For the specific heat of 1 mole of an ideal gas at constant pressure `(C_p)` and at constant volume `(C_v)` which is correct

Options

A. `C_P` of hydrogen gas is `5/2` R
B. `C_V` of hydrogen gas is `7/2R`
C. `H_2` has very small values of `C_P and C_V`
D. `C_P-C_V=1.99` cal/mol-K for `H_2`

Explanation

To solve the question regarding the specific heat of 1 mole of an ideal gas at constant pressure \( C_p \) and at constant volume \( C_v \), we will analyze the properties of a diatomic ideal gas, such as hydrogen.

### Step-by-Step Solution:

1. **Identify the Degrees of Freedom**:
For a diatomic gas, the degrees of freedom \( f \) is given as 5 (3 translational and 2 rotational).

2. **Calculate \( C_v \)**:
The formula for the specific heat at constant volume is:
\[ C_v = \frac{f}{2} R \]
Substituting \( f = 5 \):
\[ C_v = \frac{5}{2} R \]

3. **Calculate \( C_p \)**:
The formula for the specific heat at constant pressure is:
\[ C_p = C_v + R \]
Substituting the value of \( C_v \):
\[ C_p = \frac{5}{2} R + R = \frac{5}{2} R + \frac{2}{2} R = \frac{7}{2} R \]

4. **Calculate \( C_p - C_v \)**:
Now, we find the difference between \( C_p \) and \( C_v \):
\[ C_p - C_v = \left(\frac{7}{2} R\right) - \left(\frac{5}{2} R\right) = \frac{2}{2} R = R \]

5. **Convert \( R \) to Calories**:
The ideal gas constant \( R \) is approximately \( 8.31 \, \text{J/mol K} \). To convert this to calories:
\[ 1 \, \text{cal} = 4.18 \, \text{J} \implies R = \frac{8.31}{4.18} \, \text{cal/mol K} \approx 1.99 \, \text{cal/mol K} \]

6. **Evaluate the Options**:
- \( C_p \) for hydrogen gas is \( \frac{7}{2} R \) (Correct).
- \( C_v \) for hydrogen gas is \( \frac{5}{2} R \) (Correct).
- The difference \( C_p - C_v = R \approx 1.99 \, \text{cal/mol K} \) (Correct).
- The option stating various small values for \( C_p \) and \( C_v \) is incorrect.

### Final Conclusion:
The correct statements regarding the specific heats \( C_p \) and \( C_v \) for an ideal diatomic gas like hydrogen are:
- \( C_p = \frac{7}{2} R \)
- \( C_v = \frac{5}{2} R \)
- \( C_p - C_v = R \approx 1.99 \, \text{cal/mol K} \)