$$

3. Volume of the drop. $$V = \frac{M}{\rho} = \frac{1.8 \times 10^{-3}}{1000} = 1.8 \times 10^{-6} \ \text{m}^3$$

4. Volume per molecule, then spacing. $$V_1 = \frac{V}{N} = \frac{1.8 \times 10^{-6}}{6.022 \times 10^{22}} = 2.99 \times 10^{-29} \ \text{m}^3$$ $$\bar{r} = V_1^{1/3} = 3.10 \times 10^{-10} \ \text{m} = 3.1 \ \text{Å}$$

5. Compare with Example 3. In the gas the spacing was 34.4 Å; in the liquid it is 3.1 Å — about 11 times smaller, so about $11^3 \approx 1300$ times denser. That is the whole difference between a liquid and a gas in one number.

Final Answer: $6.02 \times 10^{22}$ molecules, $1.81 \times 10^{23}$ atoms, spacing $\approx 3.1$ Å.

Takeaway: In a liquid the spacing is essentially the molecular size; in a gas it is ten times larger. Every difference between the two phases — compressibility, fixed volume, intermolecular forces mattering — follows from that one factor of ten.

Part 2: The Ideal Gas Equation in All Four Forms

Six problems on the one equation that appears in four disguises. Pick the form whose symbols match the data you were handed, and check with a second form whenever you can.

Key Point: The four forms are $PV = \mu RT$ (moles), $PV = Nk_BT$ (molecules), $P = nk_BT$ (number density) and $P = \frac{\rho R T}{M_0}$ (density). They are one equation. $R = k_B N_A$ and $M_0 = m N_A$ are the only bridges you need.

Example 7: One gas sample, all four forms

A rigid 2.0-litre vessel holds 8.0 g of oxygen ($M_0 = 32$ g/mol) at 300 K. Find its pressure four separate times, once from each form of the ideal gas equation.

Solution:

Constants: $R = 8.314$ J/(mol K); $k_B = 1.38 \times 10^{-23}$ J/K; $N_A = 6.022 \times 10^{23}$ per mole.

1. The data in SI. $$V = 2.0 \ \text{L} = 2.0 \times 10^{-3} \ \text{m}^3, \qquad M = 8.0 \ \text{g} = 8.0 \times 10^{-3} \ \text{kg}$$ $$M_0 = 32 \ \text{g/mol} = 0.032 \ \text{kg/mol}$$

2. Form 1 — moles. $$\mu = \frac{M}{M_0} = \frac{8.0}{32} = 0.25 \ \text{mol}$$ $$P = \frac{\mu RT}{V} = \frac{0.25 \times 8.314 \times 300}{2.0 \times 10^{-3}} = 3.12 \times 10^{5} \ \text{Pa}$$

3. Form 2 — molecules. $$N = \mu N_A = 0.25 \times 6.022 \times 10^{23} = 1.51 \times 10^{23}$$ $$P = \frac{Nk_BT}{V} = \frac{1.51 \times 10^{23} \times 1.38 \times 10^{-23} \times 300}{2.0 \times 10^{-3}} = 3.12 \times 10^{5} \ \text{Pa}$$

4. Form 3 — number density. $$n = \frac{N}{V} = \frac{1.5055 \times 10^{23}}{2.0 \times 10^{-3}} = 7.53 \times 10^{25} \ \text{per m}^3$$ $$P = nk_BT = 7.53 \times 10^{25} \times 1.38 \times 10^{-23} \times 300 = 3.12 \times 10^{5} \ \text{Pa}$$

5. Form 4 — density. $$\rho = \frac{M}{V} = \frac{8.0 \times 10^{-3}}{2.0 \times 10^{-3}} = 4.0 \ \text{kg/m}^3$$ $$P = \frac{\rho R T}{M_0} = \frac{4.0 \times 8.314 \times 300}{0.032} = 3.12 \times 10^{5} \ \text{Pa}$$

Final Answer: $P = 3.12 \times 10^{5}$ Pa $\approx 3.08$ atm, by all four routes.

Takeaway: Choose the form whose symbols the question already gives you. Given a mass, use $\mu$ or $\rho$; given a count, use $N$ or $n$. Converting the data to fit a memorised form is where the time and the marks go.

Example 8: Molar volume, at 27°C and at 0°C

Find the volume occupied by one mole of an ideal gas at 1 atm and 27°C. Then show that the same calculation at 0°C gives the familiar 22.4 litres.

Solution:

Constants: $R = 8.314$ J/(mol K); 1 atm $= 1.013 \times 10^{5}$ Pa.

