Ideal gas law PV = nRT — derivation from kinetic theory

1.7.8 · Physics › Thermodynamics

KYA derive kar rahe hain?

Goal: yeh dikhana ki aise gas ke liye, $P V = n R T$ jahan $P$ = pressure, $V$ = volume, $n$ = number of moles, $R$ = gas constant, $T$ = absolute temperature.

KYUN collisions se shuru karein?

KAISE — Derivation scratch se

Step 1 — Ek molecule ek wall se takraana

$N$ molecules ko side $L$ (volume $V = L^{3}$ ) ke cube mein rakhein. Ek molecule lo jo mass $m$ aur velocity components $(v_{x}, v_{y}, v_{z})$ se move kar raha hai.

Wo wall consider karo jo $x$ -axis ke perpendicular hai. Molecule usse takraata hai, elastically bounce karta hai, to uski $x$ -velocity sign flip kar leti hai: $Δ p_{x} = (- m v_{x}) - (m v_{x}) = - 2 m v_{x}$

Yeh step kyun? Rigid wall ke saath elastic collision perpendicular velocity ko reverse karti hai lekin speed unchanged rehti hai; wall ko diya gaya momentum ka magnitude $2 m v_{x}$ hai.

Step 2 — Yeh kitni baar hit karta hai?

Ek $x$ -wall se bounce hone ke baad, molecule doosri wall tak aur wapas jaata hai: distance $2 L$ . Same wall par successive hits ke beech time: $Δ t = \frac{2 L}{v _{x}}$

Yeh step kyun? Kisi given wall par collisions periodic hoti hain; hit karne ki rate = $v_{x} /2 L$ per second.

Step 3 — Ek molecule se force

Sirf is ek molecule se wall par average force (Newton's 2nd law, magnitude): $F_{1} = \frac{∣Δ p _{x} ∣}{Δ t} = \frac{2 m v _{x}}{2 L / v _{x}} = \frac{m v _{x}^{2}}{L}$

Yeh step kyun? Momentum delivered per collision ÷ time per collision = average force.

Step 4 — Saare $N$ molecules add karo

$F = \frac{m}{L} \sum_{i = 1}^{N} v_{x, i}^{2} = \frac{m}{L} N ⟨ v_{x}^{2} ⟩$ jahan $⟨ v_{x}^{2} ⟩$ $v_{x}^{2}$ ka mean hai.

Yeh step kyun? Total force independent contributions ka sum hai; sum ko $N \times$ average se replace karna sirf average ki definition hai.

Step 5 — Isotropy use karo (randomness)

Motion random hai, to koi direction special nahi: $⟨ v_{x}^{2} ⟩ = ⟨ v_{y}^{2} ⟩ = ⟨ v_{z}^{2} ⟩$ Kyunki $v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ : $⟨ v^{2} ⟩ = 3 ⟨ v_{x}^{2} ⟩ \Rightarrow ⟨ v_{x}^{2} ⟩ = \frac{1}{3} ⟨ v^{2} ⟩$

Yeh step kyun? Isotropy yeh physical statement hai ki gas ki koi preferred direction nahi hai; yeh hume $⟨ v_{x}^{2} ⟩$ ko full speed se swap karne deta hai.

Step 6 — Pressure

$P = \frac{F}{area} = \frac{F}{L ^{2}} = \frac{m N ⟨ v _{x}^{2} ⟩}{L ^{3}} = \frac{m N ⟨ v ^{2} ⟩}{3 V}$

Figure — Ideal gas law PV = nRT — derivation from kinetic theory

Step 7 — Temperature laao (bridge)

Kinetic energy use karke rewrite karo. 2 se multiply aur divide karo: $P V = \frac{2}{3} N (\frac{1}{2} m ⟨ v^{2} ⟩) = \frac{2}{3} N ⟨ E_{k} ⟩$

Ideal gas ke liye temperature ki definition (equipartition, 3 translational DOF ke liye $\frac{3}{2} k_{B} T$ ): $⟨ E_{k} ⟩ = \frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$

Yeh step kyun? Temperature define ki gayi hai taaki yeh average translational kinetic energy measure kare. Yahi physical input hai jo mechanics ko thermodynamics mein badalta hai.

