3 moles of an ideal gas are contained within a cylinder by a frictionless piston and are initially at temperature T . The pressure of the gas remains constant while it is heated and its volume doubles . If R is molar gas constant , the work done by the gas in increasing its volume is

Options

A. `(3)/(2) RT` ln 2
B. `3 RT` ln 2
C. `(3)/(2)` RT
D. 3 RT

Answer

To solve the problem of finding the work done by the gas during the isobaric process, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Process**:
The problem states that the gas is heated at constant pressure while its volume doubles. This is an isobaric process.

2. **Work Done Formula**:
In an isobaric process, the work done (W) by the gas can be calculated using the formula:
\[ W = P \Delta V \]
where \( \Delta V \) is the change in volume.

3. **Determine Initial and Final Volumes**:
Let the initial volume of the gas be \( V \). Since the volume doubles, the final volume \( V_f \) will be:
\[ V_f = 2V \]
Therefore, the change in volume \( \Delta V \) is:
\[ \Delta V = V_f - V = 2V - V = V \]

4. **Calculate the Pressure**:
Using the ideal gas law, we can express the pressure of the gas. The ideal gas law is given by:
\[ PV = nRT \]
For 3 moles of gas at temperature \( T \):
\[ P \cdot V = 3RT \]
Thus, we can express the pressure \( P \) as:
\[ P = \frac{3RT}{V} \]

5. **Substitute into Work Formula**:
Now we can substitute the expression for pressure into the work done formula:
\[ W = P \Delta V = P \cdot V \]
Substituting \( P \):
\[ W = \left(\frac{3RT}{V}\right) \cdot V \]
Simplifying this gives:
\[ W = 3RT \]

6. **Final Answer**:
Therefore, the work done by the gas in increasing its volume is:
\[ W = 3RT \]