Based on the ideal gas relation:
PV = nRT
where P = pressure ; V = volume ; T = temperature
n = number of moles; R = gas constant = 0.0821 L atm/mol-K
Step 1: Find the number of moles of O2
n = PV/RT = 1 * 3.90/0.0821*273 = 0.1740 moles
Step 2: Calculate the molecules of O2
Now, 1 mole of O2 corresponds to 6.023 * 10²³ molecules of O2
Therefore, 0.1740 moles of O2 corresponds to-
0.1740 moles of O2 * 6.023*10²³ molecules of O2/1 mole of O2
= 1.048 * 10²³ molecules of O2
Hello!
How many molecules of O2 are present in a 3.90L flask at a temperature of 273K and a pressure of 1.00atm?
We have the following data:
V (volume) = 3.90 L
n (number of mols) = ?
T (temperature) = 273 K
P (pressure) = 1 atm
R (gas constant) = 0.082 atm.L / mol.K
We apply the data above to the Clapeyron equation (gas equation), let's see:
[tex]P*V = n*R*T[/tex]
[tex]1*3.90 = n*0.082*273[/tex]
[tex]3.90 = 22.386\:n[/tex]
[tex]22.386\:n = 3.90[/tex]
[tex]n = \dfrac{3.90}{22.386}[/tex]
[tex]\boxed{n \approx 0.174\:mol}[/tex]
Now, knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10²³ molecules, how many molecules of O2 are present ?
1 mol -------------------- 6.022*10²³ molecules
0.174 mol ---------------- y molecules
[tex]\dfrac{1}{0.174} = \dfrac{6.022*10^{23}}{y}[/tex]
multiply the means by the extremes
[tex]1*y = 0.174*6.022*10^{23}[/tex]
[tex]y = 1.047828*10^{23} \to \boxed{\boxed{y \approx 1.048*10^{23}\:molecules\:of\:O_2}}\end{array}}\end{array}}\qquad\checkmark[/tex]
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I Hope this helps, greetings ... Dexteright02! =)
