A fuel gas is known to contain methane, ethane, and carbon monoxide. A sample of the gas is charged into an initially evacuated 2.000-liter vessel at 25°C and 2323 mm Hg absolute. The vessel is weighed before and after being charged, and the mass difference is found to be 4.929 g. Next, the higher heating value of the gas is determined in a calorimeter to be 841.9 kJ/mol. Calculate the molar composition of the fuel gas. Calculate the lower heating value of the gas. State in your own words the physical significance of the calculated value.

Explanation:

Solution

Let n1, n2, n3 be the moles of ethane, CO and methane respectively, then total number of moles n,

n =n1+n2-n3

Then

n = PV/RT

substituting the values gives

n= 0.25 moles

total wt of gas mixture = 4.929 g

Molecular wt of mixture MWmix = 4.929.0.25 = 19.716 g/ mol

if M1 , M2, M3 be the molecular wt of  ethane, CO and methane

M1 = 30.07, M2 = 28.01, M3 = 28.01

But

MWmix = y1 * M1 + y2 * M2 + y3* M3

y1 + y2 + y3 = 1

where y1 = n1/n, y2 = n2/n, y3=n3/n

Solving gives y1 (moles of ethane) =0 .103

Solving gives y2 (moles of CO) = 0.193

Solving gives y1 (moles of Methane) =0 .704

Therefore gas mixture is

ethane = 10.3 %

CO = 19.3 %

Methane = 70.4 %

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.