Consider the sound wave travelling in ideal gases of He$\mathrm{He}$, CH₄$\mathrm{CH}_{4}$, and CO₂$\mathrm{CO}_{2}$. All the gases have the same ratio (P)/(ρ)$\frac{P}{\rho}$, where P$P$ is the pressure and ρ$\rho$ is the density. The ratio of the speed of sound through the gases vHe : v_CH₄ : v_CO₂$v_{\mathrm{He}} : v_{\mathrm{CH}_{4}} : v_{\mathrm{CO}_{2}}$ is given by

Related Formula

Core Logic

For Oxygen (O₂$O_2$) at standard temperature and pressure (S.T.P.): T = 273 K$T = 273 \text{ K}$ M = 32 g/mol = 32 × 10⁻³ kg/mol$M = 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol}$ γ = 1.4$\gamma = 1.4$ R = 8.3 J K⁻¹ mol⁻¹$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

Step 1: Calculate Velocity

Approximating to the closest given option yields 310 m/s$310 \text{ m/s}$.

Pattern Recognition

For diatomic gases around room temp or STP, velocities range roughly from 250$250$ to 350 m/s$350 \text{ m/s}$ depending on molar mass (N₂ ≈ 334$N_2 \approx 334$, O₂ ≈ 315$O_2 \approx 315$). Recognize 315$315$ is closest to option (1) due to standard approximations taken in exams.

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Kinetic Theory of Gases

Related Formula

Core Logic

For a closed organ pipe, the fundamental frequency (1st harmonic) is:

where L₁$L_1$ is the length of the closed pipe.

For an open organ pipe, the first overtone (2nd harmonic) is:

where L₂$L_2$ is the length of the open pipe (L₂ = 60 cm$L_2 = 60\mathrm{\,cm}$).

Step 2: Equating Frequencies

Given f₁ = f₂$f_1 = f_2$:

L₂ = 4L₁$L_2 = 4L_1$

Chapter Mix

Class 11 Physics: Waves