Question

300cm3 of O₂ will be collected over water at a temperature of 25°c and 1atm (101325 NM-2) What will be the volume of dry oxygen at S.T.P? (water pressure at 25c is 16630M²)

Answer 1

The volume of the gas is calculated as 1827.87cm^3

Data;

• Initial volume (v1) = 300cm^3
• Initial Pressure (p1) = 1atm = 101325 NM^-2
• Final Volume = ?
• Final Pressure (P2) = 16630NM^-2

Boyle's Law

To find the final volume of oxygen, we have to use Boyle's law which states that the volume of a gas is inversely proportional to the pressure of the gas provided that temperature remains constant.

v ∝ 1/p

$v = \frac{k}{p} \\k = pv\\p_1 v_1 = p_2 v_2$

Let's substitute the values into the equation and solve.

$p_1v_1 = p_2v_2\\101325*300 = 16630 * v_2\\30397500 = 16630v_2\\v_2 = \frac{30397500}{16630} \\v_2 = 1827.87cm^3$

The volume of the gas is calculated as 1827.87cm^3

Learn more on Boyle's law here;

https://brainly.com/question/26040104