Substitute karo: $P V = \frac{2}{3} N \cdot \frac{3}{2} k_{B} T = N k_{B} T$

Step 8 — Moles mein switch karo

$N = n N_{A}$ (moles × Avogadro), aur $R = N_{A} k_{B}$ define karo: $P V = n R T$

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Recall Quick self-test (answers chupaao)

$\frac{1}{3}$ ka factor kahan se aata hai? → isotropy se, $⟨ v_{x}^{2} ⟩ = \frac{1}{3} ⟨ v^{2} ⟩$ .
Bridge mein use ki gayi $T$ ki physical definition kya hai? → $⟨ E_{k} ⟩ = \frac{3}{2} k_{B} T$ .
$k_{B}$ ke terms mein $R$ kya hai? → $R = N_{A} k_{B}$ .
$⟨ v_{x} ⟩ = 0$ kyun hai lekin $⟨ v_{x}^{2} ⟩ \neq = 0$ kyun? → directions cancel ho jaati hain, squares nahi.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek box hai jisme tiny bouncy balls bahut tez speed se uda rahi hain. Jab bhi koi ball wall se takraati hai, wall ko thodi si push milti hai. Balls billions hain aur har waqt takraati rehti hain, to wall ko ek steady push feel hoti hai — yahi pressure hai. Agar aap gas ko heat karo, balls tez ho jaati hain, zyada zor se aur zyada baar takraati hain, to push zyada strong ho jaati hai. Box ko chhota squeeze karo to balls walls se aur zyada baar takraati hain. Un saari tiny smacks ko simple "push = kitni tez bounce karti hai × kitni baar" rules se add karo, aur magic recipe milti hai: pressure × volume = (gas ki miqdar) × (ek constant) × (kitni garam hai). Yahi $P V = n R T$ hai — yeh bouncing ke andar chupi thi.

Flashcards

Pressure microscopically kya hai?

Molecules ke wall se collide karne par per second per unit area average momentum jo deliver hota hai.

Ek molecule ke x-wall se elastically takraane par momentum change?

- 2 m v_{x}

(uski

x

-velocity reverse ho jaati hai).

Same wall par hits ke beech time?

Δ t = 2 L / v_{x}

Ek molecule ka wall par average force?

F_{1} = m v_{x}^{2} / L

⟨ v_{x}^{2} ⟩ = \frac{1}{3} ⟨ v^{2} ⟩

kyun?

Isotropy:

⟨ v_{x}^{2} ⟩ = ⟨ v_{y}^{2} ⟩ = ⟨ v_{z}^{2} ⟩

aur ye sab mila ke

⟨ v^{2} ⟩

dete hain.

Kinetic-theory pressure formula?

P = \frac{1}{3} \frac{N m ⟨ v ^{2} ⟩}{V}

Average KE ke terms mein pressure?

P V = \frac{2}{3} N ⟨ E_{k} ⟩

Bridge mein use ki gayi temperature ki definition?

⟨ E_{k} ⟩ = \frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T

R

aur

k_{B}

ka relation?

R = N_{A} k_{B} \approx 8.314

J/mol·K.

rms speed formula?

v_{r m s} = 3 R T / M = 3 k_{B} T / m

T

Kelvin mein kyun hona chahiye?

Kyunki

T

average kinetic energy ke proportional hai, jo sirf absolute zero par zero hoti hai.

⟨ v_{x} ⟩ = 0

kyun hai lekin force zero nahi?

v_{x}

ke liye directions cancel ho jaati hain, lekin force

v_{x}^{2}

par depend karta hai (hamesha positive).