A container of volume 19.4 cm3 is initially filled with air. The container is then evacuated at 0oC to a pressure of 6.0 mPa. How many molecules are in the container after evacuation if we assume that air is an ideal gas?

Answer 1

Since the gas is ideal, we can use the Ideal Gas Law:
PV = nRT
where
P = pressure in atm
V = volume in L
n = number of moles of air
R = universal gas constant = 0.0821 L-atm/mol-K
T = temperature in K

We first convert the given in the appropriate units:
V = 19.4 cm^3
cm^3 is also equal to mL. Thus there are 1000 cm^3 for every 1 L, or
V = 19.4 / 1000 = 0.0194 L

P = 6 mPa
The conversion is: 101325 Pa = 1 atm, thus
P = (6 / 1000) / 101325 =

T = 0 C
We just add 273 to make it Kelvin:
T = 0 + 273 = 273 K

Substituting to the equation:
PV = nRT
n = PV/RT
n = (5.92 10^-8)(0.0194) / (0.0821)(273)
n = 5.125 * 10^-11 moles

Note that this is only the moles. To get the number of molecules, note that 1 mol = 6.022 * 10^23 representative particles. Thus,
5.125 * 10^-11 * 5.125 * 10^-11
= 3.087 * 10^13 molecules
Check the significant figures.

Hope this helps :3

Answer 2

*lol sorry, in the last calculation that should be
5.125 * 10^-11 * 6.022 * 10^23
but the answer is still the same. :)