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 solve the problem step by step, we will analyze the information given and apply the relevant thermodynamic principles.

### Step 1: Understand the Process
In an isobaric process (constant pressure), the heat added to the system (gas X) is related to the change in enthalpy (ΔH) and the work done (W) by the system.

### Step 2: Identify the Given Values
- Heat added (Q) = 500 calories
- Work done (W) = 142.8 calories

### Step 3: Apply the First Law of Thermodynamics
According to the first law of thermodynamics, the change in internal energy (ΔU) can be expressed as:
\[ \Delta U = Q - W \]
Substituting the values:
\[ \Delta U = 500 \text{ calories} - 142.8 \text{ calories} = 357.2 \text{ calories} \]

### Step 4: Relate ΔH and ΔU
In an isobaric process, the relationship between enthalpy (ΔH) and internal energy (ΔU) is given by:
\[ \Delta H = \Delta U + P \Delta V \]
Since we are at constant pressure, we can also express this in terms of heat capacities:
\[ \Delta H = nC_p \Delta T \quad \text{and} \quad \Delta U = nC_v \Delta T \]

### Step 5: Find the Ratio of Heat Capacities
The ratio of the heat capacities can be expressed as:
\[ \frac{C_p}{C_v} = \frac{\Delta H}{\Delta U} \]
Substituting the values we calculated:
\[ \frac{C_p}{C_v} = \frac{500 \text{ calories}}{357.2 \text{ calories}} \approx 1.4 \]

### Step 6: Determine the Type of Gas
The ratio \( \frac{C_p}{C_v} \) is related to the degrees of freedom of the gas. For a diatomic gas, this ratio is typically \( \frac{7}{5} = 1.4 \). Thus, gas X is likely a diatomic gas.

### Step 7: Identify the Gas
Among common diatomic gases, oxygen (O₂) is a well-known example. Therefore, the gas X is oxygen.

### Final Answer
The gas X is **Oxygen (O₂)**.

---