When 500 calories heat is given to the gas X in an isobaric process, its work done comes out as 142.8 calories . The gas X is

Options

A. `O_2`
B. `NH_3`
C. He
D. `SO_2`

Answer

To determine the identity of gas X, we can use the information provided about the heat transfer, work done, and the relationship between heat capacities in an isobaric process.

### Step-by-Step Solution:

1. **Understanding the Process**:
- The process is isobaric, meaning it occurs at constant pressure. In this case, the heat added to the system (gas X) is given as \( Q = 500 \) calories.

2. **Work Done**:
- The work done by the gas during this process is given as \( W = 142.8 \) calories.

3. **Applying the First Law of Thermodynamics**:
- According to the first law of thermodynamics, the change in internal energy (\( \Delta U \)) is given by:
\[ \Delta U = Q - W \]
- Substituting the values:
\[ \Delta U = 500 \, \text{calories} - 142.8 \, \text{calories} = 357.2 \, \text{calories} \]

4. **Relating Heat Capacities**:
- In an isobaric process, the relationship between the heat capacities is given by:
\[ \frac{C_p}{C_v} = \gamma \]
- Where \( C_p \) is the molar heat capacity at constant pressure and \( C_v \) is the molar heat capacity at constant volume.

5. **Using the Heat Transfer**:
- We know that:
\[ Q = n C_p \Delta T \]
- And for internal energy:
\[ \Delta U = n C_v \Delta T \]
- Thus, we can express the ratio of heat capacities as:
\[ \frac{C_p}{C_v} = \frac{Q}{\Delta U} = \frac{500}{357.2} \]

6. **Calculating the Ratio**:
- Performing the calculation:
\[ \frac{C_p}{C_v} = \frac{500}{357.2} \approx 1.4 \]

7. **Identifying the Type of Gas**:
- The value \( \gamma \approx 1.4 \) is characteristic of diatomic gases (like \( O_2 \), \( N_2 \), etc.).

8. **Conclusion**:
- Therefore, gas X is a diatomic gas. Given the context, it is likely \( O_2 \) (oxygen).

### Final Answer:
Gas X is \( O_2 \) (oxygen).

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