A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole oxygen (both considered as ideal). Let `f_(1)(v) dv,` denote the fraction of molecules with speed between v and (v+ dv) with `f_(2)(v)dv`, similarly for oxygen . Then ,

Options

A. `f_(1)` (v) + `f_(2)` (v) = obeys the Maxwell's distribution law
B. `f_(1)(v),f_(2)(v)` will obey the Maxwell's disribution law separtely
C. Neither `f_(1)` (v) nor `f_(2)` (v) will obey the Maxwell's distribution law
D. `f_(2)` (v) and `f_(1)` (v) will be the same

Explanation

To solve the problem, we need to analyze the behavior of the molecules in a mixture of hydrogen and oxygen gases, both of which are considered ideal gases. The question asks us to determine how the speed distribution functions for hydrogen and oxygen relate to the Maxwell-Boltzmann distribution law.

### Step-by-Step Solution:

1. **Understanding the Maxwell-Boltzmann Distribution**:
The Maxwell-Boltzmann distribution describes the distribution of speeds of particles in an ideal gas. It is given by the formula:
\[ f(v) = \left( \frac{m}{2 \pi k T} \right)^{3/2} 4 \pi v^2 e^{-\frac{mv^2}{2kT}} \]
where \( m \) is the mass of the gas molecules, \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( v \) is the speed of the particles.

2. **Identifying the Gases**:
In this problem, we have two different gases: hydrogen (H₂) and oxygen (O₂). The molecular masses of these gases are different:
- Mass of hydrogen molecule (H₂) is approximately \( 2 \times 1.008 \, \text{g/mol} \) or \( 0.002 \, \text{kg/mol} \).
- Mass of oxygen molecule (O₂) is approximately \( 2 \times 16.00 \, \text{g/mol} \) or \( 0.032 \, \text{kg/mol} \).

3. **Speed Distribution Functions**:
Since both gases are ideal, we can write their respective speed distribution functions:
- For hydrogen: \( f_1(v) \)
- For oxygen: \( f_2(v) \)

4. **Applying the Maxwell-Boltzmann Distribution**:
Each gas will follow the Maxwell-Boltzmann distribution independently because they are different species with different masses. Therefore:
- \( f_1(v) \) will obey the Maxwell distribution law for hydrogen.
- \( f_2(v) \) will obey the Maxwell distribution law for oxygen.

5. **Conclusion**:
Since the two gases have different molecular masses, their speed distribution functions will not be the same. However, each will independently follow the Maxwell-Boltzmann distribution law. Thus, the correct conclusion is:
- **\( f_1(v) \) and \( f_2(v) \) obey the Maxwell distribution law separately.**

### Final Answer:
**The correct statement is that \( f_1(v) \) and \( f_2(v) \) will obey the Maxwell distribution law separately.**