Question

An expensive vacuum system can achieve a pressure as low as 1.00 × 10 − 8 N / m 2 at 19.5 °C . How many atoms are there in a cubic centimeter at this pressure and temperature?

Answer 1

2475042

Step-by-step explanation:

P = Pressure = [/tex]1\times 10^{-8}\ N/m^2[/tex]

V = Volume = 1 cm³

T = Temperature = 19.5 °C

R = Gas constant = 8.341 J/mol K

`N_A` = Avogadro's number =
`6.022* 10^(23)`

n = Amount of substance

From ideal gas law we have

`PV=nRT\\\Rightarrow n=(PV)/(RT)\\\Rightarrow n=(1* 10^(-8)* 1* 10^(-6))/(8.314(273.15+19.5))\\\Rightarrow n=4.11* 10^(-18)`

Number of atoms is given by

`n* N_A\\ =4.11* 10^(-18)* 6.022* 10^(23)\\ =2475042\ atoms`

The number of atoms in this vacuum is 2475042