Avogadro's law and Avogadro's number N_A = 6.022 × 10²³

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1. Avogadro's Law (the gas law)

WHAT this says: if you double the number of gas molecules while keeping $T$ and $P$ the same, the volume doubles. Gas volume doesn't care what the molecules are (H₂, O₂, CO₂) — only how many there are.

WHY it's true (derivation from the ideal gas law):

Start from the ideal gas equation (itself a summary of Boyle + Charles + Gay-Lussac): $PV = nRT$

Now solve for $V/n$: $\frac{V}{n} = \frac{RT}{P}$

• Why this step? We isolate the quantity Avogadro's law talks about ($V$ per mole).

At fixed $T$ and $P$, the right-hand side $\dfrac{RT}{P}$ is a constant. Therefore $V/n$ is constant — which is exactly Avogadro's Law.

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2. Avogadro's Number $N_A$

WHY this exact number? Historically, $N_A$ was defined so that 1 mole of carbon-12 weighs exactly 12 grams. That makes the molar mass in g/mol numerically equal to the atomic/molecular mass in u (atomic mass units) — a hugely convenient coincidence by design.

HOW $N_A$ links grams to atomic mass units: By definition $1\text{ u} = \dfrac{1}{12}\times$(mass of one C-12 atom), and 1 mol of C-12 = 12 g. So:

