The root mean square velocity of hydrogen molecule at `27^(@)C' is 'upsilon_(H)` and that of oxygen at `402^(@)C is upsilon_(0)`, then

Options

A. a.`v_0gtv_H`
B. b.`4v_0=9v_H`
C. c.`2v_0=3v_H`
D. d.`9v_0=134V_H`

Answer

To solve the problem, we need to find the relationship between the root mean square velocities of hydrogen and oxygen molecules at given temperatures. The formula for root mean square velocity (v) of a gas is given by:

\[ v = \sqrt{\frac{3kT}{m}} \]

where \(k\) is the Boltzmann constant, \(T\) is the absolute temperature in Kelvin, and \(m\) is the mass of the gas molecule.

1. **Convert the temperatures from Celsius to Kelvin**:
- For hydrogen at \(27^\circ C\):
\[ T_H = 27 + 273 = 300 \text{ K} \]
- For oxygen at \(402^\circ C\):
\[ T_O = 402 + 273 = 675 \text{ K} \]

2. **Write the expression for the root mean square velocities**:
- The root mean square velocity of hydrogen (\(v_H\)) and oxygen (\(v_O\)) can be expressed as:
\[ v_H = \sqrt{\frac{3kT_H}{m_H}} \quad \text{and} \quad v_O = \sqrt{\frac{3kT_O}{m_O}} \]

3. **Find the ratio of the root mean square velocities**:
- The ratio \( \frac{v_H}{v_O} \) can be expressed as:
\[ \frac{v_H}{v_O} = \frac{\sqrt{\frac{3kT_H}{m_H}}}{\sqrt{\frac{3kT_O}{m_O}}} = \sqrt{\frac{T_H \cdot m_O}{T_O \cdot m_H}} \]

4. **Substitute the known values**:
- For hydrogen, the molar mass \(m_H\) is approximately \(2 \text{ g/mol}\) and for oxygen \(m_O\) is approximately \(32 \text{ g/mol}\):
\[ \frac{v_H}{v_O} = \sqrt{\frac{300 \cdot 32}{675 \cdot 2}} \]

5. **Calculate the values**:
- Simplifying the expression:
\[ \frac{v_H}{v_O} = \sqrt{\frac{9600}{1350}} = \sqrt{\frac{64}{9}} = \frac{8}{3} \]

6. **Rearranging the equation**:
- We can express this as:
\[ 3v_H = 8v_O \]

7. **Final relation**:
- Thus, we have the relation between the root mean square velocities of hydrogen and oxygen:
\[ 3v_H = 8v_O \]