One mole of an ideal gas at 27 $$^\circ$$C is taken from A to B as shown in the given PV indicator diagram. The work done by the system will be ___ $$\times 10^{-1}$$ J. [Given, $$R = 8.3$$ J mole$$^{-1}$$ K, ln 2 = 0.6931] (Round off to the nearest integer)

We need to calculate the work done by the system when one mole of an ideal gas expands from state $$A$$ to state $$B$$ along the given path in the $$P\text{-}V$$ indicator diagram, using the explicit temperature condition given in the problem statement.

1. Analyze the Thermodynamic Properties Given

• Number of moles ($$n$$): $$1\text{ mole}$$
• Temperature ($$T$$): $$27^\circ\text{C} = 27 + 273.15 = 300\text{ K}$$
• Gas Constant ($$R$$): $$8.3\text{ J mole}^{-1}\text{ K}^{-1}$$
• Value of $$\ln 2$$: $$0.6931$$

From the diagram coordinates:

• Initial Volume ($$V_1$$): $$2\text{ m}^3$$
• Final Volume ($$V_2$$): $$4\text{ m}^3$$

Since the problem specifies that the gas stays at a fixed temperature of $$27^\circ\text{C}$$, the process from $$A$$ to $$B$$ is an isothermal expansion.

2. Establish the Formula for Isothermal Work Done

The standard thermodynamic work done ($$W$$) during an ideal gas isothermal expansion is:

$$W = nRT \ln\left(\frac{V_2}{V_1}\right)$$

3. Substitute Given Values and Calculate

Placing the parameters explicitly provided in the problem statement:

$$W = 1 \times 8.3 \times 300 \times \ln\left(\frac{4}{2}\right)$$

$$W = 2490 \times \ln(2)$$

$$W = 2490 \times 0.6931 = 1725.819\text{ J}$$

4. Express in the Requested Format

The question requires the answer to be written in the form of $$\text{Value} \times 10^{-1}\text{ J}$$:

$$W = 17258.19 \times 10^{-1}\text{ J}$$

Rounding off to the nearest integer gives:

$$\text{Value} \approx 17258$$

Final Answer: $$17258$$