Question bank — Speed of sound — a = √(γRT) — derivation

False. $p$ ko fixed $T$ pe raise karne se $\rho$ exactly same proportion mein badhta hai, isliye $p/\rho =RT$ unchanged rehta hai aur $a=\sqrt{\gamma RT}$ bhi unchanged — stiffness aur inertia dono saath badhte hain aur cancel ho jaate hain.

True ek ideal gas ke liye fixed composition ke saath, kyunki $a=\sqrt{\gamma RT}$ aur $\gamma ,R$ gas ke constants hain — sirf $T$ vary karta hai.

False. Thande gas mein thermal molecular motion slower hoti hai, isliye $a=\sqrt{\gamma RT}$ chhota hota hai; sound cold air mein slower hota hai (217 K pe 295 m/s vs 288 K pe 340 m/s).

True. Helium ka molar mass bahut chhota hai ($4$ g/mol) isliye uska specific $R=R_u/M$ bada hota hai; halke, faster molecules push ko jaldi relay karte hain — isliye squeaky-voice effect hota hai (dekho Ideal gas law and specific gas constant).

False. Gas law hamesha hold karta hai, lekin yahaan process constraint adiabatic (isentropic) hai: compression itna fast hota hai ki heat conduct hokar ja nahi sakti, isliye entropy constant rehta hai, temperature nahi (dekho Adiabatic vs isothermal processes).

True mathematically — lekin koi bhi real gas $\gamma =1$ nahi hota; diatomic air ke liye $\gamma =1.4$ se $a$, $\sqrt{1.4}\approx 1.18$ se badhta hai, jo Laplace correction hai jisne Newton ki 15% error fix ki.

False. Ye specific gas constant $R=R_u/M$ hai (per kilogram), air ke liye $287\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. $8.314$ use karne se ek meaningless $\sim 30$ m/s milta hai.

False. $a^2=dp/d\rho$ dene wale steps mein sirf mass aur momentum conservation use hua. Adiabatic assumption baad mein aata hai, $dp/d\rho =\gamma p/\rho$ evaluate karne ke liye.

Error ye hai ki differentiation ke dauran $T$ ko constant maan liya. Wave adiabatic hai, isliye $T$ bhi $\rho$ ke saath change hota hai; correct constraint $p\propto {\rho }^{\gamma }$ hai, jisse $dp/d\rho =\gamma RT$ milta hai.

Physically nahi kar sakte, lekin Galilean frame change legitimate hai: saari velocities se constant wave speed $a$ subtract karna unsteady problem ko steady mein convert kar deta hai bina koi physics change kiye — ye sirf ek bookkeeping choice hai.

Wo term do infinitesimals ka product hai, yaani second-order small hai, aur first-order terms ke relative vanish ho jaata hai jab disturbance $\to 0$ ho. Isko rakhna "infinitesimal pulse" ki definition ke saath contradict karta.

Ye identical hain — ideal gas law $p/\rho =RT$ unhe equal banata hai. Pehla general result hai; doosra sirf equation of state substitute karta hai.

Speed $a$ infinitesimal disturbance ke liye define ki gayi hai aur amplitude-independent hai. Finite-amplitude waves steepen hoti hain aur shocks form kar sakti hain, lekin ye linear-acoustics regime ke bahar hai (dekho Normal shock waves).

Constant entropy (isentropic) aur constant temperature ek cheez nahi hai. Adiabatic compression $T$ raise karta hai jabki $s$ fixed rehta hai; isothermal ke liye heat flow chahiye hogi, jo adiabatic ko contradict karta.

Wave frame mein flow steady hai, isliye mass aur momentum balances simple algebraic in–out statements ban jaate hain instead of time-dependent PDEs ke.

Mass aur momentum mechanical relation $a^2=dp/d\rho$ dete hain; us derivative ki value thermodynamic path pe depend karti hai, aur sirf adiabatic path $p\propto {\rho }^{\gamma }$ hi $\gamma$ inject karta hai.

Sound molecule se molecule collisions ke through pass hoti hai, isliye ye molecules se faster propagate nahi ho sakti jo ise carry kar rahe hain; dono $\sqrt{T}$ ke saath scale karte hain, isliye $a$ aur mean thermal speed saath track karte hain.

Zyada altitude ka matlab hai lower $T$, isliye lower $a=\sqrt{\gamma RT}$; $a$ chhota hone par, $M=V/a$ badhta hai chahe $V$ unchanged ho (dekho Mach number and flow regimes).

Ek infinitesimal, smooth disturbance ke liye koi finite gradients nahi hote jo dissipation cause karein, isliye adiabatic process reversible bhi hai — dono milke isentropic ka matlab hai, jo exactly $(\partial p/\partial \rho {)}_s$ hai.

$a$ ratio $p/\rho =RT$ ko track karta hai; $p$ aur $\rho$ ko equally scale karne se ratio fixed rehta hai, jabki $T$ raise karne se ratio directly badhta hai.

$a=\sqrt{\gamma RT}\to 0$: koi thermal motion nahi hogi toh molecular collisions nahi honge jo disturbance relay karein, isliye sound propagate nahi ho sakti.

Nahi. Shock ek finite, irreversible discontinuity hai jahan entropy jump karti hai; isentropic assumption wahaan fail ho jaata hai, isliye linear speed-of-sound relation use describe nahi karta (dekho Normal shock waves).

Tabhi tak jab gas continuum ki tarah behave karne ke liye dense ho; jab mean free path wavelength ke barabar ho jaata hai, collisions bahut kam ho jaate hain aur classical sound propagation break down ho jaata hai.

$V=a$, isliye $M=1$; disturbances ab object se upstream outrun nahi kar sakti, ye wo boundary hai jahan compressibility dominant ho jaati hai (dekho Compressibility and why M > 0.3 matters).

Isothermal wala, $a=\sqrt{RT}$ (Newton's value). Ye ek limiting case hai jo real audible sound kabhi satisfy nahi karta kyunki oscillations bahut zyada fast hote hain.

Monatomic gas, kyunki bada $\gamma$ se $a=\sqrt{\gamma RT}$ badhta hai; zyada compression energy pressure mein jaati hai instead of internal molecular modes ke.

Nahi — $R=R_u/M$ sirf molar mass pe depend karta hai, jo molecule ki property hai, thermodynamic state ki nahi.