Question

How many molecules of O2 are present in a 3.90L flask at a temperature of 273K and a pressure of 1.00 atm

Answer 1

 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

Answer 2

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:

$P*V = n*R*T$

$1*3.90 = n*0.082*273$

$3.90 = 22.386\:n$

$22.386\:n = 3.90$

$n = \dfrac{3.90}{22.386}$

$\boxed{n \approx 0.174\:mol}$

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

$\dfrac{1}{0.174} = \dfrac{6.022*10^{23}}{y}$

multiply the means by the extremes

$1*y = 0.174*6.022*10^{23}$

$y = 1.047828*10^{23} \to \boxed{\boxed{y \approx 1.048*10^{23}\:molecules\:of\:O_2}}\end{array}}\end{array}}\qquad\checkmark$

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I Hope this helps, greetings ... Dexteright02! =)