1. At 27°C. Kelvin first: $$T = 27 + 273.15 = 300.15 \ \text{K}$$ $$V = \frac{\mu RT}{P} = \frac{1 \times 8.314 \times 300.15}{1.013 \times 10^{5}} = 2.463 \times 10^{-2} \ \text{m}^3 = 24.63 \ \text{L}$$

2. At 0°C, which is $T = 273.15$ K: $$V = \frac{1 \times 8.314 \times 273.15}{1.013 \times 10^{5}} = 2.242 \times 10^{-2} \ \text{m}^3 = 22.42 \ \text{L}$$

3. Why the number is famous. Nothing in that line refers to which gas it is. One mole of helium, of oxygen and of carbon dioxide all occupy 22.4 L at 0°C and 1 atm — which is Avogadro's hypothesis, arriving as a consequence rather than an assumption.

4. The ratio check. The two volumes should be in the ratio of the absolute temperatures: $$\frac{24.63}{22.42} = 1.0986, \qquad \frac{300.15}{273.15} = 1.0988 \quad \checkmark$$

Final Answer: $24.63$ L at 27°C; $22.42$ L at 0°C.

Takeaway: 22.4 litres is not a property of a gas — it is a property of a temperature and a pressure. Quoting it at room temperature is a standard slip; at 27°C the molar volume is nearly 25 litres.

Example 9: What does the air in a classroom weigh?

A classroom measures 8.0 m by 6.0 m by 3.5 m. The air in it is at 1 atm and 300 K, and its mean molar mass is 28.9 g/mol. Find the density of the air, the mass of air in the room, and the number of moles.

Solution:

Constants: $R = 8.314$ J/(mol K); 1 atm $= 1.013 \times 10^{5}$ Pa.

1. Volume. $$V = 8.0 \times 6.0 \times 3.5 = 168 \ \text{m}^3$$

2. Density, from the density form. Convert the molar mass first: $$M_0 = 28.9 \ \text{g/mol} = 0.0289 \ \text{kg/mol}$$ $$\rho = \frac{P M_0}{RT} = \frac{1.013 \times 10^{5} \times 0.0289}{8.314 \times 300} = \frac{2927.6}{2494.2} = 1.17 \ \text{kg/m}^3$$

3. Mass. $$M = \rho V = 1.174 \times 168 = 197 \ \text{kg}$$

4. Cross-check through moles. $$\mu = \frac{PV}{RT} = \frac{1.013 \times 10^{5} \times 168}{8.314 \times 300} = 6823 \ \text{mol}$$ $$M = \mu M_0 = 6823 \times 0.0289 = 197 \ \text{kg} \quad \checkmark$$

Final Answer: $\rho = 1.17$ kg/m$^3$; about 197 kg of air; 6823 moles.

Takeaway: Air is not weightless — a small classroom holds about two hundred kilograms of it. You do not feel it because it pushes up on you as hard as it pushes down.

Example 10: Compressed and heated at the same time

A fixed mass of gas is at $2.0 \times 10^{5}$ Pa in a volume of 3.0 litres at 300 K. It is compressed to 1.2 litres and simultaneously heated to 400 K. Find the new pressure.

Solution:

1. Is the amount of gas fixed? Yes — nothing leaks. So $\mu$ cancels and the combined form applies: $$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

2. Both temperatures are already absolute (300 K, 400 K), so no conversion is needed. Volumes may stay in litres because they appear as a ratio.

3. Rearrange, then substitute. $$P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 2.0 \times 10^{5} \times \frac{3.0}{1.2} \times \frac{400}{300}$$ $$P_2 = 2.0 \times 10^{5} \times 2.5 \times 1.333 = 6.67 \times 10^{5} \ \text{Pa}$$

4. Sense check. Squeezing raises the pressure, heating raises it further, so the answer must exceed $2.0 \times 10^{5}$ Pa — and it does, by a factor of $\frac{10}{3}$.

Final Answer: $P_2 = 6.67 \times 10^{5}$ Pa.

Takeaway: Write the ratio form, not $PV = \mu RT$, whenever the amount of gas is unchanged. Units that appear on both sides cancel, so litres and atmospheres are perfectly safe there — but kelvin is not optional.

Example 11: A gauge that does not read what you think

A nitrogen cylinder of volume 15 litres has a pressure gauge reading 12 atm at 300 K. Find the number of moles and the mass of nitrogen inside. Molar mass of nitrogen is 28 g/mol.

Solution:

Constants: $R = 8.314$ J/(mol K); 1 atm $= 1.013 \times 10^{5}$ Pa.

