Question

An expensive vacuum system can achieve a pressure as low as 1.83 ✕ 10−7 N/m2 at 28°C. How many atoms are there in a cubic centimeter at this pressure and temperature?

Answer 1

Answer:
`44.1* 10^6` atoms

Step-by-step explanation:

According to the ideal gas equation:

`PV=nRT`

P = Pressure of the gas =
`1.83* 10^(-7)N/m^2=1.81* 10^(-12)atm`
`1N/m^2=9.87* 10^(-6)atm`

V= Volume of the gas =
`1cm^3=1ml=0.001L` (1L=1000ml)

T= Temperature of the gas = 28°C = 301 K
`0^0C=273K`

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= ?

`n=(PV)/(RT)=\frac{1.81* 10^(-12)atm* 0.001L}0.0821Latm/Kmol* 301K}=7.32* 10^(-17)moles`

Number of atoms =
`moles* {\text {avogadro's number}}=7.32* 10^(-17)mol* 6.023* 10^(23)mol^(-1)=44.1* 10^6atoms`

Thus there are
`44.1* 10^6` atoms in a cubic centimeter at this pressure and temperature.