Answer:
1 particle of the gas would occupy a volume of 3.718*10⁻²³L
Step-by-step explanation:
Hello,
1. The sample has a particle of 6.022×10²²particles
2. Volume of the sample = 22.4L
3. Temperature of the sample = 0°C = (0 +273.15)K = 273.15K
4. Pressure of the sample = 1.0atm
What volume would 1 particle of the gas occupy?
But we remember that 1 mole of any substance = 6.022×10²² molecules or particles or atoms
What would be the number of moles for 1 particule?
1 mole = 6.022×10²² particles
X moles = 1 particle
X = (1 × 1) / 6.022×10²² particles
X = 1.66×10⁻²⁴ moles
Therefore, 1 particle contains 1.66×10⁻²⁴ moles
Since we know our number of moles, we can proceed to use ideal gas equation,
Ideal gas equation holds for all ideal gas and is defined as
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles of the gas
R = ideal gas constant = 0.082 L.atm / mol.K
T = temperature of the gas
PV = nRT
Solving for V,
V = nRT/ P
We can now plug in our values into the above
equation.
V = (1.66*10⁻²⁴ × 0.082 × 273.15) / 1
V = 3.718*10⁻²³L
Therefore, 1 particule of the gas would occupy a volume of 3.718*10⁻²³L.