To solve the problem step by step, we will use the formula for work done during an isothermal process for an ideal gas.
### Step 1: Identify the given values
- Initial volume, \( V_1 = 10 \, \text{liters} = 10 \times 10^{-3} \, \text{m}^3 \)
- Final volume, \( V_2 = \frac{1}{15} V_1 = \frac{1}{15} \times 10 \, \text{liters} = \frac{10}{15} \, \text{liters} = \frac{2}{3} \, \text{liters} = \frac{2}{3} \times 10^{-3} \, \text{m}^3 \)
- Initial pressure, \( P_0 = 10^5 \, \text{Pa} \)
- Temperature, \( T = 27^\circ C = 27 + 273 = 300 \, \text{K} \)
### Step 2: Calculate the work done using the isothermal process formula
The work done \( W \) in an isothermal process is given by:
\[
W = nRT \ln\left(\frac{V_2}{V_1}\right)
\]
However, we can also express it using the initial pressure and volume:
\[
W = P_0 V_0 \ln\left(\frac{V_2}{V_1}\right)
\]
Where:
- \( V_0 = V_1 = 10 \times 10^{-3} \, \text{m}^3 \)
- \( V_2 = \frac{2}{3} \times 10^{-3} \, \text{m}^3 \)
### Step 3: Substitute the values into the equation
First, calculate the ratio \( \frac{V_2}{V_1} \):
\[
\frac{V_2}{V_1} = \frac{\frac{2}{3} \times 10^{-3}}{10 \times 10^{-3}} = \frac{2}{30} = \frac{1}{15}
\]
Now substitute into the work done formula:
\[
W = P_0 V_0 \ln\left(\frac{1}{15}\right)
\]
Substituting the known values:
\[
W = (10^5 \, \text{Pa}) \times (10 \times 10^{-3} \, \text{m}^3) \times \ln\left(\frac{1}{15}\right)
\]
\[
W = (10^5) \times (10^{-2}) \times \ln\left(\frac{1}{15}\right)
\]
\[
W = 10^3 \times \ln\left(\frac{1}{15}\right)
\]
### Step 4: Calculate \( \ln\left(\frac{1}{15}\right) \)
Using a calculator:
\[
\ln\left(\frac{1}{15}\right) = -\ln(15) \approx -2.708
\]
### Step 5: Calculate the work done
Now substitute back:
\[
W = 10^3 \times (-2.708) \approx -2708 \, \text{J}
\]
### Final Answer
The work done by the gas during the isothermal compression is approximately:
\[
W \approx -2.708 \times 10^3 \, \text{J} \, \text{or} \, -2.70 \, \text{kJ}
\]
