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 finding the work done by the gas during a reversible isothermal expansion, we can follow these steps:

### Step 1: Understand the Given Information
- Number of moles of gas (n) = 2 moles
- Temperature (T) = 300 K
- Initial volume (Vi) = 1 L
- Final pressure (Pf) = 4.926 atm

### Step 2: Calculate the Initial Pressure (Pi)
Using the ideal gas law, \( PV = nRT \):
- We need to find the initial pressure (Pi). We can rearrange the ideal gas equation to find Pi:
\[ Pi = \frac{nRT}{Vi} \]
Where:
- R (ideal gas constant) = 0.0821 L·atm/(K·mol)
- Substitute the values:
\[ Pi = \frac{(2 \text{ moles}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (300 \text{ K})}{1 \text{ L}} \]
Calculating this gives:
\[ Pi = \frac{49.26 \text{ L·atm}}{1 \text{ L}} = 49.26 \text{ atm} \]

### Step 3: Calculate the Work Done (W)
For a reversible isothermal process, the work done by the gas can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{Vf}{Vi}\right) \]
However, since we have pressures instead of volumes, we can also express it as:
\[ W = -nRT \ln\left(\frac{Pi}{Pf}\right) \]
Substituting the known values:
- \( n = 2 \)
- \( R = 8.314 \text{ J/(K·mol)} \) (convert R to J for consistency)
- \( T = 300 \text{ K} \)
- \( Pi = 49.26 \text{ atm} \)
- \( Pf = 4.926 \text{ atm} \)

### Step 4: Calculate the Natural Logarithm
Calculate the ratio:
\[ \frac{Pi}{Pf} = \frac{49.26}{4.926} \approx 10 \]
Then calculate the natural logarithm:
\[ \ln(10) \approx 2.303 \]

### Step 5: Substitute Values into the Work Formula
Now substitute all values into the work formula:
\[ W = -2 \times 8.314 \text{ J/(K·mol)} \times 300 \text{ K} \times 2.303 \]
Calculating this gives:
\[ W \approx -2 \times 8.314 \times 300 \times 2.303 \approx -11,488.285 \text{ J} \]

### Final Answer
The work done by the gas is approximately:
\[ W \approx -11,488.29 \text{ J} \]