The rms speed of oxygen at room temperature is about 500m/s. The rms speed of hydrogen at the same temperature is about

Options

A. (a)`125ms^(-1)`
B. (b)`2000ms^(-1)`
C. (c)`8000ms^(-1)`
D. (d)`31ms^(-1)`

Answer

To find the RMS speed of hydrogen at room temperature, we can use the relationship between the RMS speeds of two gases at the same temperature, which is derived from the kinetic theory of gases.

### Step-by-Step Solution:

1. **Identify Given Data:**
- RMS speed of oxygen (v1) = 500 m/s
- Molecular weight of oxygen (M1) = 32 g/mol
- Molecular weight of hydrogen (M2) = 2 g/mol

2. **Understand the Relationship:**
The RMS speed (v) of a gas is related to its molecular weight (M) and temperature (T) by the formula:
\[ v = \sqrt{\frac{3kT}{m}} \]
where \( k \) is the Boltzmann constant and \( m \) is the mass of a molecule. At the same temperature, the kinetic energy of the gases is the same.

3. **Set Up the Equation:**
Since the temperature is the same for both gases, we can equate their kinetic energies:
\[ \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 v_2^2 \]
Here, \( m_1 \) and \( m_2 \) are the masses of one mole of oxygen and hydrogen, respectively.

4. **Relate the Masses:**
The mass of one mole of a gas can be expressed in terms of its molecular weight:
\[ m_1 = M_1 \quad \text{and} \quad m_2 = M_2 \]
Thus, we can express the relationship as:
\[ M_1 v_1^2 = M_2 v_2^2 \]

5. **Substitute Known Values:**
Substitute \( M_1 = 32 \) g/mol, \( M_2 = 2 \) g/mol, and \( v_1 = 500 \) m/s into the equation:
\[ 32 \cdot (500)^2 = 2 \cdot v_2^2 \]

6. **Solve for \( v_2^2 \):**
\[ 32 \cdot 250000 = 2 \cdot v_2^2 \]
\[ 8000000 = 2 \cdot v_2^2 \]
\[ v_2^2 = \frac{8000000}{2} = 4000000 \]

7. **Calculate \( v_2 \):**
\[ v_2 = \sqrt{4000000} = 2000 \text{ m/s} \]

### Conclusion:
The RMS speed of hydrogen at room temperature is approximately **2000 m/s**.

---