1. The trap, dealt with first. A pressure gauge reads the excess over atmospheric. The gas equation wants the absolute pressure: $$P = P_{\text{gauge}} + P_{\text{atm}} = 12 + 1 = 13 \ \text{atm} = 13 \times 1.013 \times 10^{5} = 1.317 \times 10^{6} \ \text{Pa}$$

2. Volume in SI. $$V = 15 \ \text{L} = 1.5 \times 10^{-2} \ \text{m}^3$$

3. Moles. $$\mu = \frac{PV}{RT} = \frac{1.317 \times 10^{6} \times 1.5 \times 10^{-2}}{8.314 \times 300} = \frac{19753}{2494.2} = 7.92 \ \text{mol}$$

4. Mass. $$M = \mu M_0 = 7.92 \times 28 = 222 \ \text{g}$$

5. What the trap costs. Using the gauge reading of 12 atm instead would give $\mu = 7.31$ mol and $205$ g — an error of nearly 8%, and it grows as the cylinder empties.

Final Answer: $7.92$ mol, about 222 g of nitrogen.

Takeaway: Gauge pressure plus one atmosphere equals absolute pressure. Any problem that says "gauge" is testing that line and nothing else. Bubbles, tyres and cylinders are the usual settings.

Example 12: Pumping a room down to two per cent

A sealed room of volume 60 m$^3$ contains air at 1 atm and 300 K. A pump reduces the pressure to 2% of atmospheric at the same temperature. How many molecules were removed, and what mass of air is that? Take the mean molar mass of air as 28.9 g/mol.

Solution:

Constants: $R = 8.314$ J/(mol K); $k_B = 1.38 \times 10^{-23}$ J/K; $N_A = 6.022 \times 10^{23}$ per mole.

1. Moles at the start. $$\mu_1 = \frac{P_1V}{RT} = \frac{1.013 \times 10^{5} \times 60}{8.314 \times 300} = \frac{6.078 \times 10^{6}}{2494.2} = 2437 \ \text{mol}$$

2. Moles at the end. Same $V$, same $T$, so $\mu \propto P$: $$\mu_2 = 0.02 \times 2437 = 48.7 \ \text{mol}$$

3. Removed. $$\Delta\mu = 2437 - 48.7 = 2388 \ \text{mol}$$ $$\Delta N = \Delta\mu \times N_A = 2388 \times 6.022 \times 10^{23} = 1.44 \times 10^{27} \ \text{molecules}$$

4. Mass removed. $$\Delta M = \Delta\mu \times M_0 = 2388 \times 0.0289 = 69.0 \ \text{kg}$$

5. Cross-check the count directly. At the start, $$n_1 = \frac{P_1}{k_BT} = \frac{1.013 \times 10^{5}}{1.38 \times 10^{-23} \times 300} = 2.45 \times 10^{25} \ \text{per m}^3$$ $$N_1 = n_1 V = 1.47 \times 10^{27}, \qquad 0.98 N_1 = 1.44 \times 10^{27} \quad \checkmark$$

Final Answer: about $1.44 \times 10^{27}$ molecules, a mass of about 69 kg.

Takeaway: At constant volume and temperature, $\mu$, $N$, $n$ and $\rho$ are all proportional to $P$. Once you see that, "reduce the pressure to 2%" means "remove 98% of everything" without any further arithmetic.

Part 3: Leaks, Bubbles, Mixtures and Partial Pressures

Seven problems where either the amount of gas changes or there is more than one gas in the box. Both cases break the habit of reaching straight for $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.

Example 13: The oxygen cylinder after a day's use

An oxygen cylinder of volume 40 litres has an initial gauge pressure of 20 atm at 27°C. After some oxygen is used, the gauge pressure has fallen to 14 atm and the temperature to 12°C. Estimate the mass of oxygen taken out. Molar mass of oxygen is 32 g/mol.

Solution:

Constants: $R = 8.314$ J/(mol K); 1 atm $= 1.013 \times 10^{5}$ Pa.

1. Why the combined gas law is useless here. Gas has left the cylinder, so $\mu$ is not the same at the two states. We must apply $PV = \mu RT$ separately at each state and subtract.

