Co 4) what is the total mass ofoxygen gas in a room measuring 10.0 ft x 8.0 ft x
ID: 1025068 • Letter: C
Question
Co 4) what is the total mass ofoxygen gas in a room measuring 10.0 ft x 8.0 ft x 8.0 n ifthe air, which is composed of 29.95 % oxygen, is measured at STP? 5) A gas cylinder with a capacity of 105 L contains helium at a pressure of 66 atm at a temperature of 27 C. All of the gas in the cylinder is transferred to a weather balloon and at 20 miles of altitude the volume of the balloon reaches 2100 m' with an internal pressure of 2.6x10 atm. What must be the ambient temperature at 20 miles altitude?Explanation / Answer
4) volume of room = 10 X 8 X 8 = 640 ft3
1ft3 = 28.3168 liter then 640 ft3 = 640 X 28.3168 = 18122.75 liter
room contain air = 18122.75 liter
At STP 1 mole of gas occupy volume = 22.414 liter then
18122.75 liter = 18122/22.414 = 808.5 mole
total mole of gas = 808.5 mole
808.5 mole air = 100 % gas then 29.95 % oxygen =
29.95 X 808.5 / 100 = 242.1 mole
room contain 242.1 mole oxygen
molar mass of O2 = 32 g/mole
gm of compound = no. of mole X molar mass
gm of O2 = 242.1 X 32 = 7748.6 gm
total mass of oxygen in room = 7748.6 gm
5) first calculate mole of helium by using ideal gas equation
Ideal gas equation
PV = nRT where, P = atm pressure= 66 atm,
V = volume in Liter = 105 L
n = number of mole = ?
R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,
T = Temperature in K = 270C = 273.15+ 27 = 300.15 K
We can write ideal gas equation
n = PV/RT
Substitute the value
n = (66 X 105)/(0.08205 X 300.15) = 281.4 mole
gas in cylinder = 281.4 mole
now calculate tempreture of gas in baloon by using ideal gas equation
Ideal gas equation
PV = nRT where, P = atm pressure= 2.6 X10-3 atm
V = volume in Liter = 2100 m3 = 2100000 liter
n = number of mole = 281.4 mol
R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,
T = Temperature in K = ?
We can write ideal gas equation
T = PV/nR
substitute the value in above equation
T = (2.6 X10-3 X 2100000)/(281.4 X 0.08205)
T = 236.5 K
236.5K = 236.5 - 273.15 = -36.65 0C
tempreture of baloon = -36.650C
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