Question

Each of the four tires of Coach Carson's "ride" is filled with a different gas, to a tire gauge of 3.0 atm at 83 F The mass of the gas for each of the tires is as follows: Tire 1: Air (80.0 % N2 and 20.0% O2), = 116.0 grams; ire 2 Neon =80.7 grams; Tire 3: Helium = 16.0 grams; and Tire 4: an unknown gas 160.0 grams. The barometric pressure is 1.0 atm. Which tire contains the largest number of molecules/atoms? a) What is the molar mass of the unknown gas? b) What is the name of the unknown gas? c)

Answer 1

Answer - Given , Mass of Tire 1 = 116 g (Air)

Tire 2= 80.7 g(Neon)

Tire 3 =16 g (Helium)

Tire 4= 160 g (unknown)

Pressure =3.0 atm

Temperature = 83 F

for tire 2

the no. of molecules/atoms = (80.7/20.18)*6.02 *10²³ =24.07 * 10²³ atoms/ molecules

no. of moles = 3.99 moles

also PV=nRT

3 *V = 3.99 * 0.08206 * 301.48

V = 32.90 L it will be same for all tires

Now,

(a) We know that.

PV=nRT

where, P = Pressure

V = Volume (32.90L calculated above)

n = no. of moles

R = gas constant = 0.08206 L atm K^-1mol^-1

T = temperature (in kelvin) =301.483

3*32.90 = n *0.08206* 301.483

n =3.989 moles

So no. of atoms in 0.124 moles = 3.989 * 6.02 * 10²³ = 24.02 * 10²³ atoms

the no. of atoms/ molecules is 24.02 *10²³ atoms/ molecules and is same for all the tires.

(b) for tire 4 we have to find the molecular mass of the unknown gas

The no. of atoms/ molecules is same i.e. 24.02 *10²³ atoms/ molecules

  

We know that,

No. of atoms/ molecules = (mass/ molecular mass) * 6.02 *10²³

Put all the values

24.02* 10²³ = (160/ molecular mass) * 6.02 * 10²³

molecular mass = (160 * 6.02) / 24.02

= 40.09 u

So, the molecular mass of unkown gas is 40.09u

(c) The molecular mass of Argon is 39.948u which almost equal to our answer 40.09u. So, the unknown gas in the Tire 4 must be ARGON.