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

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 = (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

`(1)/(0.174) = (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! =)

Answer 2

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