Question

Two monoatomic ideal gases \(A\) and \(B\) of molecular masses \(m_1\) and \(m_2\) respectively, are enclosed in separate containers kept at the same temperature. The ratio of speed of sound in gas \(A\) to that in gas \(B\) is given by

• A. \( \dfrac{m_1}{m_2} \)
• B. \( \sqrt{\dfrac{m_2}{m_1}} \)
• C. \( \sqrt{\dfrac{m_1}{m_2}} \)
• D. \( \dfrac{m_2}{m_1} \)

Hint:

At constant temperature, speed of sound in a gas is inversely proportional to the square root of its molecular mass.

Answer 1

The Correct Option is B

Step 1: Formula for speed of sound in a gas.
Speed of sound in an ideal gas is given by:
\[ v = \sqrt{\frac{\gamma RT}{M}} \] where \(M\) is the molecular mass.
Step 2: Same temperature and same type of gas.
Since both gases are monoatomic and at the same temperature, \(\gamma\), \(R\), and \(T\) are the same for both gases.
Step 3: Ratio of speeds of sound.
\[ \frac{v_A}{v_B} = \sqrt{\frac{M_B}{M_A}} \]
Step 4: Substituting given molecular masses.
\[ \frac{v_A}{v_B} = \sqrt{\frac{m_2}{m_1}} \]
Step 5: Conclusion.
The required ratio is \( \sqrt{\dfrac{m_2}{m_1}} \).