Given:
- Mass of silver (Ag) deposited = 108 g
- The quantity of electricity passed is the same for both the deposition of Ag and liberation of O2

To find:
Volume of O2 liberated at 1 bar and 273K

Concept:
- According to Faraday's law of electrolysis, the mass of a substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
- The volume of a gas liberated during electrolysis can be determined using Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules.

Calculation:
1. Calculate the moles of Ag deposited:
- The molar mass of Ag is 108 g/mol (from the periodic table).
- Moles of Ag deposited = Mass of Ag deposited / Molar mass of Ag
= 108 g / 108 g/mol
= 1 mol

2. Calculate the moles of O2 liberated:
- Since the quantity of electricity passed is the same for both Ag deposition and O2 liberation, the moles of O2 liberated will also be 1 mol.

3. Calculate the volume of O2 liberated at STP:
- The molar volume of any gas at STP (Standard Temperature and Pressure) is 22.4 L/mol.
- Volume of O2 liberated = Moles of O2 liberated × Molar volume of gas at STP
= 1 mol × 22.4 L/mol
= 22.4 L

4. Convert the volume to the given conditions (1 bar and 273K):
- According to the Ideal Gas Law, the volume of a gas is inversely proportional to pressure and directly proportional to temperature.
- Using the formula: (P1 × V1) / T1 = (P2 × V2) / T2
where P1 = 1 bar, V1 = 22.4 L, T1 = 273K, P2 = unknown, V2 = unknown, T2 = 273K
- Rearranging the formula: (1 bar × 22.4 L) / 273K = (P2 × V2) / 273K
- Solving for V2: V2 = (1 bar × 22.4 L) / P2

5. Substitute the given values:
- V2 = (1 bar × 22.4 L) / 1 bar
= 22.4 L

Final Answer:
The volume of O2(g) liberated at 1 bar and 273K by the same quantity of electricity is 22.4 L.