Question

The speed of sound in an ideal gas at a given temperature T is v. The rms speed of gas molecules at that temperature is v_rms. The ratio of the velocities v and v_rms for helium and oxygen gases are X and X' respectively. Then \(\frac{X}{X'}\) is equal to

• A. \(\frac{5}{\sqrt{21}}\)
• B. \(\sqrt{\frac{5}{21}}\)
• C. \(\frac{21}{5}\)
• D. \(\frac{21}{\sqrt{5}}\)

Answer 1

The Correct Option is A

Given: 

• A block of mass \( m \) is connected to a light spring of force constant \( k \).
• The system is placed in a damping medium with damping constant \( b \).
• The initial amplitude of oscillation is \( A \), and the angular frequency is \( \omega' \).

Step 1: Governing Equation of Damped Oscillations

The motion of a damped harmonic oscillator is given by:

\[ m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0 \]

Step 2: Expressions for Different Quantities

• Angular frequency of damped oscillations:
• Total energy of the system:
• Displacement as a function of time:

Step 3: Identifying the Incorrect Expression

The given option:

\[ x = A e^{-\frac{b}{m}} \cos(\omega' t + \phi) \]

is incorrect because the exponent should be \( -\frac{b}{2m} t \), not \( -\frac{b}{m} \).

Answer: The incorrect option is D.

Speed of Sound and RMS Velocity

Given:

• Speed of sound in an ideal gas at temperature \( T \) is \( v \).
• RMS speed of gas molecules at the same temperature is \( v_{rms} \).
• Ratios of these velocities for helium and oxygen are \( X \) and \( X' \) respectively.

Step 1: Expression for the Ratio

The ratio \( \frac{X}{X'} \) is given by:

\[ \frac{X}{X'} = \frac{5}{\sqrt{21}} \]

Answer: The correct option is A.

Answer 2

The speed of sound \( v \) in an ideal gas is related to the temperature \( T \) and the molar mass \( M \) by the formula: \[ v = \sqrt{\frac{\gamma RT}{M}} \] where: - \( \gamma \) is the adiabatic index (ratio of specific heats), - \( R \) is the universal gas constant, - \( T \) is the temperature, - \( M \) is the molar mass of the gas. The root-mean-square (rms) velocity of the gas molecules \( v_{\text{rms}} \) is given by: \[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \] where: - \( k \) is the Boltzmann constant, - \( m \) is the mass of one molecule of the gas. Now, for the gases in question: - For helium (a monatomic gas), \( \gamma = \frac{5}{3} \) and \( M_{\text{He}} \) is the molar mass of helium, - For oxygen (a diatomic gas), \( \gamma = \frac{7}{5} \) and \( M_{\text{O}_2} \) is the molar mass of oxygen. The ratio \( \frac{X}{X'} \), which is the ratio of the velocities, simplifies as follows: \[ \frac{v_{\text{He}}}{v_{\text{O}_2}} = \frac{\sqrt{\frac{\gamma_{\text{He}} RT}{M_{\text{He}}}}}{\sqrt{\frac{\gamma_{\text{O}_2} RT}{M_{\text{O}_2}}}} = \sqrt{\frac{\gamma_{\text{He}} M_{\text{O}_2}}{\gamma_{\text{O}_2} M_{\text{He}}}} \] Substituting the values: - \( \gamma_{\text{He}} = \frac{5}{3} \), - \( \gamma_{\text{O}_2} = \frac{7}{5} \), - \( M_{\text{He}} = 4 \, \text{g/mol} \), - \( M_{\text{O}_2} = 32 \, \text{g/mol} \), we get: \[ \frac{X}{X'} = \sqrt{\frac{\frac{5}{3} \times 32}{\frac{7}{5} \times 4}} = \sqrt{\frac{5 \times 32}{3 \times 7 \times 4}} = \sqrt{\frac{160}{84}} = \sqrt{\frac{40}{21}} = \frac{\sqrt{40}}{\sqrt{21}} = \frac{21}{5} \] Thus, the correct answer is (C).