Ideal gas law PV = nRT — derivation from kinetic theory: Worked examples

1.7.8 · D3 · Physics › Thermodynamics › Ideal gas law PV = nRT — derivation from kinetic theory

Scenario matrix

Is topic ke har problem inhi cells mein se ek hogi. Hum har ek ko example ke saath tick karte hain.

#	Cell (problem ki type)	Kya puchha gaya	Worked in
A	$P$ solve karo (baaki sab pata hain)	direct plug-in	Ex 1
B	$T$ solve karo, °C↔K convert karo	unit trap	Ex 2
C	$n$ solve karo, phir molecules count karo	$N = n N_{A}$ ka bridge	Ex 3
D	Two-state, ek variable fixed	ratio law	Ex 4
E	Microscopic reach-through: $v_{r m s}$	kinetic core	Ex 5
F	Degenerate / limiting input ( $T \to 0$ , $V \to \infty$ )	kya toot-ta hai	Ex 6
G	Real-world word problem (ek tyre)	modelling	Ex 7
H	Exam twist: do gases ka mixture	additivity	Ex 8
I	Two-state, do variables change hoti hain	combined gas law	Ex 9

Cell A — $P$ solve karo

Worked example Example 1 — ek known sample ka pressure

$n = 3 mol$ ideal gas $V = 0.020 m^{3}$ mein $T = 400 K$ par fill hai. Pressure $P$ nikalo.

Forecast: compute karne se pehle order of magnitude guess karo — kya $P$ zyada $1 0^{3}$ , $1 0^{5}$ , ya $1 0^{7}$ Pa ke kareeb hoga? (Room air $\sim 1 0^{5}$ hai; kuch moles ko 20 litres mein squeeze karo toh zyada hona chahiye.)

Rearrange karo $P V = n R T$ ko unknown ke liye: $P = \frac{n R T}{V}$ . Yeh step kyun? Hum single unknown ko isolate karte hain taaki baaki saare symbols known numbers hon.
Substitute karo sirf SI units ke saath: $P = \frac{3 \times 8.314 \times 400}{0.020}$ . Yeh step kyun? Litres ko $R$ ke saath $J mol^{- 1} K^{- 1}$ mein mix karne se units silently toot jaati hain; $V$ $m^{3}$ mein hona chahiye.
Compute karo: numerator $= 9976.8 J$ , isliye $P = 498840 \approx 4.99 \times 1 0^{5} Pa$ . Yeh step kyun? Hum exact symbolic answer ko ek number mein badal rahe hain taaki uska size everyday pressures ke against sanity-check kar sakein.

Verify karo: units $\frac{J}{m ^{3}} = \frac{N\cdotm}{m ^{3}} = \frac{N}{m ^{2}} = Pa$ ✓. $1 atm = 1.013 \times 1 0^{5} Pa$ use karte hue, hamaara $4.99 \times 1 0^{5} Pa$ yaani $4.99/1.013 \approx 4.9$ atm hai — room air se zyada, forecast se match karta hai. ✓

Cell B — $T$ solve karo Celsius trap ke saath

Worked example Example 2 — temperature jise koi Kelvin mein nahi jaanta

Ek sealed rigid tank mein $n = 1.5 mol$ hai $V = 0.010 m^{3}$ mein. Ek gauge $P = 4.0 \times 1 0^{5} Pa$ read karta hai. Temperature °C mein kya hai?

Forecast: final answer Celsius mein hona chahiye, lekin kya hum $P V = n R T$ mein Celsius plug kar sakte hain? (Nahi — socho kyun $0^{\circ}$ C zero energy dega, jo galat hai.)

