For hydrogen gas `C_(p) - C_(v) = a` and for oxygen gas `C_(p) - C_(v) = b`. So, the relation between a and b is given by

Options

A. `a = 16` b
B. `16 a = b `
C. `a = 4 b`
D. `a = b `

Explanation

To solve the problem, we need to find the relationship between \( a \) and \( b \), where \( a = C_p - C_v \) for hydrogen gas and \( b = C_p - C_v \) for oxygen gas.

### Step-by-Step Solution:

1. **Understand the relationship**: The difference between specific heats at constant pressure and volume for any ideal gas can be expressed as:
\[ C_p - C_v = \frac{R}{M} \]
where \( R \) is the universal gas constant and \( M \) is the molar mass of the gas.

2. **Apply the formula for hydrogen**: For hydrogen gas, we denote its molar mass as \( M_H \). Thus, we can write:
\[ a = C_p - C_v = \frac{R}{M_H} \]
The molar mass of hydrogen \( M_H \) is approximately 2 g/mol. Therefore:
\[ a = \frac{R}{2} \]
(Equation 1)

3. **Apply the formula for oxygen**: For oxygen gas, we denote its molar mass as \( M_O \). The molar mass of oxygen \( M_O \) is approximately 32 g/mol. Thus, we can write:
\[ b = C_p - C_v = \frac{R}{M_O} = \frac{R}{32} \]
(Equation 2)

4. **Find the relationship between \( a \) and \( b \)**: Now, we can find the relationship between \( a \) and \( b \) by dividing Equation 1 by Equation 2:
\[ \frac{a}{b} = \frac{\frac{R}{2}}{\frac{R}{32}} = \frac{R}{2} \cdot \frac{32}{R} \]
The \( R \) cancels out:
\[ \frac{a}{b} = \frac{32}{2} = 16 \]

5. **Express \( a \) in terms of \( b \)**: Rearranging gives us:
\[ a = 16b \]

### Final Result:
The relation between \( a \) and \( b \) is:
\[ a = 16b \]