How to solve this?

Question

Calculate the density of Helium, He, in grams per liter at 21 degrees C and 752 mmHg. The density of air under these conditions is 1.188 g/L. What is the difference in mass between 1 liter of air and liter of helium? (This mass difference is equivalent to the buoyant, or lifting foce of helium per liter.)

Answer 1

Density is g/L.

 

Assume you have exactly one mole of He; determine the mass and the volume independently, then divide to get density.

 

You can figure out the mass in grams of one mole of He if you multiply by the molar mass (4.0026 g/mol) from the periodic table.

 

m = (1 mol) x (4.0026 g/mol) = 4.0026 g

 

You can figure out the volume in liters by using the ideal gas law (PV=nRT), solving for volume and plugging in the pressure and temperature data, the one mole, and the gas law constant.  It's important to use units in this calculation, because they're easily mixed up.  I'll use R = 0.08206 L atm / K mol.  To match the units in R, the temperature needs to be converted from Celsius to Kelvin and the pressure from mm Hg to atm.  We can do these calculations separately or put them all into the rearranged gas law equation.  I'll do the former.  

 

T = 21 C + 273 = 294 K 

 

P = 752 mm Hg x (1 atm / 760 mm Hg) = 0.989 atm

 

V = nRT/P = (1 mole) x (0.08206 L atm / K mol) x (294 K) / (0.989 atm) = 24.38 L

 

Now we can use the mass and volume to compute the density.

 

D = m/V = (4.0026 g) / (24.38 L) = 0.164 g/L

 

In these calculations, it's best not to round anything until the end. The final answer will have three sig figs, because the temperature and pressure limit it to three. (An easy mistake is to think the temperature is only 2 sig figs. Review the sig-fig rule for addition and subtraction to see why it's three.)

 

The difference between the masses of helium and air in one liter is:

 

1.188 g - 0.164 g = 1.024 g

 

Four significant figure are justified in this answer, again, because of the rule for addition and subtraction.

 

 

Answer 2

Use avogadro principle

 

at STP (273K, 1 atm, 760mm Hg)

 

1 molecular gram wt = 22.4 liters

 

At 21 degree C and 752 mm Hg volume will be different , we can calculate that as follows:

 

P1 V1/T1  = P2 V2/T2

 

760 x 22.4/273 = 752 x V2/(273+21)

 

V2 = 24.38

 

Molecular gram wt of helium = 4

 

density = 4/24.38 = 0.1641 g/liter

 

Difference of mass = 1.188 - 0.1641 = 1.0239 grams

:)