Question

Calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273 K and 3 atm pressure.

Answer 1

Steps and Understanding :

1) Assuming Oxygen gas is behaving like Ideal.
We have,
Volume of gas in container ,V= 224 mL = 0.224L
Pressure Exerted by it,P = 3 atm
Temperature of gas,T = 273 K
Universal Gas Constant, R = 0.0821 Latm/mol/K

2) By applying Ideal Gas Equation ,

PV = nRT
=> n= PV /RT
$= > n = \frac{3 \times 0.224}{0.0821 \times 273} \\ \\ = > n \: \: \: = \: \: 30 \times {10}^{ - 3}$

3) No. of Molecules = No. of moles * Avogadro Number of particles

$= 30 \times {10}^{ - 3} \times 6.022 \times {10}^{23} \\ = 1.81 \times {10}^{22}$
Therefore, No. of molecules of Oxygen gas is 1.81 * 10^(22).