The `r.m.s`. velocity of hydrogen at `27^(@)C, R= 8.314 J mol^(-1)K^(-1)` is:

Options

A. `1.934 m//s`
B. `19.34 m//s`
C. `193.4 m//s`
D. `1934 m//s`

Explanation

To find the root mean square (r.m.s) velocity of hydrogen at 27°C, we can use the formula:

\[ V_{rms} = \sqrt{\frac{3RT}{M}} \]

where:
- \( V_{rms} \) = root mean square velocity
- \( R \) = universal gas constant = 8.314 J mol\(^{-1}\) K\(^{-1}\)
- \( T \) = temperature in Kelvin
- \( M \) = molar mass of the gas in kg/mol

### Step 1: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
For 27°C:
\[ T = 27 + 273.15 = 300.15 \text{ K} \approx 300 \text{ K} \]

### Step 2: Determine the molar mass of hydrogen
The molar mass of hydrogen (H₂) is approximately 2 g/mol. To use it in our calculations, we need to convert it to kg/mol:
\[ M = 2 \text{ g/mol} = 2 \times 10^{-3} \text{ kg/mol} \]

### Step 3: Substitute the values into the r.m.s velocity formula
Now we can substitute the values of \( R \), \( T \), and \( M \) into the formula:
\[ V_{rms} = \sqrt{\frac{3 \times 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \times 300 \text{ K}}{2 \times 10^{-3} \text{ kg/mol}}} \]

### Step 4: Calculate the numerator
Calculate the product of \( 3 \), \( R \), and \( T \):
\[ 3 \times 8.314 \times 300 = 7482.6 \text{ J/mol} \]

### Step 5: Calculate the division
Now divide by the molar mass:
\[ \frac{7482.6}{2 \times 10^{-3}} = 3741300 \text{ m}^2/\text{s}^2 \]

### Step 6: Take the square root
Now take the square root to find \( V_{rms} \):
\[ V_{rms} = \sqrt{3741300} \approx 1935.2 \text{ m/s} \]

### Final Answer
The r.m.s velocity of hydrogen at 27°C is approximately:
\[ V_{rms} \approx 1935 \text{ m/s} \]