Consider the figure provided. 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at $$18°C$$. If the piston is moved to position B, keeping the temperature unchanged, then '$$x$$' L atm work is done in this reversible process. $$x =$$ ______ L atm. (nearest integer) [Given: Absolute temperature $$= °C + 273.15$$, $$R = 0.08206 \text{ L atm mol}^{-1} \text{K}^{-1}$$]

Given Data:

• Number of moles ($$n$$) = $$1\text{ mol}$$
• Initial volume ($$V_1$$) at position A = $$10\text{ L}$$
• Final volume ($$V_2$$) at position B = $$10\text{ L} + 90\text{ L} = 100\text{ L}$$
• Temperature ($$T$$) = $$18^\circ\text{C} + 273.15 = 291.15\text{ K}$$
• Gas constant ($$R$$) = $$0.08206\text{ L atm mol}^{-1}\text{K}^{-1}$$

Formula and Calculation:

For an isothermal reversible expansion, the magnitude of work done ($$W$$) is given by:

$$W = 2.303 \, nRT \log_{10}\left(\frac{V_2}{V_1}\right)$$

Substitute the given values into the formula:

$$x = 2.303 \times 1 \times 0.08206 \times 291.15 \times \log_{10}\left(\frac{100}{10}\right)$$

Since $$\log_{10}(10) = 1$$:

$$x = 2.303 \times 0.08206 \times 291.15 \times 1$$

$$x \approx 55.02\text{ L atm}$$

Rounding to the nearest integer, $$x = 55$$.