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, we need to find the work done by the gas when it expands at constant pressure and doubles its volume. Here’s a step-by-step solution:

### Step 1: Understand the Work Done Formula
The work done \( W \) by a gas during an isobaric (constant pressure) process is given by the formula:
\[ W = P \Delta V \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.

### Step 2: Determine the Change in Volume
Let the initial volume of the gas be \( V \). Since the volume doubles, the final volume \( V_f \) 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
From the ideal gas law, we know:
\[ PV = nRT \]
where \( n \) is the number of moles, \( R \) is the molar gas constant, and \( T \) is the temperature.

### Step 4: Calculate the Initial Pressure
Using the ideal gas law for the initial state:
\[ P = \frac{nRT}{V} \]
Substituting \( n = 3 \) moles and \( T = T \):
\[ P = \frac{3RT}{V} \]

### Step 5: Substitute into the Work Done Formula
Now, substituting \( P \) and \( \Delta V \) into the work done formula:
\[ W = P \Delta V = \left(\frac{3RT}{V}\right) \cdot V \]
The volume \( V \) cancels out:
\[ W = 3RT \]

### Final Answer
Thus, the work done by the gas in increasing its volume is:
\[ \boxed{3RT} \]