The ammount of heat required to raise the temperature of 1 mole of diatomic gas by `1^(@)C` at constant pressure is `60cal` . The amount of heat which goes as internal energy of the gas is nearly.

Options

A. 60 cal
B. 30 cal
C. 42.8 cal
D. 49.8 cal

Explanation

To solve the problem, we need to determine the amount of heat that goes into the internal energy of a diatomic gas when the temperature is raised by \(1^\circ C\) at constant pressure.

### Step-by-Step Solution:

1. **Identify Given Information**:
- The heat required to raise the temperature of 1 mole of diatomic gas by \(1^\circ C\) at constant pressure (\(C_p\)) is given as 60 cal.

2. **Understanding Heat Capacities**:
- For a diatomic gas, the relationship between heat capacities at constant pressure (\(C_p\)) and constant volume (\(C_v\)) is given by:
\[ \frac{C_p}{C_v} = \gamma \]
- For diatomic gases, \(\gamma\) (the heat capacity ratio) is approximately 1.4.

3. **Calculate \(C_v\)**:
- Rearranging the formula gives us:
\[ C_v = \frac{C_p}{\gamma} \]
- Substituting the values we have:
\[ C_v = \frac{60 \text{ cal}}{1.4} \approx 42.857 \text{ cal} \]
- Rounding this, we can say \(C_v \approx 42.8 \text{ cal}\).

4. **Calculate Change in Internal Energy (\( \Delta U \))**:
- The change in internal energy for 1 mole of gas when the temperature changes by \(dT\) is given by:
\[ \Delta U = n C_v \Delta T \]
- Here, \(n = 1\) (1 mole), \(C_v \approx 42.8 \text{ cal}\), and \(\Delta T = 1^\circ C\).
- Thus:
\[ \Delta U = 1 \times 42.8 \text{ cal} \times 1 = 42.8 \text{ cal} \]

5. **Final Answer**:
- The amount of heat that goes into the internal energy of the gas is approximately **42.8 cal**.