# Nitrogen balloon: volume, number of moles, mass of nitrogen

A round balloon had a radius of 0.30 m & contains nitrogen (N2) at 30 degrees Celsius and one atmosphere total pressure. a) What is the volume of the nitrogen in the balloon?

b) How many moles of nitrogen are in the balloon?

c) What is the mass of the nitrogen in the balloon?

d) If the balloon is then submerged in a body of fresh water to a depth of 25 meters where the water temperature is 7 degrees Celsius, what is the volume of the nitrogen in the balloon at the depth of 25 meters?

https://brainmass.com/physics/ideal-gas-law/nitrogen-balloon-volume-number-of-moles-mass-of-nitrogen-40529

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Volume of Nitrogen:

radius of balloon= 0.3 m

Volume of balloon= (4/ 3 ) pi r ^3 = 0.1131 m^3

Therefore Volume of Nitrogen in the balloon= 0.1131 m^3

No of moles of Nitrogen:

Volume=V= 0.1131 m^3

Pressure=P= 1 atm= 101350 N / m^2

Temperature= T= 30 degrees C= 303.15 K =273.15+30

Ideal gas Law is PV=nRT

or n=PV/RT

R= Gas constant= 8.3145 J.mol ^-1K ^-1

Therefore

No of moles=n= 4.55 moles =(101350*0.1131)/(8.3145*303.15)

There are 4.55 moles of Nitrogen in the balloon

Mass of Nitrogen:

No of moles = 4.55 moles

1 mole of Nitrogen (N2)weighs 28 gms

Therefore 4.55 moles of Hydrogen weigh 127.4 gms =4.55*28

or 0.1274 kg

Volume of the nitrogen in the balloon at the depth of 25 meters:

Pressure at this depth = Atmospheric pressure + Pressure of water= h rhog

h= 25 m

rho= 1000 kg/m^3

g= 9.81 m/s^2

Pressure of water= 245250 N/m^2 =25*1000*9.81

Total pressure= 346600 N/m^2 =101350+245250

no of moles=n= 4.5500 moles

Pressure=P= 346600 N / m^2

Temperature= T= 7 degrees C= 280.15 K =273.15+7

Ideal gas Law is PV=nRT

or V=nRT/P

R= Gas constant= 8.3145 J.mol ^-1K ^-1

Therefore

Volume=V= 0.03060 m^3 =(4.55*8.3145*280.15)/346600

Volume of nitrogen= 0.03060 m^3

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