At 400K, 5 moles of an ideal gas expands isothermally and reversibly from `10dm^3` to `20 dm^3`. Calculate the work done by the gas. 

Answer

To calculate the work done by the gas during isothermal and reversible expansion, we can use the formula for work done in an isothermal process for an ideal gas:

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

Where:
- \( W \) = work done by the gas
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin
- \( V_f \) = final volume
- \( V_i \) = initial volume

### Step 1: Identify the given values
- Number of moles, \( n = 5 \) moles
- Temperature, \( T = 400 \) K
- Initial volume, \( V_i = 10 \) dm³
- Final volume, \( V_f = 20 \) dm³
- Universal gas constant, \( R = 8.314 \) J/(mol·K)

### Step 2: Convert volumes to the same unit (if necessary)
In this case, both volumes are already in dm³, so we can proceed without conversion.

### Step 3: Calculate the ratio of final volume to initial volume
\[ \frac{V_f}{V_i} = \frac{20 \, \text{dm}^3}{10 \, \text{dm}^3} = 2 \]

### Step 4: Calculate the natural logarithm of the volume ratio
Using the properties of logarithms:
\[ \ln(2) \approx 0.693 \]

### Step 5: Substitute the values into the work done formula
\[ W = -nRT \ln\left(\frac{V_f}{V_i}\right) \]
\[ W = -5 \times 8.314 \, \text{J/(mol·K)} \times 400 \, \text{K} \times 0.693 \]

### Step 6: Calculate the work done
\[ W = -5 \times 8.314 \times 400 \times 0.693 \]
\[ W = -5 \times 8.314 \times 400 \times 0.693 \approx -11528.4 \, \text{J} \]

### Step 7: Convert the work done from Joules to kilojoules
Since \( 1 \, \text{kJ} = 1000 \, \text{J} \):
\[ W \approx -11.5284 \, \text{kJ} \]

### Final Answer
The work done by the gas during the isothermal expansion is approximately:
\[ W \approx -11.53 \, \text{kJ} \]

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