In a class room demonstration, a teacher inflated a balloon to 2.70 L by adding
ID: 1049672 • Letter: I
Question
In a class room demonstration, a teacher inflated a balloon to 2.70 L by adding 0.243g of hydrogen gas to the balloon. She then added 0.0040 mol of O2 to it
a/ What will be the final volume of the balloon? State any assumptions insolving this problem
b/ If the balloon was filled under STP conditions, what would the volume be on the next day when the conditions had changed to 30.0 celsius and 0.97 atm and the number of gas molescules has decreased by 0.05 mole due to effusion
c/ In the above conditions, what is pressure exerted by O2 ? by H2
Explanation / Answer
Calculate the moles of H2
Moles of H2 = mass of H2 in g / molar mass of H2
Molar mass of H2 = 2.0158 g/mol
0.243/2.0158=0.1205 mol
a) calculation of final volume
V1/V2 = n1/n2
V1 = 2.70 L , n1 = 0.1205
V2 = unknown , n2 = mol of O2 + mol H2 = 0.0040+0.1206=0.1245
V2 = V1 x (n2/n1) = 2.70*0.1245/0.1205=2.79 L
b) At STP
Temperature is 273.15 K , pressure is 1 atm.
V1 = 2.79 L , n1 = 0.1246 mol
V2 = unknown
T2 = 30.0+273.15=303.15 K
n2= 0.1246-0.05=0.0746 mol
P2=0.97 atm
We know P1V1/P2V2 = (n1RT1/n2RT2)
(1*2.79)/(0.97*V2) = (0.1246*273.15)/(0.0746*303.15)
V2 = 1.911 L
C) pressure of the H2 at STP
Mol = 0.1205 , V=2.79 L , T = 273.15
Partial pressure of H2 = nRT/V
= 0.1206*0.08206*273.15/2.79=0.96889 atm
Pressure of H2 = 0.969 atm
Pressure of O2
= 0.0040*0.08206*273.15/2.79=0.03214 atm