Question #2e2b3

2 Answers
May 15, 2015

The number of moles is #~~1.10# and the mass is #~~4.36# #g#

That's a Chemistry question and you need to use the ideal gas law to find the number of moles #n#.

#pV = nRT#, where #p# is the pression in Pascal, #V# the volume in #m^3#, #R# is the ideal gas constant (#=8.134[J/(K*mol)]#), #T# the temperature in Kelvin.

Here, we have a ballon with a radius of #r = 18 cm = 0.18m#.
The volume is given by #V = 4/3 pi r^3~~0.0078pi# #m^3#

The pression #p = 1.05# #atm#. We know that #1# #atm = 101325# #Pa#.

Therefore, by the Rule of Three :

#p = (1.05*101325)/1 = 106391.25# #Pa#

For the temperature, #0°# is #273.15# #K#, so #20°# is #293.15# #K#.

#n = (pV)/(RT) = (106391.25*0.0078pi)/(8.134*293.15) ~~ 1.10# #mol#.

For the mass, #m# #[g] = n * M#, where M is the molar mass.

The molar mass of #He# is #4 [g/(mol)]#.

Therefore, #m ~~ 1.10*4~~4.36# #g#.

May 15, 2015

The number of moles of helium will be equal to 1.1 and the mass of helium will be 4.4 g.

SIDE NOTE Small correction - helium cannot exist as a diatomic molecule, so the correct notation is #"He"#, not #"He"_2#.

There's nothing wrong with the method used in the other answer, but I want to show you how you can get the number of moles of helium without doing that many conversions.

You that your balloon has a radius of 18 cm. You can determine its volume in #"cm"^3# by

#V_"balloon" = 4/3 * pi * "radius"^3#

#V_"balloon" = 4/3 * pi * ("18 cm")^3 = "24429 cm"^3#

Use the fact that #"1 L" = "1 dm"^3# to express the volume in liters

#24429cancel("cm"^3) * ("1 L")/(10^3cancel("cm"^3)) = "24.429 L"#

Now use the ideal gas law equation to solve for #n# by using #R# expressed in L atm/mol K

#PV = nRT => n = (PV)/(RT)#

#n = (1.05cancel("atm") * 24.429cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 20)cancel("K")) = color(green)("1.1 moles")#

To get the mass of helium, use its molar mass, which expresses the mass of 1 mole of helium

#1.1cancel("moles He") * "4 g"/(1cancel("mole He")) = color(green)("4.4 g")#