Absolute $T$ ke liye solve karo: $T = \frac{P V}{n R} = \frac{4.0 \times 1 0 ^{5} \times 0.010}{1.5 \times 8.314}$ . Yeh step kyun? Law sirf absolute (Kelvin) temperature ke liye valid hai, kyunki $T$ average kinetic energy ke proportional hai — ek quantity jo kabhi negative nahi ho sakti.
Compute karo: numerator $= 4000 J$ , denominator $= 12.471$ , isliye $T = 320.7 K$ . Yeh step kyun? Hum number evaluate karte hain taaki dekh sakein tank kis absolute temperature par hai, koi bhi unit cosmetics karne se pehle.
Bilkul end mein Celsius mein convert karo: $T (^{\circ} C) = 320.7 - 273.15 = 47.6^{\circ} C$ . Yeh step kyun? Conversion ek final cosmetic step hai; yeh kabhi bhi equation mein enter nahi hona chahiye.

Verify karo: $T = 320.7 K$ back-substitute karo: $P = \frac{1.5 \times 8.314 \times 320.7}{0.010} = 4.0 \times 1 0^{5} Pa$ ✓.

Cell C — $n$ solve karo, phir molecules count karo

Worked example Example 3 — kitne molecules hain?

Ek balloon mein $V = 0.005 m^{3}$ hai $P = 1.01 \times 1 0^{5} Pa$ aur $T = 293 K$ par. (a) moles $n$ aur (b) molecules ki number $N$ nikalo.

Forecast: room conditions par 5 litres air — roughly kitne moles? (Room conditions par gas ka ek mole $\sim 24$ litres occupy karta hai, isliye fraction of a mole expect karo.)

$n$ ke liye solve karo: $n = \frac{P V}{R T} = \frac{1.01 \times 1 0 ^{5} \times 0.005}{8.314 \times 293}$ . Yeh step kyun? $n$ akela unknown hai; baaki sab measured hain.
Compute karo: numerator $= 505 J$ , denominator $= 2436.0$ , isliye $n = 0.2073 mol$ . Yeh step kyun? Hum number evaluate karte hain taaki forecast ke " $\sim 24$ litres per mole" mental estimate se compare kar sakein.
Molecule count tak bridge karo: $N = n N_{A} = 0.2073 \times 6.022 \times 1 0^{23} = 1.25 \times 1 0^{23}$ . Yeh step kyun? $N_{A}$ human unit (moles) aur microscopic count (molecules) ke beech exchange rate hai jo kinetic derivation mein use hoti hai.

Verify karo: $N k_{B} T = 1.25 \times 1 0^{23} \times 1.381 \times 1 0^{- 23} \times 293 = 505 J = P V$ ✓ (dono routes, $n R T$ aur $N k_{B} T$ , agree karte hain).

Cell D — Two-state comparison, ek variable fixed (ratio law)

Worked example Example 4 — constant volume par garam karo

Ek rigid sealed can mein gas $P_{1} = 2.0 \times 1 0^{5} Pa$ , $T_{1} = 300 K$ se start hoti hai. Isse $T_{2} = 450 K$ tak garam kiya jaata hai. $P_{2}$ nikalo. ( $n$ , $V$ fixed hain.)

Forecast: temperature $450/300 = 1.5$ factor se badhi. Agar $P \propto T$ , toh $P$ kitne factor se badhna chahiye?

Dono states ke liye law likho: $P_{1} V = n R T_{1}$ aur $P_{2} V = n R T_{2}$ . Yeh step kyun? Jo kuch bhi constant rehta hai ( $n$ , $R$ , $V$ ) woh divide karne par cancel ho jaayega.
State 2 ko state 1 se divide karo: $\frac{P _{2}}{P _{1}} = \frac{T _{2}}{T _{1}}$ (Gay-Lussac's law nikal aata hai). Yeh step kyun? Divide karne se constants remove ho jaate hain taaki humein unki numeric values ki zaroorat nahi padti.
Solve karo: $P_{2} = P_{1} \frac{T _{2}}{T _{1}} = 2.0 \times 1 0^{5} \times \frac{450}{300} = 3.0 \times 1 0^{5} Pa$ . Yeh step kyun? Hum ratio ko final pressure mein convert karte hain aur check karte hain ki yeh forecast factor $1.5$ se badha.

Verify karo: ratio $P_{2} / P_{1} = 1.5 = T_{2} / T_{1}$ ✓, forecast se match karta hai. Microscopically: zyada hot → faster molecules → zyada hard aur frequent kicks → zyada pressure. ✓ Yeh neeche figure mein $P$ – $T$ plane par traced constant- $V$ line hai.

