To solve the problem of calculating the work done by a litre of dry air at STP when it expands isothermally to a volume of 3 litres, we will use the formula for work done during an isothermal process for an ideal gas.
### Step-by-Step Solution:
1. **Identify the Formula**:
The work done \( W \) during an isothermal expansion of an ideal gas is given by:
\[
W = nRT \ln\left(\frac{V_2}{V_1}\right)
\]
where:
- \( n \) = number of moles of gas
- \( R \) = universal gas constant (approximately \( 8.314 \, \text{J/(mol K)} \))
- \( T \) = absolute temperature in Kelvin
- \( V_1 \) = initial volume
- \( V_2 \) = final volume
2. **Convert Volume to Cubic Meters**:
The initial volume \( V_1 = 1 \, \text{L} = 1 \times 10^{-3} \, \text{m}^3 \) and the final volume \( V_2 = 3 \, \text{L} = 3 \times 10^{-3} \, \text{m}^3 \).
3. **Calculate the Number of Moles \( n \)**:
At STP (Standard Temperature and Pressure), 1 mole of an ideal gas occupies 22.4 L. Therefore, the number of moles \( n \) in 1 L of air is:
\[
n = \frac{1 \, \text{L}}{22.4 \, \text{L/mol}} = \frac{1 \times 10^{-3} \, \text{m}^3}{22.4 \times 10^{-3} \, \text{m}^3/\text{mol}} \approx 0.04464 \, \text{mol}
\]
4. **Determine the Temperature \( T \)**:
At STP, the temperature is \( 0^\circ C = 273.15 \, \text{K} \).
5. **Substitute Values into the Formula**:
We can now substitute \( n \), \( R \), \( T \), \( V_1 \), and \( V_2 \) into the work formula:
\[
W = nRT \ln\left(\frac{V_2}{V_1}\right)
\]
Using \( \ln(3) \approx 1.1 \):
\[
W = (0.04464 \, \text{mol}) \times (8.314 \, \text{J/(mol K)}) \times (273.15 \, \text{K}) \times 1.1
\]
6. **Calculate the Work Done**:
First, calculate \( nRT \):
\[
nRT \approx 0.04464 \times 8.314 \times 273.15 \approx 101.325 \, \text{J}
\]
Now, substituting this into the work equation:
\[
W = 101.325 \times 1.1 \approx 111.46 \, \text{J}
\]
7. **Final Answer**:
The work done by the air during the isothermal expansion is approximately:
\[
W \approx 111.1 \, \text{J}
\]
