In an isothermal reversible expansion, if the volume of 96 gm of oxygen at `27^(@)C` is increased from 70 litres to 140 litres , then the work done by the gas will be

Answer

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

### Step 1: Identify the parameters
We are given:
- Mass of oxygen (m) = 96 g
- Initial volume (V_i) = 70 L = 0.07 m³
- Final volume (V_f) = 140 L = 0.14 m³
- Temperature (T) = 27°C = 300 K (since we convert Celsius to Kelvin by adding 273)

### Step 2: Calculate the number of moles of oxygen
To find the number of moles (n), we use the formula:
\[ n = \frac{m}{M} \]
where \( M \) is the molar mass of oxygen. The molar mass of oxygen (O₂) is approximately 32 g/mol.

Calculating the number of moles:
\[ n = \frac{96 \text{ g}}{32 \text{ g/mol}} = 3 \text{ moles} \]

### Step 3: Use the work done formula for isothermal processes
The work done (W) during an isothermal expansion can be calculated using the formula:
\[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]

### Step 4: Substitute the values into the formula
We know:
- \( n = 3 \) moles
- \( R = 8.31 \, \text{J/(mol K)} \) (universal gas constant)
- \( T = 300 \, \text{K} \)
- \( V_f = 0.14 \, \text{m³} \)
- \( V_i = 0.07 \, \text{m³} \)

Substituting these values into the work done formula:
\[ W = 3 \times 8.31 \times 300 \times \ln\left(\frac{0.14}{0.07}\right) \]

### Step 5: Calculate the logarithm
Calculating the logarithm:
\[ \frac{0.14}{0.07} = 2 \]
Thus,
\[ \ln(2) \approx 0.693 \]

### Step 6: Calculate the work done
Now substituting back:
\[ W = 3 \times 8.31 \times 300 \times 0.693 \]

Calculating:
1. \( 3 \times 8.31 = 24.93 \)
2. \( 24.93 \times 300 = 7479 \)
3. \( 7479 \times 0.693 \approx 5182.07 \, \text{J} \)

### Step 7: Final result
The work done by the gas during the isothermal expansion is approximately:
\[ W \approx 5182.07 \, \text{J} \]

### Conclusion
Thus, the work done by the gas is approximately 5182 J or 5.18 kJ.

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