An ideal gas occupies a volume of 1cm^ atv20°c and atmospheric pressure determine the number of molecules of gas in the container

By Charles Law the volume of 1 cm^3 at standard conditions is

V = V0 * 273 / 293 = 1 cm^3 * .932 cm^3

Reducing the temperature to 0 deg C results in a volume of .932 cm^3

1 mole of gas contains 22.4 L of gas

1 L = 1000 cm^3

1000 cm^3 / L / (.932 cm^3 / container) = 1073   containers / L

So 1 mole of gas = 1073 containers * 22.4 L /container =  24,305 L

6.02E23 molecules / mole * 1 / 24305 molecules / mole = 2.48E19

number of molecules present = 2.48 * 10^19

(Easier to calculate volume of 1 cc

1 cc = 6.02E23 / (1000 * 22.4) = 2.69E19 molecules in 1 cc

Vol of 1cc = 273 / 293 = .932

2.69E19 * .932 = 2.51 * 10^19)

be subject to a force andchange its motion

measuring things 1 litre of water