What is the work done by 0.2 mole of a gas at room temperature to double its volume during isobaric process? (Take R=2 cal `mol^(-1).^(@)C^(-1)`)

Options

A. 30 cal
B. 40 cal
C. 120 cal
D. 160 cal

Answer

To solve the problem of calculating the work done by 0.2 moles of a gas during an isobaric process where the volume doubles, we can follow these steps:

### Step 1: Understand the Isobaric Process
In an isobaric process, the pressure remains constant while the volume changes. We need to find the work done by the gas when its volume doubles.

### Step 2: Define Initial and Final Volumes
Let the initial volume be \( V \). Therefore, the final volume when it doubles will be:
\[ V_f = 2V \]

### Step 3: Calculate the Change in Volume
The change in volume (\( \Delta V \)) can be calculated as:
\[ \Delta V = V_f - V_i = 2V - V = V \]

### Step 4: Use the Work Done Formula for Isobaric Process
The work done (\( W \)) in an isobaric process is given by:
\[ W = P \Delta V \]
Substituting \( \Delta V \):
\[ W = P \cdot V \]

### Step 5: Apply the Ideal Gas Law
Using the ideal gas equation:
\[ PV = nRT \]
where:
- \( n \) = number of moles = 0.2 moles
- \( R \) = gas constant = 2 cal/mol·°C
- \( T \) = temperature in Kelvin = 300 K (room temperature)

### Step 6: Calculate \( PV \)
Substituting the values into the ideal gas equation:
\[ PV = nRT = 0.2 \, \text{mol} \times 2 \, \text{cal/mol·°C} \times 300 \, \text{K} \]
Calculating this gives:
\[ PV = 0.2 \times 2 \times 300 = 120 \, \text{calories} \]

### Step 7: Conclusion
Thus, the work done by the gas during the isobaric process is:
\[ W = 120 \, \text{calories} \]

### Final Answer
The work done by 0.2 moles of gas at room temperature to double its volume during an isobaric process is **120 calories**.

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