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

KYA chahiye: hume $P$ chahiye, aur $n,T,V$ already pata hain — $PV=nRT$ ke har symbol ki value pata hai sirf $P$ ke alawa. YE tool kyun: ideal gas law exactly inhi chaar quantities ke beech direct relation hai; koi kinetic detail zaroorat nahi. $P=\frac{nRT}{V}=\frac{3\times 8.314\times 400}{0.02}=\frac{9976.8}{0.02}\approx 4.99\times 10^5\text{Pa}$ Units check: $\frac{\text{mol}\cdot ({\text{J mol}}^{-1}{\text{K}}^{-1})\cdot \text{K}}{{\text{m}}^3}=\frac{\text{J}}{{\text{m}}^3}=\text{Pa}$. ✓

• $P$ — pressure — pascal (Pa = N/m²)
• $V$ — volume — cubic metre (m³)
• $n$ — amount of substance — mole (mol)
• $R$ — universal gas constant — J mol⁻¹ K⁻¹
• $T$ — absolute temperature — kelvin (K)

Ek important point: $T$ Kelvin mein hota hai, kabhi Celsius mein nahi (neeche L1 mistake dekho).

KYA: $PV=nRT$ ko $V$ ke liye solve karo. KYU: chaar mein se teen unknowns diye hue hain, ek target hai. $V=\frac{nRT}{P}=\frac{0.5\times 8.314\times 320}{1.5\times 10^5}=\frac{1330.24}{1.5\times 10^5}\approx 8.87\times 10^{-3}{\text{m}}^3$ Ye lagbhag $8.9$ litres hai — ek reasonable lab flask.

KYA: per-mole speed formula use karo. YE tool kyun: rms speed tab milti hai jab per mole translational energy $\frac{3}{2}RT$ ko $\frac{1}{2}M⟨v^2⟩$ ke barabar rakhte hain — ye scaling parent ke Example 1 mein ki gayi hai. $v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.314\times 300}{0.032}}=\sqrt{\frac{7482.6}{0.032}}=\sqrt{2.338\times 10^5}\approx 483.4\text{m/s}$ Parent ke $N_2$ (517 m/s) se bhaari hai kyunki $M$ zyada hai → slower. ✓ Is value ke aas-paas speeds ke poore spread ke liye dekho Maxwell-Boltzmann Speed Distribution.

KYA: $PV=N{k}_{B}T$ (molecule form) use karo. YE form kyun: question directly $N$ maang raha hai, isliye $R$ nahi $k_B$ use karo. $N=\frac{PV}{k_{B}T}=\frac{1.0\times 10^5\times 2.0\times 10^{-3}}{1.381\times 10^{-23}\times 273}=\frac{200}{3.770\times 10^{-21}}\approx 5.31\times 10^{22}$ $N_A$ se cross-check: pehle moles nikalo, $n=PV/(RT)=200/(8.314\times 273)\approx 0.0881$ mol. Phir Avogadro's number se molecules mein convert karo: $N=n{N}_A=0.0881\times 6.022\times 10^{23}\approx 5.31\times 10^{22}$. ✓ Dono routes agree karte hain, jo exactly $R=N_A{k}_B$ ka statement hai.

KYA: usi gas ke do states ko relate karo. RATIO kyun: $n,V,R$ nahi badle, isliye $P/T=nR/V$ constant hai. Dono states ko divide karne se jo hum nahi jaante wo cancel ho jaata hai. $\frac{P_2}{P_1}=\frac{T_2}{T_1}\Rightarrow P_2=P_1\frac{T_2}{T_1}=2.0\times 10^5\times \frac{450}{300}=3.0\times 10^5\text{Pa}$ Microscopic reading: $⟨v^2⟩\propto T$, isliye $1.5\times$ temperature par molecules walls se zyada zor se aur zyada baar takraate hain → pressure $1.5\times$ badh jaata hai.

KYA: dono rms speeds ko barabar set karo aur $T_{\text{He}}$ ke liye solve karo. SQUARES kyun matter karte hain: $v_{rms}^2=3RT/M$, isliye barabar speeds ka matlab hai barabar $T/M$ ratios. $\frac{3R{T}_{\text{He}}}{M_{\text{He}}}=\frac{3R{T}_{N_2}}{M_{N_2}}\Rightarrow T_{\text{He}}=T_{N_2}\frac{M_{\text{He}}}{M_{N_2}}=300\times \frac{0.004}{0.028}\approx 42.9\text{K}$ Helium $7\times$ halka hai, isliye wo same speed $\frac{1}{7}$ temperature par reach karta hai. Thanda helium utna hi fast hai jitna warm nitrogen — surprising hai par correct hai.