2. State 1, in absolute units. $$P_1 = (20 + 1) \ \text{atm} = 21 \times 1.013 \times 10^{5} = 2.127 \times 10^{6} \ \text{Pa}$$ $$T_1 = 27 + 273.15 = 300.15 \ \text{K}, \qquad V = 40 \ \text{L} = 4.0 \times 10^{-2} \ \text{m}^3$$ $$\mu_1 = \frac{P_1V}{RT_1} = \frac{2.127 \times 10^{6} \times 4.0 \times 10^{-2}}{8.314 \times 300.15} = \frac{85092}{2495.4} = 34.10 \ \text{mol}$$

3. State 2. $$P_2 = (14 + 1) \times 1.013 \times 10^{5} = 1.520 \times 10^{6} \ \text{Pa}, \qquad T_2 = 12 + 273.15 = 285.15 \ \text{K}$$ $$\mu_2 = \frac{1.520 \times 10^{6} \times 4.0 \times 10^{-2}}{8.314 \times 285.15} = \frac{60780}{2370.6} = 25.64 \ \text{mol}$$

4. Convert both to mass. With $M_0 = 32$ g/mol: $$M_1 = 34.10 \times 32 = 1091 \ \text{g}, \qquad M_2 = 25.64 \times 32 = 820 \ \text{g}$$

5. Subtract. $$\Delta M = 1091 - 820 = 271 \ \text{g}$$

Final Answer: about $271$ g of oxygen was withdrawn.

Takeaway: A leak makes $\mu$ a variable. The tell is a question that asks for a mass or a number rather than a pressure — you cannot get either out of a ratio equation in which $\mu$ has already cancelled.

Example 14: The bubble that rises 25 metres

An air bubble of volume 2.0 cm$^3$ forms at the bottom of a lake 25 m deep, where the water is at 15°C. It rises to the surface, where the temperature is 30°C. Find its volume there. Take the density of water as 1000 kg/m$^3$, $g = 9.8$ m/s$^2$ and atmospheric pressure as $1.013 \times 10^{5}$ Pa.

Solution:

1. The pressure at the bottom is atmospheric plus the water column above: $$P_1 = P_{\text{atm}} + \rho g h = 1.013 \times 10^{5} + 1000 \times 9.8 \times 25$$ $$P_1 = 1.013 \times 10^{5} + 2.45 \times 10^{5} = 3.463 \times 10^{5} \ \text{Pa} \ (= 3.42 \ \text{atm})$$

2. The pressure at the surface is just atmospheric: $$P_2 = 1.013 \times 10^{5} \ \text{Pa}$$

3. Kelvin, both ends. $$T_1 = 15 + 273.15 = 288.15 \ \text{K}, \qquad T_2 = 30 + 273.15 = 303.15 \ \text{K}$$

4. The air inside the bubble is a fixed amount, so the combined form is the right tool: $$V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 2.0 \times \frac{3.463 \times 10^{5}}{1.013 \times 10^{5}} \times \frac{303.15}{288.15}$$ $$V_2 = 2.0 \times 3.419 \times 1.052 = 7.19 \ \text{cm}^3$$

5. Which factor did the work? The pressure ratio contributed $\times 3.42$ and the temperature ratio only $\times 1.05$. The bubble grows because it is escaping the water, not because it is warming.

Final Answer: the bubble grows to about $7.2$ cm$^3$, roughly $3.6$ times its original size.

Takeaway: A rising bubble is a pressure problem with a small temperature correction. Every 10.3 m of water adds one atmosphere, so a bubble doubles in volume in the first ten metres and grows more slowly after that.

Example 15: Helium and oxygen sharing one flask

A 5.0-litre flask at 300 K contains 4.0 g of helium ($M_0 = 4$ g/mol) and 16 g of oxygen ($M_0 = 32$ g/mol). Find the partial pressure of each gas, the total pressure, and the mole fraction of helium.

Solution:

Constants: $R = 8.314$ J/(mol K); 1 atm $= 1.013 \times 10^{5}$ Pa.

1. Moles of each. $$\mu_{He} = \frac{4.0}{4} = 1.00 \ \text{mol}, \qquad \mu_{O_2} = \frac{16}{32} = 0.50 \ \text{mol}$$

2. Partial pressure is the pressure each gas would exert alone in the whole 5.0 L: $$P_{He} = \frac{\mu_{He} RT}{V} = \frac{1.00 \times 8.314 \times 300}{5.0 \times 10^{-3}} = 4.99 \times 10^{5} \ \text{Pa}$$ $$P_{O_2} = \frac{0.50 \times 8.314 \times 300}{5.0 \times 10^{-3}} = 2.49 \times 10^{5} \ \text{Pa}$$

3. Dalton's law — just add them. $$P_{\text{total}} = 4.99 \times 10^{5} + 2.49 \times 10^{5} = 7.48 \times 10^{5} \ \text{Pa} = 7.39 \ \text{atm}$$

4. Mole fraction of helium. $$x_{He} = \frac{\mu_{He}}{\mu_{He} + \mu_{O_2}} = \frac{1.00}{1.50} = 0.667$$ and indeed $\frac{P_{He}}{P_{\text{total}}} = \frac{4.99}{7.48} = 0.667$, which is the same number — partial pressures are in the ratio of the mole numbers.

