Question

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

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

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