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

Answer 1

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