(a) An ideal gas occupies a volume of 1.8 cm3 at 20°C and atmospheric pressure. Determine the number of molecules of gas in the container.
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PV = nRT
n = PV/RT
P = 1.00 atm
V = 1.8 cm³ = 1.8 cm³ x (1 mL / 1 cm³) x (1 L / 1000 mL) = 0.0018 L
R = 0.0821 Latm/moleK
T = 20C = 20+273.15 = 293.15K
n = (1.00 atm) x (0.0018 L) / [ (0.0821 Latm/moleK) x (293.15K) ]
n = 7.48x10^-5 moles..
7.48x10^-5 moles x (6.022x10^23 molecules / 1 mole) = 4.50x10^19 molecules.
you should probably only have 2 sig figs since 1.8 has 2 and 20 has at most 2...so call the final answer....
= 4.5x10^19 molecules
N = 1.6×10⁻¹¹ Pa ∙ 2.4×10⁻⁶ m³ / (1.38065 ×10⁻²³J/K ∙ 293.15K) = 9488
Both parts or one?
let me redo it.
(a) you use the ideal gas law
PV=nRT=NRT/Na where N is the number of molecules
1 atm = 101,325 Pa , V = 1.8 cm3 = 1.8*10^(-6) m3 , T=293 K
N=4.51 x 10^19
(b) using the same formula, we get N=1.07 x 10^4
Hope this helps
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