Question

Two moles of Helium are mixed with \(n\) moles of Hydrogen. The rms speed of the gas molecules in the mixture is \(\sqrt{2}\) times the speed of sound in the mixture. The value of \(n\) is

• A. \(1\)
• B. \(3\)
• C. \(2\)
• D. \(\dfrac{3}{2}\) \bigskip

Hint:

For any ideal gas, \[ \frac{v_{\text{rms}}}{v_s} = \sqrt{\frac{3}{\gamma}}. \] If \(v_{\text{rms}}=\sqrt{2}\,v_s\), then \[ \gamma=\frac{3}{2}. \] For gas mixtures, first calculate the total \(C_P\) and \(C_V\), then use \[ \gamma=\frac{C_P}{C_V}. \]

Answer 1

The Correct Option is C

Concept: For an ideal gas, \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] and the speed of sound is \[ v_s = \sqrt{\frac{\gamma RT}{M}}. \] Hence, \[ \frac{v_{\text{rms}}}{v_s} = \sqrt{\frac{3}{\gamma}}. \]

Step 1: Use the given condition. Given, \[ v_{\text{rms}} = \sqrt{2}\,v_s. \] Therefore, \[ \sqrt{\frac{3}{\gamma}} = \sqrt{2}. \] Squaring both sides, \[ \frac{3}{\gamma}=2. \] \[ \gamma=\frac{3}{2}. \]

Step 2: Find the effective \(\gamma\) of the mixture. For Helium (monoatomic), \[ C_{V,\text{He}}=\frac{3}{2}R, \qquad C_{P,\text{He}}=\frac{5}{2}R. \] For Hydrogen (diatomic), \[ C_{V,\text{H}_2}=\frac{5}{2}R, \qquad C_{P,\text{H}_2}=\frac{7}{2}R. \] For the mixture, \[ C_P = 2\left(\frac{5}{2}R\right) + n\left(\frac{7}{2}R\right) = \frac{10+7n}{2}R. \] \[ C_V = 2\left(\frac{3}{2}R\right) + n\left(\frac{5}{2}R\right) = \frac{6+5n}{2}R. \] Thus, \[ \gamma = \frac{C_P}{C_V} = \frac{10+7n}{6+5n}. \]

Step 3: Equate \(\gamma\) to \(\dfrac{3}{2}\). \[ \frac{10+7n}{6+5n} = \frac{3}{2}. \] \[ 2(10+7n) = 3(6+5n). \] \[ 20+14n = 18+15n. \] \[ n=2. \]

Step 4: Write the final answer. \[ \boxed{n=2} \] \[ \boxed{\text{Answer = (C)}} \]