For hydrogen gas `C_p-C_v=a` and for oxygen gas `C_p-C_v=b`,`C_p` and `C_v` being molar specific heats. The relation between a and b is

Options

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

Explanation

To solve the problem, we need to establish the relationship between the specific heats \(C_p\) and \(C_v\) for hydrogen and oxygen gases. The specific heat difference is given as \(C_p - C_v = a\) for hydrogen and \(C_p - C_v = b\) for oxygen.

### Step-by-Step Solution:

1. **Understand the relationship for any ideal gas:**
The difference between the molar specific heats at constant pressure (\(C_p\)) and constant volume (\(C_v\)) 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 relationship to hydrogen gas:**
For hydrogen gas, the molar mass \(M_H\) is approximately 2 g/mol. Therefore, we can write:
\[ C_p - C_v = \frac{R}{M_H} = \frac{R}{2} \]
According to the problem, this is equal to \(a\):
\[ a = \frac{R}{2} \]

3. **Apply the relationship to oxygen gas:**
For oxygen gas, the molar mass \(M_O\) is approximately 32 g/mol. Thus, we have:
\[ C_p - C_v = \frac{R}{M_O} = \frac{R}{32} \]
According to the problem, this is equal to \(b\):
\[ b = \frac{R}{32} \]

4. **Relate \(a\) and \(b\):**
Now, we can relate \(a\) and \(b\) using the expressions we derived:
\[ a = \frac{R}{2} \quad \text{and} \quad b = \frac{R}{32} \]
To find the relationship between \(a\) and \(b\), we can express \(a\) in terms of \(b\):
\[ a = 16b \]
This is because:
\[ \frac{R}{2} = 16 \cdot \frac{R}{32} \]

### Final Relationship:
Thus, the relationship between \(a\) and \(b\) is:
\[ a = 16b \]