Exercises — Speed of sound — a = √(γRT) — derivation

Sirf Temperature. $a=\sqrt{\gamma RT}$ se, ek hi variable hai $T$; $\gamma$ aur $R$ kisi bhi given gas ke liye fixed hote hain. Halanki $p$ aur $\rho$ form $\sqrt{\gamma p/\rho }$ mein appear karte hain, unka ratio $p/\rho =RT$ poori tarah $T$ se set hota hai. Dekho Ideal gas law and specific gas constant.

Adiabatic (actually isentropic). Oscillation itni fast hoti hai ki heat ek cycle mein wave ke across conduct nahi ho sakti. Is choice ko record karne wala symbol hai $\gamma =c_p/c_v$. Agar process isothermal hoti toh hame milta $a=\sqrt{RT}$ bina $\gamma$ ke. Dekho Adiabatic vs isothermal processes.

$a^2={\left(\frac{\partial p}{\partial \rho }\right)}_s$ Subscript $s$ ka matlab hai "constant entropy par" — derivative ek isentropic (reversible adiabatic) path ke along li jaati hai. Dekho Isentropic relations p ∝ ρ^γ.

$a=\sqrt{\gamma RT}=\sqrt{1.4\times 287\times 300}=\sqrt{120,540}\approx 347m/s$ Humne kya kiya: $T$ ko seedha $\sqrt{\gamma RT}$ mein plug kiya. Kyun: humein temperature di gayi thi, toh yahi face use karna tha.

Pehle convert karo: $T=-50+273=223\mathrm{K}$ ($-50$ directly kabhi use mat karo — negative ka square root yahan meaningless hai). $a=\sqrt{1.4\times 287\times 223}=\sqrt{89,597}\approx 299m/s$ Kyun sea level se kam: thandi air → slow molecules → slow sound.

$M=\frac{V}{a}=\frac{272}{340}=0.80$ Kyunki $0.3<M<0.8$, yeh high subsonic hai aur compressibility matter karti hai ($M>0.3$). Dekho Mach number and flow regimes aur Compressibility and why M > 0.3 matters.

Tarika 1 ($p,\rho$ directly given): $a=\sqrt{\gamma \frac{p}{\rho }}=\sqrt{1.4\times \frac{60,000}{0.80}}=\sqrt{1.4\times 75,000}=\sqrt{105,000}\approx 324m/s$ Tarika 2 (pehle $T$ nikalo): $p=\rho RT$ se, $T=\frac{p}{\rho R}=\frac{60,000}{0.80\times 287}=261.3\mathrm{K},a=\sqrt{1.4\times 287\times 261.3}\approx 324m/s$ Dono match karte hain, kyunki $\gamma p/\rho =\gamma RT$ usi number ke do alag likhne ke tarike hain. Dono kyun karo: yeh prove karta hai ki tum samajhte ho ki $p/\rho$ hai hi $RT$, jo "$a$ sirf $T$ par depend karta hai" ka dil hai.

Sea level: $a=\sqrt{1.4\times 287\times 288}=\sqrt{115,718}\approx 340m/s$, toh $M=250/340\approx 0.735$. Altitude: $a=\sqrt{1.4\times 287\times 217}=\sqrt{87,175}\approx 295m/s$, toh $M=250/295\approx 0.847$. Altitude wale aircraft ka Mach number zyada hai. Same speed, lekin thandi air mein sound zyada slowly travel karti hai, toh same $V$ $a$ ka ek bada fraction hai. Yehi reason hai ki cruising jets apna Mach limit dekhte hain, true airspeed nahi.

Pehle specific constant: $R_{\text{He}}=R_u/M=8314/4=2078.5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ ($R_u=8314\mathrm{J}\mathrm{kmo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ aur $M=4g/mol=4kg/kmol$ use karte hue). $a=\sqrt{1.667\times 2078.5\times 288}=\sqrt{997,860}\approx 999m/s$ Yeh air mein speed ka lagbhag $999/340\approx 2.9\times$ hai. Kyun: helium molecules bahut halki hoti hain (chhota $M$ → bada $R$), toh same temperature par woh zyada tezi se ghoomti hain aur pressure pulse ko zyada jaldi aage pass karti hain. Yeh squeaky-voice effect ke peeche ki physics hai.

Newton (isothermal): $a_N=\sqrt{RT}=\sqrt{287\times 288}=\sqrt{82,656}\approx 287.5m/s$. Correct (adiabatic): $a=\sqrt{\gamma RT}=\sqrt{1.4}\times a_N=1.1832\times 287.5\approx 340m/s$. Error: $\frac{a_N-a}{a}=\frac{287.5-340.2}{340.2}\approx -0.155=-15.5\%$ Newton lagbhag $15.5\%$ kam tha. Poora discrepancy missing factor $\sqrt{\gamma }=\sqrt{1.4}$ ki wajah se hai. Fix kyun kaam karta hai: real compression apni heat trap karta hai (adiabatic), jo gas ko isothermal model ki assumption se zyada stiff banata hai — zyada stiff medium mein sound faster travel karta hai. Dekho Adiabatic vs isothermal processes.