5. Note what is equal and what is not. Both gases are at 300 K, so their molecules have the same average translational kinetic energy; but helium's molecules are 8 times lighter, so they move $\sqrt{8} = 2.83$ times faster.

Final Answer: $P_{He} = 4.99 \times 10^{5}$ Pa, $P_{O_2} = 2.49 \times 10^{5}$ Pa, total $7.48 \times 10^{5}$ Pa; $x_{He} = \frac{2}{3}$.

Takeaway: Equal masses are not equal moles, and only moles decide pressure. Four grams of helium beats sixteen grams of oxygen two to one, because a helium atom is eight times lighter.

Example 16: Adding a second gas without letting the first out

A 10-litre vessel holds nitrogen at 2.0 atm and 300 K. Now 8.0 g of helium ($M_0 = 4$ g/mol) is pumped in, the temperature being held at 300 K throughout. Find the new total pressure.

Solution:

Constants: $R = 8.314$ J/(mol K); 1 atm $= 1.013 \times 10^{5}$ Pa.

1. The nitrogen does not care. Its volume and temperature are unchanged and none of it left, so its partial pressure is still $$P_{N_2} = 2.0 \ \text{atm} = 2.026 \times 10^{5} \ \text{Pa}$$ (For interest, $\mu_{N_2} = \frac{PV}{RT} = \frac{2.026 \times 10^{5} \times 10^{-2}}{8.314 \times 300} = 0.812$ mol.)

2. The helium's partial pressure, computed as though it were alone in the 10 litres: $$\mu_{He} = \frac{8.0}{4} = 2.00 \ \text{mol}$$ $$P_{He} = \frac{\mu_{He}RT}{V} = \frac{2.00 \times 8.314 \times 300}{1.0 \times 10^{-2}} = 4.99 \times 10^{5} \ \text{Pa} = 4.92 \ \text{atm}$$

3. Add. $$P_{\text{total}} = 2.0 + 4.92 = 6.92 \ \text{atm} = 7.01 \times 10^{5} \ \text{Pa}$$

4. Check by total moles. $$\mu_{\text{total}} = 0.812 + 2.00 = 2.812 \ \text{mol}$$ $$P = \frac{2.812 \times 8.314 \times 300}{1.0 \times 10^{-2}} = 7.01 \times 10^{5} \ \text{Pa} \quad \checkmark$$

Final Answer: $P_{\text{total}} = 7.01 \times 10^{5}$ Pa, about 6.9 atm.

Takeaway: Adding a non-reacting gas never changes the pressure of the gas already there. Each species behaves as if the others were absent, which is Dalton's law and, molecularly, a consequence of the molecules not interacting.

Example 17: Forty per cent leaks out while the tank warms up

A rigid tank contains gas at $4.0 \times 10^{5}$ Pa and 300 K. A slow leak removes 40% of the molecules, and by the time it is noticed the temperature has risen to 350 K. Find the new pressure.

Solution:

1. Which quantities changed? $V$ is fixed. $N$ fell to $0.60$ of its value. $T$ rose from 300 K to 350 K. Both temperatures are already absolute.

2. Use the form with $N$ in it, because $N$ is what changed: $$P = \frac{Nk_BT}{V} \quad\Longrightarrow\quad \frac{P_2}{P_1} = \frac{N_2}{N_1}\times\frac{T_2}{T_1}$$

3. Substitute. $$\frac{P_2}{P_1} = 0.60 \times \frac{350}{300} = 0.60 \times 1.1667 = 0.70$$ $$P_2 = 0.70 \times 4.0 \times 10^{5} = 2.80 \times 10^{5} \ \text{Pa}$$

4. Reading the answer. The leak alone would have taken the pressure to $2.40 \times 10^{5}$ Pa; the warming pushed it back up to $2.80 \times 10^{5}$ Pa. A leak can be masked by a temperature rise, which is exactly why gas cylinders are checked at a stated temperature.

Final Answer: $P_2 = 2.80 \times 10^{5}$ Pa.

Takeaway: When the amount of gas changes, put $N$ (or $\mu$) into the ratio explicitly: $\frac{P_2}{P_1} = \frac{N_2}{N_1}\cdot\frac{V_1}{V_2}\cdot\frac{T_2}{T_1}$. Dropping the first factor is the standard error.

Example 18: Taking the water vapour out of moist air

A closed vessel at 40°C holds moist air at a total pressure of 1.0 atm. At that temperature the partial pressure of the water vapour is 7.4 kPa. Find the partial pressure of the dry air, and the fraction of the molecules in the vessel that are water molecules.

