if 120 ml of oxygen is collected over water at 27 degrees celsius & 740 mm pressure what will the volume of the dry gas be at STP

V₂ = 106.33 ml

Explanation:

Data Given:

Initial Temperature = T₁ = 27 °C + 273.15 = 300.15 K

Initial Pressure = P₁ = 740 mmHg

Initial Volume = V₁ = 120 ml

Final Temperature = T₂ = 0 °C + 273.15 = 273.15 K

Final Pressure = P₂ = 760 mmHg

Fnal Volume = V₂ = ?

Formula Used:

Let's assume that the Oxygen gas is acting as an Ideal gas, the according to Ideal Gas Equation,

P₁ V₁ / T₁ = P₂ V₂ / T₂

Solving for V₂,

V₂ = P₁ V₁ T₂ / T₁ P₂

Putting values,

V₂ = 740 mmHg * 120 ml * 273.15 K / 300.15 K * 760 mmHg

V₂ = 106.33 ml