Scaling: ek gas ke liye $\gamma R$ constant hai, toh $a=\sqrt{\gamma R}\sqrt{T}\Rightarrow a\propto \sqrt{T}$. Isliye $\frac{a_2}{a_1}=\sqrt{\frac{T_2}{T_1}}.$ Apply karo: $a_2=340\times \sqrt{\frac{350}{288}}=340\times \sqrt{1.2153}=340\times 1.1024\approx 375m/s$ Yeh kyun powerful hai: ratio trick $\gamma$ aur $R$ ko poori tarah cancel kar deti hai, toh yeh kisi bhi gas ke liye kaam karta hai jab tak woh same gas rehti hai. $T$ mein $22\%$ rise se sirf $\sqrt{1.22}-1\approx 10\%$ rise in $a$ hoti hai — square root ise tame karta hai.

$V=Ma=M\sqrt{\gamma RT}$. Sea level: $a_1=\sqrt{1.4\times 287\times 288}\approx 340.2m/s$, toh $V_1=0.85\times 340.2\approx 289m/s$. Altitude: $a_2=\sqrt{1.4\times 287\times 223}\approx 299.3m/s$, toh $V_2=0.85\times 299.3\approx 254m/s$. Comment: thandi, patli air mein same Mach number rakhne ke liye aircraft ko actually true airspeed mein slower fly karna padta hai ($289$ se $254m/s$ tak), kyunki $a$ drop ho gayi hai. Mach number, raw speed nahi, jo shock/compressibility behaviour govern karta hai. Dekho Mach number and flow regimes.

Step A — isentropic law differentiate karo. $C$ constant ke saath, $\frac{dp}{d\rho }=C\gamma {\rho }^{\gamma -1}=\gamma \frac{C{\rho }^{\gamma }}{\rho }=\gamma \frac{p}{\rho }.$ Kyun: $a^2=dp/d\rho$ mass aur momentum se aaya tha; ab hum woh process ($p\propto {\rho }^{\gamma }$) supply karte hain jo batata hai ki $p$ actually $\rho$ ke saath kaise move karta hai. Step B — $a^2$ mein insert karo. $a^2=\frac{dp}{d\rho }=\gamma \frac{p}{\rho }$. Step C — ideal gas law use karo $p=\rho RT\Rightarrow p/\rho =RT$: $a^2=\gamma RT\Rightarrow a=\sqrt{\gamma RT}.$ $250\mathrm{K}$ par evaluate karo: $a=\sqrt{1.4\times 287\times 250}=\sqrt{100,450}\approx 317m/s$.

Upstream: $a_1=\sqrt{1.4\times 287\times 250}=\sqrt{100,450}\approx 316.9m/s$. Downstream: $a_2=\sqrt{1.4\times 287\times 421.9}=\sqrt{169,519}\approx 411.7m/s$. Ratio: $a_2/a_1=411.7/316.9\approx 1.299$. Check: $\sqrt{T_2/T_1}=\sqrt{421.9/250}=\sqrt{1.6876}\approx 1.299$. ✓ Kyun match karna zaroori hai: shock ke across gas abhi bhi air hai ($\gamma ,R$ unchanged), toh $a\propto \sqrt{T}$ hold karta hai chahe flow violently non-isentropic ho. Sound-speed formula local state ($T$) par depend karta hai, is baat par nahi ki gas wahan kaise pahunchi. Dekho Normal shock waves.

$R_{\mathrm{C}{\mathrm{O}}_2}=R_u/M=8314/44=188.95\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. $a=\sqrt{\gamma RT}$ invert karo: $T=\frac{a^2}{\gamma R}=\frac{300^2}{1.289\times 188.95}=\frac{90,000}{243.55}\approx 369.5\mathrm{K}.$ Air ko $a=300m/s$ tak pahunchane ke liye: $T_{\text{air}}=300^2/(1.4\times 287)=90,000/401.8\approx 224\mathrm{K}$. CO₂ ko bahut zyada hot gas chahiye ($\approx 370$ K vs $224$ K). Kyun: CO₂ molecules heavy hoti hain (bada $M$ → chhota $R$) aur kam springy hoti hain (lower $\gamma$), toh pulse ko same speed par travel karane ke liye unhe tab tak heat karna padta hai jab tak unki thermal motion match na kare. Heavy, floppy medium → same sound speed ke liye zyada temperature chahiye.

• $T$ given ho toh formula ka kaun sa face use karo? ::: $a=\sqrt{\gamma RT}$
• $p$ aur $\rho$ given ho toh kaun sa face use karo? ::: $a=\sqrt{\gamma p/\rho }$
• $a$ temperature ke saath kaise scale karta hai? ::: $a\propto \sqrt{T}$
• Same $V$, thandi air → Mach number kya karta hai? ::: Badhta hai (kyunki $a$ drop hoti hai)
• Kya normal shock $a=\sqrt{\gamma RT}$ ko invalidate karta hai? ::: Nahi — yeh ek local state property hai; har side ko alag alag evaluate karo