The temperature at which the root mean square speed of `SO_2`molecule is same as that of methane at `27^@` C.

Options

A. 600 K
B. 900 K
C. 1200 K
D. 400 K

Answer

To find the temperature at which the root mean square speed of sulfur dioxide (SO₂) is the same as that of methane (CH₄) at 27°C, we can follow these steps:

### Step 1: Understand the Root Mean Square Speed Formula
The root mean square speed (Vrms) of a gas is given by the formula:
\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin
- \( M \) = molar mass of the gas in kg/mol

### Step 2: Set Up the Equation
Since we want the root mean square speeds of SO₂ and CH₄ to be equal, we can set up the equation:
\[ V_{rms, SO_2} = V_{rms, CH_4} \]
This leads to:
\[ \sqrt{\frac{3RT_1}{M_1}} = \sqrt{\frac{3RT_2}{M_2}} \]
Squaring both sides, we get:
\[ \frac{3RT_1}{M_1} = \frac{3RT_2}{M_2} \]
The \( 3R \) cancels out, simplifying to:
\[ \frac{T_1}{M_1} = \frac{T_2}{M_2} \]

### Step 3: Identify Molar Masses
Next, we need to identify the molar masses:
- Molar mass of SO₂ (M₁) = 64 g/mol = 0.064 kg/mol
- Molar mass of CH₄ (M₂) = 16 g/mol = 0.016 kg/mol

### Step 4: Convert Temperature to Kelvin
The temperature of methane (T₂) is given as 27°C. To convert this to Kelvin:
\[ T_2 = 27 + 273 = 300 \text{ K} \]

### Step 5: Substitute Values into the Equation
Now we can substitute the values into our equation:
\[ \frac{T_1}{0.064} = \frac{300}{0.016} \]

### Step 6: Solve for T₁
Cross-multiplying gives:
\[ T_1 \cdot 0.016 = 300 \cdot 0.064 \]
Calculating the right side:
\[ T_1 \cdot 0.016 = 19.2 \]
Now, solve for \( T_1 \):
\[ T_1 = \frac{19.2}{0.016} = 1200 \text{ K} \]

### Final Answer
The temperature at which the root mean square speed of sulfur dioxide is the same as that of methane at 27°C is **1200 K**.

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