KYA: kinetic pressure ke density form ko invert karo. YE form kyun: yahan hume $\rho =Nm/V$ (mass per volume) diya gaya hai, $M$ ya $N$ nahi — density form $P=\frac{1}{3}\rho ⟨v^2⟩$ exactly isi ke liye bana hai. $⟨v^2⟩=\frac{3P}{\rho }=\frac{3\times 10^5}{1.25}=2.4\times 10^5{\text{m}}^2/{\text{s}}^2$ $v_{rms}=\sqrt{2.4\times 10^5}\approx 489.9\text{m/s}$

Step A — volume. $V=L^3=(0.10{)}^3=1.0\times 10^{-3}{\text{m}}^3$. Step B — pressure (pure mechanics, temperature abhi nahi): $P=\frac{1}{3}\frac{Nm⟨v^2⟩}{V}=\frac{1}{3}\cdot \frac{(1.0\times 10^{22})(4.65\times 10^{-26})(2.4\times 10^5)}{1.0\times 10^{-3}}$ Andar numerator: $1.0\times 10^{22}\times 4.65\times 10^{-26}=4.65\times 10^{-4}$; $\times 2.4\times 10^5=111.6$. Phir $\frac{1}{3}\times 111.6/10^{-3}$: $P=\frac{37.2}{10^{-3}}=3.72\times 10^4\text{Pa}$ Step C — EK wall par force (figure ki red wall dekho, area $L^2$): $F=P\cdot L^2=3.72\times 10^4\times (0.10{)}^2=3.72\times 10^4\times 0.01=372\text{N}$ Step D — bridge se temperature: $T=\frac{PV}{N{k}_B}=\frac{3.72\times 10^4\times 10^{-3}}{1.0\times 10^{22}\times 1.381\times 10^{-23}}=\frac{37.2}{0.1381}\approx 269.4\text{K}$ Toh ye gas freezing ke paas hai — ek modest speed ke saath consistent hai. Derivation ke har arrow ko start se end tak use kiya gaya.

KYA: total number of moles conserved hai; temperature fixed hai. MOLES kyun count karein: $PV=nRT$ kehta hai $n=PV/(RT)$, aur moles tab bas add ho jaate hain jab gases merge hoti hain. Pehle moles: $n_A=\frac{P_A{V}_A}{RT}$, $n_B=\frac{P_B{V}_B}{RT}$. Valve kholne ke baad, total volume $V_f=V_A+V_B$, total moles $n_A+n_B$, same $T$: $P_f(V_A+V_B)=(n_A+n_B)RT=P_A{V}_A+P_B{V}_B$ $RT$ cancel ho jaata hai — beautiful. Solve karte hain: $P_f=\frac{P_A{V}_A+P_B{V}_B}{V_A+V_B}=\frac{(3.0\times 10^5)(2.0)+(1.0\times 10^5)(3.0)}{2.0+3.0}$ Unit consistency note: agar is ratio mein litre factors cancel ho jaate hain, phir bhi hamesha poori calculation mein ek consistent unit system rakho. Agar SI mein convert karo, to $V_A=2.0\times 10^{-3}{\text{m}}^3$ aur $V_B=3.0\times 10^{-3}{\text{m}}^3$ use karo; $10^{-3}$ tab upar aur neeche cancel ho jaayega, aur identical answer milega. Kabhi ek term mein litres aur doosre mein m³ mix mat karo — agar kisi problem mein units cancel na hon, to wo silent mismatch ek classic wrong answer deta hai. $P_f=\frac{6.0\times 10^5+3.0\times 10^5}{5.0}=\frac{9.0\times 10^5}{5.0}=1.8\times 10^5\text{Pa}$ Ye bas dono pressures ka volume-weighted average hai.

