A gas expands isothermally from `10 dm^(3)` to `20 dm^(3)` at `27^(@)C` and work obtained is 4.620 kJ.Calculate the number of moles of the gas.

Explanation

To solve the problem step by step, we will use the formula for work done during an isothermal expansion of an ideal gas. The formula is given by:

\[ W = -nRT \log\left(\frac{V_2}{V_1}\right) \]

Where:
- \( W \) is the work done (in Joules)
- \( n \) is the number of moles of the gas
- \( R \) is the ideal gas constant (8.314 J/(mol·K))
- \( T \) is the absolute temperature (in Kelvin)
- \( V_1 \) is the initial volume (in dm³)
- \( V_2 \) is the final volume (in dm³)

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature is given as \( 27^\circ C \). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

So,

\[ T = 27 + 273 = 300 \, K \]

### Step 2: Identify the volumes
The initial volume \( V_1 \) is \( 10 \, dm³ \) and the final volume \( V_2 \) is \( 20 \, dm³ \).

### Step 3: Convert work done from kJ to J
The work done is given as \( 4.620 \, kJ \). To convert this to Joules:

\[ W = 4.620 \, kJ \times 1000 = 4620 \, J \]

### Step 4: Substitute values into the work formula
We can now substitute the known values into the work formula:

\[ -4620 = -n \cdot 8.314 \cdot 300 \cdot \log\left(\frac{20}{10}\right) \]

### Step 5: Calculate the logarithm
Calculate the logarithm:

\[ \log\left(\frac{20}{10}\right) = \log(2) \approx 0.301 \]

### Step 6: Substitute the logarithm into the equation
Now substitute this value back into the equation:

\[ -4620 = -n \cdot 8.314 \cdot 300 \cdot 0.301 \]

### Step 7: Simplify the equation
Now calculate the right side:

\[ -4620 = -n \cdot 8.314 \cdot 300 \cdot 0.301 \]
\[ -4620 = -n \cdot 8.314 \cdot 90.3 \]
\[ -4620 = -n \cdot 750.78 \]

### Step 8: Solve for \( n \)
Now, divide both sides by \( -750.78 \):

\[ n = \frac{4620}{750.78} \approx 6.15 \]

### Step 9: Final answer
Thus, the number of moles of the gas is approximately:

\[ n \approx 6.15 \, moles \]