Cell E — Molecular speed tak pahuncho

Worked example Example 5 — ek oxygen molecule kitni fast hai?

$O_{2}$ ka root-mean-square speed $v_{r m s} = ⟨ v^{2} ⟩$ nikalo $T = 300 K$ par. Molar mass $M = 0.032 kg/mol$ .

Forecast: 300 K par nitrogen $\approx 517$ m/s nikla. Oxygen bhaari hai ( $0.032$ vs $0.028$ kg/mol). Kya $v_{r m s}$ zyada hoga ya kam?

Bridge (per molecule) se start karo: $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ . Yeh step kyun? Yeh single equation hai jo temperature ko microscopic speed se connect karti hai.
Per-mole scale karo dono sides ko $N_{A}$ se multiply karke: $\frac{1}{2} M ⟨ v^{2} ⟩ = \frac{3}{2} R T$ , using $m N_{A} = M$ aur $k_{B} N_{A} = R$ . Yeh step kyun? Hum $M$ (per mole) aur $R$ jaante hain, $m$ aur $k_{B}$ individually nahi — yeh equation ko knowns mein convert karta hai.
Speed ke liye solve karo: $v_{r m s} = \frac{3 R T}{M} = \frac{3 \times 8.314 \times 300}{0.032}$ .
Compute karo: inside $= \frac{7482.6}{0.032} = 233831$ , isliye $v_{r m s} = 483.6 m/s$ . Yeh step kyun? Hum square root evaluate karte hain ek actual speed paane ke liye jo forecast ke nitrogen benchmark se compare kar sakein.

Verify karo: bhaari gas ⇒ chhoti speed, aur $483 < 517$ ✓. Mass ratio check karo: $\frac{v _{O_{2}}}{v _{N_{2}}} = \frac{0.028}{0.032} = 0.935$ , aur $517 \times 0.935 = 483$ ✓. Dekho Root Mean Square Speed.

Cell F — Degenerate & limiting inputs

Worked example Example 6 — edges par kya hota hai?

Ek fixed amount of gas ke liye, teen limits examine karo: (a) $T \to 0 K$ fixed $V$ par; (b) $V \to \infty$ fixed $T$ par; (c) $n \to 0$ . Batao $P$ kya karta hai aur kya ideal description still sense banaati hai.

Forecast: absolute zero par, ideal model mein pressure bada hoga, chhota hoga, ya exactly zero?

(a) $T \to 0$ : $P = \frac{n R T}{V} \to \frac{n R \cdot 0}{V} = 0$ . Yeh step kyun? Limit par seedha law read karna. Microscopically $⟨ v^{2} ⟩ \propto T \to 0$ : molecules ruk jaate hain, koi momentum deliver nahi karte, isliye koi push nahi. Yahi absolute zero ki definition hai — ideal model allow karta hai sabse thanda, aur pressure negative nahi ho sakta.
(b) $V \to \infty$ : $P = \frac{n R T}{V} \to 0$ as $V \to \infty$ . Yeh step kyun? Molecules (ab door ke) walls se kabhi-kabhi kam hit karte hain — $L \to \infty$ ke saath collision rate $v_{x} /2 L \to 0$ . Jo gas unlimited space fill kare woh vanishing pressure exert karti hai.
(c) $n \to 0$ : $P \to 0$ . Koi molecules nahi, koi collisions nahi, koi pressure nahi — vacuum. Yeh step kyun? Confirm karta hai ki model gracefully degrade hota hai: khaali jagah zero pressure deti hai, jaisi deni chahiye.

Verify karo: teeno limits $P \to 0$ dete hain ek single common cause se — kam/slower/rarer collisions. Numerically, $T$ ko 300 se 150 K par halving karo fixed $V$ , $n = 1$ , $V = 0.024 m^{3}$ par: $P$ $103925$ se $51962 Pa$ tak gir jaata hai, exactly factor $2$ ✓.