Solution:

Constants: 1 atm $= 1.013 \times 10^{5}$ Pa.

1. Dalton's law, rearranged. The total is the sum of the parts: $$P_{\text{dry}} = P_{\text{total}} - P_{\text{water}} = 1.013 \times 10^{5} - 7.4 \times 10^{3} = 9.39 \times 10^{4} \ \text{Pa}$$

2. Fraction of molecules. At a common $V$ and $T$, $P \propto N$, so the mole fraction and the pressure fraction are the same number: $$x_{\text{water}} = \frac{P_{\text{water}}}{P_{\text{total}}} = \frac{7.4 \times 10^{3}}{1.013 \times 10^{5}} = 0.0731 = 7.3\%$$

3. What this does not depend on. Nothing here needed the molar mass of water or of air. Partial pressures count molecules, not mass — the 7.3% is a fraction by number, and because water (18 g/mol) is lighter than air (28.9 g/mol) the fraction by mass is smaller, about 4.6%.

Final Answer: dry air $9.39 \times 10^{4}$ Pa; water vapour is about 7.3% of the molecules.

Takeaway: A partial-pressure fraction is a fraction by number of molecules. Convert it to a mass fraction only by bringing in the molar masses; the two are equal only if the gases happen to have the same $M_0$.

Example 19: Three gases, one flask, one mean molar mass

A 20-litre vessel at 400 K holds 2.0 g of hydrogen, 28 g of nitrogen and 44 g of carbon dioxide. Find the total pressure, the mean molar mass of the mixture, and check the answer through the density form of the gas equation. Molar masses: $H_2$ 2, $N_2$ 28, $CO_2$ 44 g/mol.

Solution:

Constants: $R = 8.314$ J/(mol K); 1 atm $= 1.013 \times 10^{5}$ Pa.

1. Moles of each — each comes out to a whole number here. $$\mu_{H_2} = \frac{2.0}{2} = 1, \qquad \mu_{N_2} = \frac{28}{28} = 1, \qquad \mu_{CO_2} = \frac{44}{44} = 1$$ $$\mu_{\text{total}} = 3.0 \ \text{mol}$$

2. Total pressure. $$P = \frac{\mu_{\text{total}}RT}{V} = \frac{3.0 \times 8.314 \times 400}{2.0 \times 10^{-2}} = 4.99 \times 10^{5} \ \text{Pa} = 4.92 \ \text{atm}$$ Each gas contributes exactly one third of this, $1.66 \times 10^{5}$ Pa, because each supplies one third of the moles.

3. Mean molar mass is total mass over total moles: $$M_{0,\text{mean}} = \frac{2.0 + 28 + 44}{3.0} = \frac{74}{3.0} = 24.7 \ \text{g/mol} = 0.0247 \ \text{kg/mol}$$

4. Check through the density form. $$\rho = \frac{M}{V} = \frac{74 \times 10^{-3}}{2.0 \times 10^{-2}} = 3.70 \ \text{kg/m}^3$$ $$P = \frac{\rho R T}{M_{0,\text{mean}}} = \frac{3.70 \times 8.314 \times 400}{0.0247} = 4.99 \times 10^{5} \ \text{Pa} \quad \checkmark$$

Final Answer: $P = 4.99 \times 10^{5}$ Pa; mean molar mass $24.7$ g/mol.

Takeaway: A mixture obeys the gas equation with a mean molar mass defined as total mass divided by total moles, $M_{0,\text{mean}} = \frac{\sum \mu_i M_{0,i}}{\sum \mu_i}$. It is a mole-weighted average, not a plain one. Here the moles happened to be equal so the two coincide; take 4 g of hydrogen instead of 2 g and the mean drops to $19$ g/mol while the plain average of $2$, $28$ and $44$ stays put.

Part 4: The Pressure Formula, Forwards and Backwards

Five problems on $P = \frac{1}{3}nm\overline{v^2} = \frac{1}{3}\rho\,\overline{v^2}$ — the one result that connects a mechanical push on a wall to the motion of individual molecules. Half the questions give you speeds and want a pressure; the other half do the reverse.

Key Point: $P = \frac{1}{3}\rho\,\overline{v^2}$ contains the mean of the squares $\overline{v^2}$, never the square of the mean. Rearranged, $\sqrt{\overline{v^2}} = v_{rms} = \sqrt{\frac{3P}{\rho}}$ — which is why a pressure and a density alone are enough to get an rms speed, with no thermometer anywhere.

Example 20: Pressure from a density and a speed

Carbon dioxide has a density of 1.98 kg/m$^3$ and its molecules have an rms speed of 393 m/s. Find the pressure of the gas.