\quad\Rightarrow\quad 1\ \text{u} = \frac{1\ \text{g}}{N_A} = \frac{1}{6.022\times10^{23}}\ \text{g} = 1.66\times10^{-24}\ \text{g}$$ - *Why this step?* Dividing 12 g among $N_A$ atoms gives one atom's mass; taking 1/12 of it gives 1 u. ![[1.1.11-Avogadro's-law-and-Avogadro's-number-N_A-=-6.022-×-1023.png]] --- ## 3. Worked examples > [!example] (a) How many molecules in 18 g of water? > **Given:** $m=18$ g, $M(\text{H}_2\text{O}) = 2(1)+16 = 18$ g/mol. > $$n = \frac{m}{M} = \frac{18}{18} = 1\ \text{mol} > \quad\Rightarrow\quad N = n N_A = 6.022\times10^{23}\ \text{molecules}$$ > *Why:* first grams → moles ($m/M$), then moles → count ($\times N_A$). > **Bonus:** atoms of hydrogen $= 2N$ (each molecule has 2 H) $=1.204\times10^{24}$. > [!example] (b) Count atoms in 0.5 mol of $\text{CO}_2$ > $$N_{\text{molecules}} = 0.5 \times 6.022\times10^{23} = 3.011\times10^{23}$$ > Each CO₂ has 3 atoms (1 C + 2 O), so total atoms $= 3 \times 3.011\times10^{23} = 9.03\times10^{23}$. > *Why:* $N_A$ counts *whatever entity you named* — here molecules — then multiply by atoms per molecule. > [!example] (c) Mass of a single sodium atom > $M(\text{Na}) = 23$ g/mol. > $$m_{\text{atom}} = \frac{M}{N_A} = \frac{23}{6.022\times10^{23}} = 3.82\times10^{-23}\ \text{g}$$ > *Why:* one mole (23 g) shared among $N_A$ atoms. > [!example] (d) Gas-law style: volume comparison > 8 g of O₂ at some $T,P$ occupies 5.0 L. What volume does 8 g of He occupy at the same $T,P$? > $n(\text{O}_2)=8/32=0.25$ mol; $n(\text{He})=8/4=2.0$ mol. > By Avogadro's law $V\propto n$: > $$V_{\text{He}} = 5.0 \times \frac{2.0}{0.25} = 40\ \text{L}$$ > *Why:* same $T,P$ ⇒ volume scales purely with mole count, not mass or identity. --- ## 4. Common mistakes (steel-manned) > [!mistake] "Equal *masses* of gases have equal volumes." > **Why it feels right:** Avogadro's law says volume depends only on amount, and mass feels like 'amount'. > **The fix:** Avogadro's law is about **number of molecules (moles)**, *not* mass. 8 g of H₂ (4 mol) has far more molecules than 8 g of O₂ (0.25 mol), so it occupies 16× the volume. Convert to *moles* first. > [!mistake] "$N_A$ molecules = $N_A$ atoms." > **Why it feels right:** we sloppily say "$N_A$ particles." > **The fix:** $N_A$ counts the entity you *specify*. 1 mol of O₂ = $N_A$ *molecules* = $2N_A$ *atoms*. Always ask: "moles of WHAT?" > [!mistake] "Molar volume is always 22.4 L." > **Why it feels right:** it's a memorized number. > **The fix:** 22.4 L is for the *old* STP (1 atm, 273 K). The modern IUPAC STP (1 bar, 273 K) gives **22.7 L**. And *neither* applies to non-STP conditions or non-ideal gases — always fall back to $PV=nRT$. > [!mistake] Using $M$ in kg/mol with $R=8.314$ and expecting grams. > **Why it feels right:** SI wants kg. > **The fix:** Keep units consistent. If $M$ is in g/mol, $m$ must be in g. For $PV=nRT$ use $R=8.314$ J/(mol·K) with $P$ in Pa, $V$ in m³. --- ## 5. Flashcards #flashcards/chemistry Avogadro's Law (statement) ::: Equal volumes of gases at the same T and P contain equal numbers of molecules; i.e. $V\propto n$ at fixed $T,P$. Value of Avogadro's number ::: $6.022\times10^{23}$ (per mole). Derive $V/n=$ const from ideal gas law ::: $PV=nRT \Rightarrow V/n = RT/P$, constant at fixed $T,P$. Formula for number of particles from moles ::: $N = n\,N_A$. Formula linking mass and count ::: $N = \dfrac{m}{M}\,N_A$. Molar volume of ideal gas at STP (1 bar, 273 K) ::: $22.7$ L (old STP at 1 atm gives 22.4 L). Molecules in 18 g of water ::: $6.022\times10^{23}$ (18 g = 1 mol H₂O). Mass of 1 atomic mass unit in grams ::: $1/N_A = 1.66\times10^{-24}$ g. Atoms in 1 mol of CO₂ ::: $3N_A = 1.807\times10^{24}$ (3 atoms per molecule). Why molar mass in g/mol equals atomic mass in u ::: Because $N_A$ was defined so 1 mol of C-12 weighs exactly 12 g. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine marbles so tiny you could never count them by hand. So we invent a "counting box" called the **mole** — every box holds exactly the same huge number of marbles: about 602,200,000,000,000,000,000,000 of them ($6.022\times10^{23}$). Because chemists chose this number cleverly, a box of carbon marbles weighs 12 grams, a box of water molecules weighs 18 grams, and so on — so weighing something tells you how many boxes (moles) you have. And here's a cool gas trick: if you have the same number of *any* gas marbles at the same temperature and squeeze, they take up the same room — a box of helium and a box of oxygen puff up to the same size. That's Avogadro's law. > [!mnemonic] Remember it > **"Same T, same P, same *count*, same size."** — that's the gas law. > For the number: **6-0-2-2** like a phone number — "$6.022$, times $10$ to the twenty-three." Or: *"Six oh two two, particles per mole for me and you."* ## Connections - [[The mole concept]] — $N_A$ *is* the definition of the mole's size. - [[Ideal gas law PV=nRT]] — Avogadro's law is the $T,P$-constant slice of it. - [[Molar mass and atomic mass unit]] — why g/mol = u numerically. - [[Empirical and molecular formulas]] — mole ratios come from counting via $N_A$. - [[Stoichiometry]] — moles are the currency of every balanced-equation calculation. - [[Boyle's law]] and [[Charles's law]] — the other components of $PV=nRT$. ## 🖼️ Concept Map ```mermaid flowchart TD IGL[Ideal gas law PV=nRT] -->|solve for V/n| AL[Avogadro's Law V/n constant] AL -->|same T,P equal molecules| EQ[Equal volumes equal N] AL -->|set n=1 at STP| MV[Molar volume 22.7 L] MOLE[Mole] -->|counts| NA[Avogadro's number 6.022e23] NA -->|defined via C-12 = 12 g| MM[Molar mass M g/mol] NA -->|N = n × N_A| PART[Particle count N] MM -->|n = m/M| NMOL[Moles n] NMOL -->|multiply by N_A| PART MASS[Mass m in grams] -->|divide by M| NMOL AL -.->|both named Avogadro| NA ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, yaha do alag cheezein hain jinke naam mein "Avogadro" aata hai — inhe mat mila do. Pehla hai **Avogadro's Law**, jo gases ke baare mein hai: agar temperature aur pressure same rakho, to *equal volume* ke andar *equal number of molecules* hote hain — chahe gas H₂ ho, O₂ ho ya CO₂. Matlab volume sirf molecules ki *ginti* (moles) pe depend karta hai, unke *type* ya *mass* pe nahi. Isko derive karna easy hai: $PV=nRT$ se $V/n = RT/P$, aur fixed $T,P$ pe RHS constant hai — bas ho gaya, $V \propto n$. > > Dusra hai **Avogadro's Number**, $N_A = 6.022\times10^{23}$. Atoms itne chhote hote hain ki hum unhe ginti nahi kar sakte, isliye chemists ne ek "counting box" banaya jise **mole** kehte hain. Har mole mein exactly $6.022\times10^{23}$ particles hote hain. Ye number aise choose kiya gaya ki 1 mole carbon-12 ka weight exactly 12 gram aaye — isliye kisi bhi cheez ka **molar mass (g/mol)** aur uska **atomic mass (u)** number mein same aate hain. Bahut hi convenient trick! > > Real problems mein bas do formula yaad rakho: pehle grams ko moles banao $n = m/M$, phir moles ko particles $N = n \times N_A$. Jaise 18 g paani = 1 mole = $6.022\times10^{23}$ molecules. Aur gas problems mein, same $T,P$ pe $V_1/n_1 = V_2/n_2$ laga do. > > Sabse common galti: "equal mass = equal volume" samajh lena. Nahi bhai! 8 g H₂ (4 mol) aur 8 g O₂ (0.25 mol) ka volume bilkul alag hoga kyunki moles alag hain. Hamesha pehle **moles nikaalo**, mass se seedha mat jaao. Ye ek chhoti si aadat tumhe half chemistry ke numericals mein bacha legi. ![[audio/1.1.11-Avogadro's-law-and-Avogadro's-number-N_A-=-6.022-×-1023.mp3]]