Step A — mechanics se shuru: $PV=\frac{1}{3}Nm⟨v^2⟩$. Step B — energy introduce karo $\frac{1}{2}$ ka factor daalkaar: 2 se multiply aur divide karo, $PV=\frac{2}{3}N\left(\frac{1}{2}m⟨v^2⟩\right)=\frac{2}{3}N⟨E_k⟩.$ Step C — measured law $PV=N{k}_{B}T$ ke barabar rakho: $\frac{2}{3}N⟨E_k⟩=N{k}_{B}T\Rightarrow \boxed{⟨E_k⟩=\frac{3}{2}{k}_{B}T}.$ YE sirf $T$ par kyun depend karta hai: mass $m$ aur speed cancel ho gayi; molecule ke type ke baare mein kuch bhi nahi bachta — sirf $T$ bachta hai. Ye exactly Equipartition Theorem ka result hai 3 translational degrees of freedom ke liye. 300 K par numeric: $⟨E_k⟩=\frac{3}{2}\times 1.381\times 10^{-23}\times 300=6.215\times 10^{-21}\text{J}.$ Consequence: 300 K par helium aur nitrogen ka $⟨E_k⟩=6.22\times 10^{-21}$ J identical hai. Halka helium same energy carry karne ke liye bas faster move karta hai — ye Temperature and Internal Energy se connect karta hai.

KYA: ideal $P_{\text{ideal}}=nRT/V$ ko volume-corrected $P_{\text{corr}}=nRT/(V-nb)$ se compare karo. $(V-nb)$ kyun: point particles pura box use kar sakte the, lekin real molecules space occupy karte hain, isliye kam room free hoti hai — dekho Real Gases and Van der Waals Equation. Ratio lene par $nRT$ cancel ho jaata hai: $\frac{P_{\text{corr}}}{P_{\text{ideal}}}=\frac{V}{V-nb}=\frac{1.0\times 10^{-3}}{1.0\times 10^{-3}-1\times 3.2\times 10^{-5}}=\frac{1.0\times 10^{-3}}{9.68\times 10^{-4}}\approx 1.03306$ Toh corrected pressure lagbhag $3.31\%$ zyada hai: $\text{underestimate}=\left(\frac{V}{V-nb}-1\right)\times 100\%=\frac{nb}{V-nb}\times 100\%\approx 3.31\%.$ Ideal law yahan pressure under-predict karta hai kyunki wo galti se assume karta hai ki molecules pura volume access kar sakte hain.

Pehle symbols. Ek molecule ki velocity ke teen perpendicular components hain $(v_x,v_y,v_z)$, ek box ke har axis ke along. $v_x$ bas itna hai ki wo $x$-direction mein kitni fast move karta hai; $⟨v_x^2⟩$ har molecule par us component ke square ka average hai. Full speed obey karta hai $v^2=v_x^2+v_y^2+v_z^2$. KYA: full-speed spread ko one-axis spread se relate karo. $\sqrt{3}$ kyun: isotropy ne diya tha $⟨v^2⟩=3⟨v_x^2⟩$ parent derivation ke Step 5 mein, kyunki koi bhi direction special nahi hai ($⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩$). Square roots lo: $v_{rms}=\sqrt{⟨v^2⟩}=\sqrt{3⟨v_x^2⟩}=\sqrt{3}\sqrt{⟨v_x^2⟩}.$ Oxygen ke liye, $v_{rms}\approx 483.4$ m/s (Ex 2.2), isliye $\sqrt{⟨v_x^2⟩}=\frac{v_{rms}}{\sqrt{3}}=\frac{483.4}{1.732}\approx 279.1\text{m/s}.$ Har single axis total speed spread ka $1/\sqrt{3}$ carry karta hai — factor $\frac{1}{3}$ ka geometric heart yahi hai.

Gas law ka kaunsa form molecule count $N$ use karta hai? ::: $PV=N{k}_{B}T$ ($k_B$ ke saath pair hota hai). Fixed $V,n$ par $T$ double karne se $P$ ka kya hota hai? ::: double ho jaata hai ($P\propto T$). $v_{rms}$ double karne ke liye temperature kitne factor se badhni chahiye? ::: chaar guna ($v_{rms}\propto \sqrt{T}$). Temperature $T$ par per molecule average translational KE? ::: $\frac{3}{2}{k}_{B}T$, gas type se independent. Fixed $T$ par do tanks merge karna: kya conserved hota hai? ::: total moles, yaani $PV$ add hota hai.