##### (a) An ideal gas occupies a volume of 1.8 cm3 at 20°C and atmospheric pressure.

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(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.

molecules

(b) If the pressure of the 1.8-cm3 volume is reduced to 2.4 ✕ 10−11 Pa (an extremely good vacuum) while the temperature remains constant, how many moles of gas remain in the container?
mol
Jun 21st, 2015

a) 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..

finally...
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

b)  N = 1.6×10⁻¹¹ Pa ∙ 2.4×10⁻⁶ m³ / (1.38065 ×10⁻²³J/K ∙ 293.15K) = 9488

Jun 21st, 2015

wrong

Jun 21st, 2015

Both parts or one?

Jun 21st, 2015

let me redo it.

Jun 21st, 2015

yes both

Jun 21st, 2015

(a) you use the ideal gas law

PV=nRT=NRT/Na where N is the number of molecules

so N=PVNa/(RT)

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

Jun 21st, 2015

Jun 21st, 2015
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Jun 21st, 2015
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Jun 21st, 2015
Dec 7th, 2016
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