Question

Calculate the volume occupied by 10 ki power 22 molecules of N2 at 27 degree Celsius one atmospheric pressure

Answer 1

Explanation:

Given => pressure p = 1atm , temperature T = 27+273 = 300K .

Now ,number if moles of N2 is

6.02 × 10^23/10^22 = 60.2 mol ≈ 60mol

as we know

PV = nRT

=> 1 × V = 60 × 0.0821 × 300

=> V = 1477.8 L (approx)

thanks

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

V = 1477.8 L (approx)

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given that :

Pressure (p) = 1atm

Temperature (T) = 300K.      (27+273 = 300K)

Solution :

★ Number of moles of N2 is :

$\sf \longrightarrow 6.02\times \dfrac{ 10^2^3}{10^2^2 }\\\\\sf \longrightarrow 60.2\; mol\; approx\; 60mol$

We know that :

★ $\boxed{\sf P\times V = n\times R\times T}$

Put the given values :

=> 1 × V = 60 × 0.0821 × 300

=> V = 1477.8 L (approx)

Hence,

V = 1477.8 L (approx)