A mixture of two ideal gases A and B is in thermal equilibrium at 600 K.A molecule of A has one-fourth the mass of a molecule of B and the rms speed of molecules of A is 400 m/s. Determine the rms speed of molecules of B.

Options

A. `100 m//s`
B. `200 m//s`
C. `400 m//s`
D. `800 m//s`

Answer

To determine the root mean square (rms) speed of molecules of gas B, we can follow these steps:

### Step 1: Understand the relationship between rms speed and molecular mass
The root mean square speed (Vrms) of an ideal gas is given by the formula:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas.

### Step 2: Set up the relationship between gases A and B
Given that:
- The rms speed of gas A (\( V_{A} \)) is 400 m/s,
- The mass of a molecule of gas A is one-fourth that of gas B.

Let \( M_B \) be the molar mass of gas B. Then, the molar mass of gas A (\( M_A \)) can be expressed as:
\[ M_A = \frac{1}{4} M_B \]

### Step 3: Write the rms speed equations for gases A and B
For gas A:
\[ V_A = \sqrt{\frac{3RT}{M_A}} = 400 \, \text{m/s} \]
For gas B:
\[ V_B = \sqrt{\frac{3RT}{M_B}} \]

### Step 4: Relate the rms speeds of gases A and B
Since both gases are in thermal equilibrium at the same temperature \( T \), we can relate their rms speeds:
\[ \frac{V_A}{V_B} = \sqrt{\frac{M_B}{M_A}} \]

Substituting \( M_A = \frac{1}{4} M_B \):
\[ \frac{V_A}{V_B} = \sqrt{\frac{M_B}{\frac{1}{4} M_B}} = \sqrt{4} = 2 \]

### Step 5: Solve for \( V_B \)
Rearranging the equation gives:
\[ V_B = \frac{V_A}{2} \]
Substituting \( V_A = 400 \, \text{m/s} \):
\[ V_B = \frac{400}{2} = 200 \, \text{m/s} \]

### Conclusion
The rms speed of molecules of gas B is:
\[ \boxed{200 \, \text{m/s}} \]

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