Solution:

1. The formula, in the form the data fits. We were handed a density, so use $$P = \frac{1}{3}\rho\,\overline{v^2}$$

2. Which speed is this? The formula wants the mean square speed, and $v_{rms}$ is by definition its square root, so $\overline{v^2} = v_{rms}^2$. Substituting $\bar{v}$ or $v_{mp}$ here would be wrong. $$\overline{v^2} = (393)^2 = 1.544 \times 10^{5} \ \text{m}^2/\text{s}^2$$

3. Substitute. $$P = \frac{1}{3} \times 1.98 \times 1.544 \times 10^{5} = 1.02 \times 10^{5} \ \text{Pa}$$

4. Sense check. That is $1.01$ atm — and 1.98 kg/m$^3$ is indeed close to the density of carbon dioxide at atmospheric pressure and room temperature, so the numbers hang together.

Final Answer: $P \approx 1.02 \times 10^{5}$ Pa, essentially one atmosphere.

Takeaway: $\overline{v^2}$ means $v_{rms}^2$. Any speed offered to this formula must be the rms one; if a problem gives you $\bar{v}$ instead, convert first using $v_{rms} = \sqrt{\frac{3\pi}{8}}\,\bar{v} = 1.085\,\bar{v}$.

Example 21: Running the formula backwards to name the gas

A gas at $1.5 \times 10^{5}$ Pa has a density of 2.4 kg/m$^3$ and a temperature of 300 K. Find the rms speed of its molecules and identify the gas.

Solution:

Constants: $R = 8.314$ J/(mol K).

1. Which speed? The formula $P = \frac{1}{3}\rho\,\overline{v^2}$ contains the mean square speed, so rearranging it yields $v_{rms}$ and nothing else. Speed straight from pressure and density — no temperature needed for this step: $$v_{rms} = \sqrt{\frac{3P}{\rho}} = \sqrt{\frac{3 \times 1.5 \times 10^{5}}{2.4}} = \sqrt{1.875 \times 10^{5}} = 433 \ \text{m/s}$$

2. Now bring in the temperature to get the molar mass. Since $v_{rms} = \sqrt{\frac{3RT}{M_0}}$, $$M_0 = \frac{3RT}{v_{rms}^2} = \frac{3 \times 8.314 \times 300}{1.875 \times 10^{5}} = \frac{7482.6}{1.875 \times 10^{5}} = 0.0399 \ \text{kg/mol}$$

3. Convert back to the familiar unit. $$M_0 = 0.0399 \ \text{kg/mol} = 39.9 \ \text{g/mol}$$ That is argon.

4. Round trip. Feeding $M_0 = 0.0399$ kg/mol back in: $$v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{0.0399}} = 433 \ \text{m/s} \quad \checkmark$$

Final Answer: $v_{rms} = 433$ m/s; the gas is argon, $M_0 = 39.9$ g/mol.

Takeaway: A molar mass in kg/mol comes out as a number like $0.040$, and a molar mass in g/mol as a number like $40$. If your rearranged formula produces $39.9$ where kg/mol is expected, you have dropped a factor of 1000 somewhere.

Example 22: Which average does the pressure formula want?

Five molecules have speeds 300, 400, 500, 600 and 700 m/s. Find the mean speed, the mean square speed, the rms speed, and the square of the mean speed. Which of these belongs in $P = \frac{1}{3}\rho\,\overline{v^2}$?

Solution:

1. Mean speed — add and divide: $$\bar{v} = \frac{300 + 400 + 500 + 600 + 700}{5} = \frac{2500}{5} = 500 \ \text{m/s}$$

2. Mean square speed — square first, then average: $$\overline{v^2} = \frac{300^2 + 400^2 + 500^2 + 600^2 + 700^2}{5} = \frac{1{,}350{,}000}{5} = 2.70 \times 10^{5} \ \text{m}^2/\text{s}^2$$

3. Root mean square speed. $$v_{rms} = \sqrt{2.70 \times 10^{5}} = 519.6 \ \text{m/s}$$

4. Square of the mean, for contrast: $$(\bar{v})^2 = 500^2 = 2.50 \times 10^{5} \ \text{m}^2/\text{s}^2$$

5. Compare. $\overline{v^2} = 2.70 \times 10^5$ but $(\bar{v})^2 = 2.50 \times 10^5$ — using the square of the mean falls short of the true mean square by $\frac{2.70 - 2.50}{2.70} = 7.4\%$, and $v_{rms}$ exceeds $\bar{v}$ by 3.9%. They are equal only if every molecule has the same speed; any spread at all makes $v_{rms} > \bar{v}$.

