Question

An expensive vacuum system can achieve a pressure as low as 1.00×10^{-7} N/m^2 at 20ºC . How many atoms are there in a cubic centimeter at this pressure and temperature? (answer in ×10^{7} atoms)

Answer 1

Answer:

$24.70818432141\times 10^7\ atoms$

Explanation:

P = Pressure = $1\times 10^{-7}\ N/m^2$

V = Volume = 1 cm³

n = Amount of substance

N = Number of atoms

$N_A$ = Avogadro's constant = $6.022\times 10^{23}\ /mol$

R = Gas constant = 8.314 J/k mol

T = Temperature = 273.15+20 = 293.15 K

From the ideal gas law

$PV=nRT\\\Rightarrow n=\frac{PV}{RT}$

$n=\frac{N}{N_A}$

$\frac{N}{N_A}=\frac{PV}{RT}\\\Rightarrow N=N_A\times \frac{PV}{RT}\\\Rightarrow N=\frac{1\times 10^{-7}\times 1\times 10^{-6}}{8.314\times 293.15}\times 6.022\times 10^{23}\\\Rightarrow N=24708184.32141\ atoms=24.70818432141\times 10^7\ atoms$

The number of atoms is $24.70818432141\times 10^7\ atoms$