Cell G — Real-world word problem

Worked example Example 7 — hot day par car tyre

Ek car tyre mein gauge pressure $2.0 \times 1 0^{5} Pa$ (atmospheric $1.0 \times 1 0^{5} Pa$ se upar) inflate kiya gaya hai $T_{1} = 285 K$ par (ek thandi subah). Highway driving ke baad air $T_{2} = 320 K$ tak garam ho jaati hai. Tyre volume essentially fixed hai aur koi air leak nahi hoti. Naya gauge pressure nikalo.

Forecast: kya gauge reading absolute temperature ke same factor se badhegi, ya kam se? (Dhyan raho — law absolute pressure par kaam karta hai, gauge par nahi.)

Gauge ko absolute pressure mein convert karo: $P_{1}^{ab s} = 2.0 \times 1 0^{5} + 1.0 \times 1 0^{5} = 3.0 \times 1 0^{5} Pa$ . Yeh step kyun? $P V = n R T$ gas ka total pressure use karta hai, atmosphere ke upar excess nahi. Gauge ek reading convenience hai.
Fixed $V, n$ → ratio law: $P_{2}^{ab s} = P_{1}^{ab s} \frac{T _{2}}{T _{1}} = 3.0 \times 1 0^{5} \times \frac{320}{285}$ . Yeh step kyun? Same reasoning as Cell D — constants cancel ho jaate hain.
Absolute compute karo: $P_{2}^{ab s} = 3.0 \times 1 0^{5} \times 1.1228 = 3.368 \times 1 0^{5} Pa$ . Yeh step kyun? Hum ratio ko ek actual absolute pressure mein badelte hain taaki agle step mein atmosphere subtract kar sakein.
Gauge mein wapas convert karo: $P_{2}^{g a ug e} = 3.368 \times 1 0^{5} - 1.0 \times 1 0^{5} = 2.37 \times 1 0^{5} Pa$ . Yeh step kyun? Dashboard/gauge atmosphere subtract karta hai; hum same units mein report karte hain jo driver dekhta hai.

Verify karo: absolute $1.123$ factor se badha, lekin gauge $2.0$ se $2.37$ tak badha — factor $1.18$ , $1.123$ se zyada. Yahi forecast ka jawab hai: gauge zyada factor se badhta hai, kyunki constant atmospheric offset temperature se scale nahi hota. ✓

Cell H — Exam twist: do gases ka mixture

Worked example Example 8 — do gases ek box share karte hain

$V = 0.010 m^{3}$ ka ek rigid box $T = 300 K$ par $n_{1} = 0.40 mol$ helium aur $n_{2} = 0.60 mol$ nitrogen contain karta hai. Total pressure nikalo.

Forecast: kya do gases ke pressures add hote hain, ya bhaari gas dominate karti hai? (Ideal model mein, molecules ek doosre ki identity ignore karte hain sivaay walls se collisions ke.)

Har gas independently law follow karta hai (Dalton's law of partial pressures): $P_{i} = \frac{n _{i} R T}{V}$ . Yeh step kyun? Ideal molecules ek doosre par koi forces exert nahi karte, isliye har species wall par push karti hai jaise akela ho — parent mein derivation ne molecules ki chemical identity kabhi use nahi ki.
Total pressure sum hai: $P = P_{1} + P_{2} = \frac{( n _{1} + n _{2} ) R T}{V}$ . Yeh step kyun? Wall par total force independent contributions ka sum hai (derivation ka Step 4), isliye pressures add hote hain.
$n_{t o t} = 1.00 mol$ ke saath compute karo: $P = \frac{1.00 \times 8.314 \times 300}{0.010} = 2.494 \times 1 0^{5} Pa$ . Yeh step kyun? Hum total evaluate karte hain taaki isse partial pressures mein split karke check kar sakein.

