Two moles of an ideal gas is contained in a cylinder fitted with a frictionless movable piston, exposed to the atmosphere, at an initial temperature `T_(0)`. The gas is slowly heated so that its volume becomes four times the initial value. The work done by gas is

Options

A. zero
B. `2 RT_(0)`
C. `4 RT_(0)`
D. `6 RT_(0)`

Explanation

To solve the problem, we need to calculate the work done by the gas when it expands from an initial volume \( V_0 \) to a final volume \( V_2 = 4V_0 \). We will use the ideal gas law and the formula for work done during an expansion.

### Step-by-Step Solution:

1. **Identify Initial and Final Volumes:**
- Let the initial volume of the gas be \( V_0 \).
- The final volume after heating is \( V_2 = 4V_0 \).

2. **Calculate the Change in Volume:**
- The change in volume \( \Delta V \) is given by:
\[ \Delta V = V_2 - V_0 = 4V_0 - V_0 = 3V_0 \]

3. **Use the Ideal Gas Law to Find Initial Pressure:**
- The ideal gas law states:
\[ PV = nRT \]
- For the initial state:
\[ P_0 = \frac{nRT_0}{V_0} \]
- Given that \( n = 2 \) moles, we can substitute:
\[ P_0 = \frac{2RT_0}{V_0} \]

4. **Calculate Work Done by the Gas:**
- The work done \( W \) by the gas during expansion at constant pressure can be calculated using the formula:
\[ W = P \Delta V \]
- Substitute \( P_0 \) and \( \Delta V \):
\[ W = P_0 \cdot \Delta V = \left(\frac{2RT_0}{V_0}\right) \cdot (3V_0) \]
- Simplifying this gives:
\[ W = 2RT_0 \cdot 3 = 6RT_0 \]

5. **Final Result:**
- The work done by the gas is:
\[ W = 6RT_0 \]

### Conclusion:
The work done by the gas when it expands to four times its initial volume is \( 6RT_0 \).