Question:

Two monoatomic ideal gases '1' and '2' of molecular masses \(m_1\) and \(m_2\) respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas '1' to that in gas '2' is

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Speed of sound in a gas is proportional to root of T over molar mass.
Updated On: Oct 1, 2026
  • \(\sqrt{\frac{m_1}{m_2}}\)
  • \(\sqrt{\frac{m_2}{m_1}}\)
  • \(\frac{m_1}{m_2}\)
  • \(\frac{m_2}{m_1}\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept
The speed of sound in an ideal gas is \(v=\sqrt{\dfrac{\gamma RT}{M}}\). Both gases are monatomic, so \(\gamma=\dfrac53\) for each. The temperature is the same.

Step 2: Ratio
\[ \frac{v_1}{v_2}=\sqrt{\frac{M_2}{M_1}}=\sqrt{\frac{m_2}{m_1}} \]

Step 3: Meaning
A heavier gas has a smaller speed of sound. Gas 1 with larger molecular mass \(m_1\) is slower, so the ratio has \(m_2\) on top.

Step 4: Check the options
Option (A) is the inverse. Options (C) and (D) lack the square root. So the answer is (B).

Final Answer:
At equal temperature the speed varies as one over root of molecular mass, so the ratio is root of m2 over m1, option (B). \[ \boxed{\sqrt{\frac{m_2}{m_1}}} \]
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