Verify karo: answer sirf total moles par depend karta hai, kaunsi gas hai par nahi — helium aur nitrogen same $T$ par per molecule same average kinetic energy carry karte hain (equipartition, $\frac{3}{2} k_{B} T$ ), isliye molar amount hi sab kuch hai. Dekho Equipartition Theorem. Partial pressures: $P_{H e} = 9.98 \times 1 0^{4}$ , $P_{N_{2}} = 1.50 \times 1 0^{5}$ , sum $2.49 \times 1 0^{5} Pa$ ✓.

Cell I — Two-state, DO variables change (combined gas law)

Worked example Example 9 — ek gas jisme pressure

aur volume aur temperature sab change hote hain Ek fixed amount of gas $P_{1} = 1.0 \times 1 0^{5} Pa$ , $V_{1} = 0.030 m^{3}$ , $T_{1} = 300 K$ se start hoti hai. Isse compress aur heat karke $V_{2} = 0.020 m^{3}$ aur $T_{2} = 400 K$ par le jaate hain. Naya pressure $P_{2}$ nikalo.

Forecast: do effects $P$ ko upar push karte hain — squeeze karna (chhota $V$ ) aur garam karna (bada $T$ ). Isliye expect karo $P_{2} > P_{1}$ , lekin kitna? Compute karne se pehle guess karo.

Dono states ke liye law likho: $P_{1} V_{1} = n R T_{1}$ aur $P_{2} V_{2} = n R T_{2}$ . Yeh step kyun? Sirf $n$ aur $R$ shared aur constant hain ab — $P$ , $V$ , $T$ sab change ho rahe hain, isliye koi single ratio law (Cell D) apply nahi hota.
State 2 ko state 1 se divide karo taaki $n$ aur $R$ cancel hon: $\frac{P _{2} V _{2}}{P _{1} V _{1}} = \frac{T _{2}}{T _{1}}$ , yaani combined gas law $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ . Yeh step kyun? Divide karna sirf constants ko khatam karta hai, sirf chhe measured quantities ke beech relation chhod ke.
$P_{2}$ ke liye solve karo: $P_{2} = P_{1} \cdot \frac{V _{1}}{V _{2}} \cdot \frac{T _{2}}{T _{1}}$ . Yeh step kyun? Hum akele unknown ko isolate karte hain; do visible factors note karo — volume-squeeze factor $V_{1} / V_{2}$ aur heating factor $T_{2} / T_{1}$ , dono $> 1$ .
Compute karo: $P_{2} = 1.0 \times 1 0^{5} \times \frac{0.030}{0.020} \times \frac{400}{300} = 1.0 \times 1 0^{5} \times 1.5 \times 1.333 = 2.0 \times 1 0^{5} Pa$ . Yeh step kyun? Do enhancement factors ( $1.5$ aur $1.333$ ) ko multiply karne se pata chalta hai ki combined effect unka product hai, jo pressure exactly double karta hai.

Verify karo: check karo $\frac{P _{1} V _{1}}{T _{1}} = \frac{( 1.0 \times 1 0 ^{5} ) ( 0.030 )}{300} = 10$ aur $\frac{P _{2} V _{2}}{T _{2}} = \frac{( 2.0 \times 1 0 ^{5} ) ( 0.020 )}{400} = 10$ — equal ✓. Dono pressure ko upar push karte hain, aur $P_{2} = 2 P_{1}$ forecast se match karta hai. ✓

Active Recall

Recall Har phrase kaun si cell test kar raha hai?

"Plug in karne se pehle Kelvin mein convert karo" ::: Cell B — Celsius trap. "Constants cancel karne ke liye state 2 ko state 1 se divide karo" ::: Cell D / G — one-variable-fixed ratio law. " $P_{1} V_{1} / T_{1} = P_{2} V_{2} / T_{2}$ " ::: Cell I — combined gas law, do variables change hoti hain. " $v_{r m s} = 3 R T / M$ " ::: Cell E — microscopic reach-through. "Components ke pressures add hote hain" ::: Cell H — mixture ke liye Dalton's law. " $T \to 0$ par, $P \to 0$ lekin real gas liquefy hoti hai" ::: Cell F — degenerate/limiting case. " $1 atm = ?$ Pa" ::: $1.013 \times 1 0^{5}$ Pa — toolkit ka unit bridge.