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1.)A 4.12 gram sample of argon gas has a volume of 922 milliliters at a pressure

ID: 1039714 • Letter: 1

Question

1.)A 4.12 gram sample of argon gas has a volume of 922 milliliters at a pressure of 3.96 atm. The temperature of the Ar gas sample is ___________°C.

2.)
A mixture of carbon dioxide and oxygen gases, in a 9.44 L flask at 81 °C, contains 11.9 grams of carbon dioxide and 2.92 grams of oxygen. The partial pressure of oxygen in the flask is___________________atm and the total pressure in the flask is ___________________atm.

3.)


The stopcock connecting a 3.19 L bulb containing argon gas at a pressure of 2.96 atm, and a 4.56 L bulb containing carbon dioxide gas at a pressure of 1.69 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is _______________atm.

Explanation / Answer

1)

Molar mass of Ar = 39.95 g/mol


mass(Ar)= 4.12 g

use:
number of mol of Ar,
n = mass of Ar/molar mass of Ar
=(4.12 g)/(39.95 g/mol)
= 0.1031 mol

Given:
P = 3.96 atm
V = 922.0 mL
= (922.0/1000) L
= 0.922 L
n = 0.1031 mol

use:
P * V = n*R*T
3.96 atm * 0.922 L = 0.1031 mol* 0.08206 atm.L/mol.K * T
T = 431 K
T = (431 -273 ) oC
T = 158 oC
Answer: 158 oC

2)
a)find partial pressure of oxygen:

Molar mass of O2 = 32 g/mol


mass(O2)= 2.92 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(2.92 g)/(32 g/mol)
= 9.125*10^-2 mol

Given:
V = 9.44 L
n = 0.0912 mol
T = 81.0 oC
= (81.0+273) K
= 354 K

use:
P * V = n*R*T
P * 9.44 L = 0.0912 mol* 0.08206 atm.L/mol.K * 354 K
P = 0.2806 atm
Answer: 0.281 atm

b)
find partial pressure of CO2:

Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol


mass(CO2)= 11.9 g

use:
number of mol of CO2,
n = mass of CO2/molar mass of CO2
=(11.9 g)/(44.01 g/mol)
= 0.2704 mol

Given:
V = 9.44 L
n = 0.2704 mol
T = 81.0 oC
= (81.0+273) K
= 354 K

use:
P * V = n*R*T
P * 9.44 L = 0.2704 mol* 0.08206 atm.L/mol.K * 354 K
P = 0.832 atm

Total pressure = partial pressure of O2 + partial pressure of CO2
= 0.281 atm + 0.832 atm
= 1.11 atm

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