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. 10 cal
D. 160 cal

Answer

To solve the problem of calculating the work done by 0.2 moles of a gas at room temperature to double its volume during an isobaric process, we can follow these steps:

### Step 1: Understand the Isobaric Process
In an isobaric process, the pressure (P) remains constant while the volume (V) changes. The work done (W) by the gas during this process can be calculated using the formula:

\[ W = P \Delta V \]

where \( \Delta V \) is the change in volume.

### Step 2: Define Initial and Final Volume
Let the initial volume be \( V \). Since the volume is doubled, the final volume will be:

\[ V_f = 2V \]

Thus, the change in volume \( \Delta V \) is:

\[ \Delta V = V_f - V = 2V - V = V \]

### Step 3: Use the Ideal Gas Law
We can relate pressure, volume, and temperature using the ideal gas law:

\[ PV = nRT \]

where:
- \( n \) = number of moles of gas (0.2 moles)
- \( R \) = gas constant (2 cal/mol·°C)
- \( T \) = temperature in °C (room temperature, which we take as 25°C)

### Step 4: Calculate the Work Done
From the ideal gas law, we can express pressure \( P \) as:

\[ P = \frac{nRT}{V} \]

Substituting this into the work done formula:

\[ W = P \Delta V = P \cdot V \]

Now substituting for \( P \):

\[ W = \left(\frac{nRT}{V}\right) \cdot V = nRT \]

Now substituting the values:

- \( n = 0.2 \) moles
- \( R = 2 \) cal/(mol·°C)
- \( T = 25 \) °C

Thus:

\[ W = 0.2 \times 2 \times 25 \]

Calculating this gives:

\[ W = 0.2 \times 50 = 10 \text{ cal} \]

### Conclusion
The work done by the gas to double its volume during the isobaric process is **10 calories**.