Two moles of a gas expand reversibly and isothermally at temperature of 300K. Initial volume of the gas is 1 L while the final pressure is 4.926 atm . The work done by gas is

Options

A. `-11488.28J`
B. `-15036.28J`
C. `-22488.28 J`
D. `-33488.28J`

Answer

To solve the problem of calculating the work done by the gas during its isothermal and reversible expansion, we can follow these steps:

### Step 1: Identify the Given Data
- Number of moles (n) = 2 moles
- Temperature (T) = 300 K
- Initial volume (V_i) = 1 L
- Final pressure (P_f) = 4.926 atm

### Step 2: Calculate Initial Pressure (P_i)
Using the ideal gas equation \( PV = nRT \), we can find the initial pressure \( P_i \).

\[ P_i = \frac{nRT}{V_i} \]

Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \) (ideal gas constant)

Substituting the known values:

\[ P_i = \frac{2 \, \text{mol} \times 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 300 \, \text{K}}{1 \, \text{L}} \]

Calculating this gives:

\[ P_i = \frac{49.26 \, \text{atm}}{1} = 49.26 \, \text{atm} \]

### Step 3: Calculate Work Done (W)
The work done by the gas during an isothermal expansion can be calculated using the formula:

\[ W = -nRT \ln\left(\frac{V_f}{V_i}\right) \]

However, we need to express \( \frac{V_f}{V_i} \) in terms of pressures. From the ideal gas law, we know:

\[ P_i V_i = P_f V_f \implies \frac{V_f}{V_i} = \frac{P_i}{P_f} \]

Substituting the values:

\[ \frac{V_f}{V_i} = \frac{49.26 \, \text{atm}}{4.926 \, \text{atm}} \approx 10 \]

### Step 4: Substitute Values into the Work Done Formula
Now substituting back into the work done formula:

\[ W = -nRT \ln\left(\frac{P_i}{P_f}\right) \]

Calculating \( \ln(10) \):

\[ W = -2 \times 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 300 \, \text{K} \times \ln(10) \]

Using \( \ln(10) \approx 2.303 \):

\[ W = -2 \times 0.0821 \times 300 \times 2.303 \]

Calculating this gives:

\[ W \approx -2 \times 0.0821 \times 300 \times 2.303 \approx -11,488.28 \, \text{J} \]

### Final Answer
The work done by the gas is approximately:

\[ W \approx -11,488.28 \, \text{J} \]