the molecules of O2 that are present in 3.90 L flask at a temperature of 273 K and a pressure of 1.00 atm is 1.047 x 10^23 molecules of O2
calculation
Step 1: used the ideal gas equation to calculate the moles of O2
that is Pv=n RT where;
P(pressure)= 1.00 atm
V(volume) =3.90 L
n(number of moles)=?
R(gas constant) = 0.0821 L.atm/mol.K
T(temperature) = 273 k
by making n the subject of the formula by dividing both side by RT
n= Pv/RT
n=[( 1.00 atm x 3.90 L) /(0.0821 L.atm/mol.k x273)]=0.174 moles
Step 2: use the Avogadro's law constant to calculate the number of molecules
that is according to Avogadro's law
1 mole = 6.02 x10^23 molecules
0.174 moles=? molecules
by cross multiplication
the number of molecules
= (0.174 moles x 6.02 x10^23 molecules)/ 1 mole =1.047 x 10^23 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! =)
