At what temperature will hydrogen molecules have the same root mean square speed as nitrogen molecules at `27^(@)C` ?

Options

A. `21.43^(@)C`
B. `42.86K`
C. `21.43K`
D. `42.86^(@)C`

Explanation

To solve the problem of finding the temperature at which hydrogen molecules have the same root mean square speed as nitrogen molecules at 27°C, we can follow these steps:

### Step 1: Convert the temperature of nitrogen from Celsius to Kelvin
The temperature of nitrogen is given as 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
So,
\[ T(N_2) = 27 + 273 = 300 \, K \]

### Step 2: Write the formula for root mean square speed (rms speed)
The root mean square speed (v_rms) of a gas is given by the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( M \) is the molar mass of the gas in kg/mol.

### Step 3: Calculate the rms speed of nitrogen
For nitrogen (\( N_2 \)):
- Molar mass (\( M_{N_2} \)) = 28 g/mol = 0.028 kg/mol
- Temperature (\( T_{N_2} \)) = 300 K

Using the formula:
\[ v_{rms(N_2)} = \sqrt{\frac{3R \cdot 300}{0.028}} \]

### Step 4: Write the expression for the rms speed of hydrogen
For hydrogen (\( H_2 \)):
- Molar mass (\( M_{H_2} \)) = 2 g/mol = 0.002 kg/mol
- Let the unknown temperature be \( T_{H_2} \).

Using the formula:
\[ v_{rms(H_2)} = \sqrt{\frac{3RT_{H_2}}{0.002}} \]

### Step 5: Set the rms speeds equal to each other
Since we want the rms speeds of hydrogen and nitrogen to be equal:
\[ v_{rms(N_2)} = v_{rms(H_2)} \]
Thus,
\[ \sqrt{\frac{3R \cdot 300}{0.028}} = \sqrt{\frac{3RT_{H_2}}{0.002}} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{3R \cdot 300}{0.028} = \frac{3RT_{H_2}}{0.002} \]

### Step 7: Cancel out common terms
Since \( 3R \) is common on both sides, we can cancel it:
\[ \frac{300}{0.028} = \frac{T_{H_2}}{0.002} \]

### Step 8: Solve for \( T_{H_2} \)
Rearranging gives:
\[ T_{H_2} = 300 \cdot \frac{0.002}{0.028} \]

Calculating this:
\[ T_{H_2} = 300 \cdot \frac{2}{28} = 300 \cdot \frac{1}{14} = \frac{300}{14} \approx 21.43 \, K \]

### Final Answer
The temperature at which hydrogen molecules will have the same root mean square speed as nitrogen molecules at 27°C is approximately **21.43 K**.

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