In the Earth’s atmosphere, the concentration of rare gas argon (Ar) is 0.934 % v
ID: 895055 • Letter: I
Question
In the Earth’s atmosphere, the concentration of rare gas argon (Ar) is 0.934 % volume.
The volume of our classroom has been estimated to be about 600 m3.
Assuming argon behaves as a perfect gas, and assuming the pressure to be 1.00 atm, and the temperature to be 22 °C, calculate the mass of argon in our classroom. Atomic mass of argon is 39.95 amu ( 1 point)
What is the partial pressure of Ar in our classroom, expressed in Pa? (1 point)
Considering argon to behave as a real gas, calculate its partial pressure in our classroom, under the conditions stated in (a), using van der Waals’ equation.
(van der Waals coefficients for argon are : a = 1.337 atm dm6 mol-2 , and b= 3.20 x 10-2 dm3 mol-1 ) (1 point)
Explanation / Answer
a) Volume of argon in classroom = 5.604 m3
Pressure = 1 atm
T = 22 ºC = 295 K
m = ?
Molar mass = 39.95 g/mol
n = PV/RT = (1atm)*(5604L) / (0.0821 L-atm/mole-K)*(295K) = 231.38 moles
mass = moles * molar mass = 231.38 moles * 39.95 g/mol = 9243.78 grams
b) total moles = (1atm)*(600000L) / (0.0821 L-atm/mole-K)*(295K) = 24773.42637 moles
molar fraction of argon = 231.38/24773.42 = .00934
1 atm * .00934 = 0.00934 atm = 946.3755 Pa
c) Van der Waals
[P + a(n/V)2] * [V - nb] = nRT
[1atm + 1.337(n/600000 dm3)2] * [600000 - n(3.2x10-2)] = n(0.0821 dm3 atm/K mol) * (295K)
[1atm + 1.337(n/600000 dm3)2] * [600000 - n(3.2x10-2)] = 24.2195 n
[1atm + 3.71389x10-12 n2] * [600000 - n(.032)] = 24.2195 n
total moles = 24797.2
[1atm + 1.337(n/5604 dm3)2] * [5604 - n(3.2x10-2)] = 24.2195 n
argon moles = 231.605
Partial Pressure = 231.605/24797.2 * 1 = 0.0093399 atm = 1006.56 Pa