6. The answer to the question asked. The pressure formula and the energy formula both want $\overline{v^2}$, so $P$ here would use $2.70 \times 10^{5}$ m$^2$/s$^2$.

Final Answer: $\bar{v} = 500$ m/s, $\overline{v^2} = 2.70 \times 10^{5}$ m$^2$/s$^2$, $v_{rms} = 519.6$ m/s, $(\bar{v})^2 = 2.50 \times 10^{5}$ m$^2$/s$^2$. Use $\overline{v^2}$.

Takeaway: $\overline{v^2} \neq (\bar{v})^2$, always $\overline{v^2} > (\bar{v})^2$. Squaring rewards the fast molecules twice over. Using $\frac{1}{2}m(\bar{v})^2$ for the average kinetic energy underestimates it — here by 7.4%, and for a real Maxwellian gas by exactly $1 - \frac{8}{3\pi} = 15.1\%$. An underestimate is always measured against the true, larger value, so the denominator is $\overline{v^2}$, never $(\bar{v})^2$.

Example 23: Squeezing the gas without warming it

Air at $1.0 \times 10^{5}$ Pa has a density of 1.29 kg/m$^3$. It is compressed isothermally to half its volume. Find the rms speed before and after, and the new pressure.

Solution:

1. Before. $$v_{rms} = \sqrt{\frac{3P}{\rho}} = \sqrt{\frac{3 \times 1.0 \times 10^{5}}{1.29}} = \sqrt{2.326 \times 10^{5}} = 482 \ \text{m/s}$$

2. What halving the volume does. The mass of gas is unchanged, so $$\rho_2 = 2\rho_1 = 2.58 \ \text{kg/m}^3$$ and at constant temperature Boyle's law gives $P_2 = 2P_1 = 2.0 \times 10^{5}$ Pa.

3. After. $$v_{rms} = \sqrt{\frac{3 \times 2.0 \times 10^{5}}{2.58}} = \sqrt{2.326 \times 10^{5}} = 482 \ \text{m/s}$$ Identical, to every digit.

4. Why it had to be. $v_{rms} = \sqrt{\frac{3RT}{M_0}}$ depends on temperature and molar mass and on nothing else. The compression was isothermal, so the speeds could not have changed. The pressure doubled purely because the molecules are hitting the walls twice as often, not because they are hitting harder.

Final Answer: $v_{rms} = 482$ m/s both before and after; $P_2 = 2.0 \times 10^{5}$ Pa.

Takeaway: Pressure has two ingredients — how hard each hit is, and how many hits per second. Isothermal compression changes only the second. A question that says "compressed isothermally" and then asks about molecular speed is asking you to answer "unchanged".

Example 24: The translational energy of the air in a room

A room of volume 60 m$^3$ holds air at 1 atm and 300 K. Find the total translational kinetic energy of all the air molecules in it.

Solution:

Constants: $R = 8.314$ J/(mol K); 1 atm $= 1.013 \times 10^{5}$ Pa.

1. The shortcut. Kinetic theory gives $PV = \frac{2}{3}E$, where $E$ is the total translational kinetic energy. So $$E = \frac{3}{2}PV$$

2. Substitute — no temperature, no molar mass, no molecule count required. $$E = \frac{3}{2} \times 1.013 \times 10^{5} \times 60 = 9.12 \times 10^{6} \ \text{J}$$

3. Check by the long road. $$\mu = \frac{PV}{RT} = \frac{1.013 \times 10^{5} \times 60}{8.314 \times 300} = 2437 \ \text{mol}$$ $$E = \frac{3}{2}\mu RT = \frac{3}{2} \times 2437 \times 8.314 \times 300 = 9.12 \times 10^{6} \ \text{J} \quad \checkmark$$

4. Is it a big number? It is about the energy of a 1500 kg car travelling at 110 m/s. It is completely unavailable as useful work, because it is disordered — which is the entire subject of the previous chapter.

5. A warning about $E$ and $U$. Air is mostly diatomic, so its molecules also rotate. Its internal energy is $U = \frac{5}{2}\mu RT = 1.52 \times 10^{7}$ J, larger than $E$ by a factor of $\frac{5}{3}$. The question asked for translational energy, so $9.12 \times 10^{6}$ J is the answer.

Final Answer: $E = 9.12 \times 10^{6}$ J of translational kinetic energy.

Takeaway: $E = \frac{3}{2}PV$ needs only a pressure and a volume. But $E$ is translational only — for anything other than a monatomic gas, $U > E$, and reading one symbol for the other is a favourite trap.