Can I get detailed answer for this one Thanks Show that for

Can I get detailed answer for this one? Thanks

Show that, for a mixture of Ideal Gases, the mol fraction of each component of the mixture is equal to the partial pressure of that component divided by the total pressure of the mixture. Petrol (C_8 H_15) is burned at an air:fuel ratio (AFR) of 15:1. Assuming perfect combustion, calculate the mass of CO_2 produced per gram of fuel burned. A passenger car consumes 5.6 litres of petrol (C_8H_15) per 100 km travelled. Taking the density of petrol to be 740 grams per litre, calculate the vehicle \'s CO_2 emissions in grams of CO_2 per km travelled. Briefly explain the role and operating principle of a three-way catalytic converter (TWC) in the exhaust system of a spark-ignition engine.

Solution

1. first of all , we need to consider the Dalton\'s law of patial pressure statement:

P_t is the total pressure of mixture of gases

P₁,P₂...... are the partial pressure of the mixture of gases.

P_t=P₁+P₂+P₃.....................(1)

if the gas is alone or independent then we can consider ideal gas law,

pv=n*R*t..................(2)

therefore for first component we need to take number of mole =n1,n2,n3....

P₁V₁₌n₁*R*T₁..............(3)

P₁₌n_1*R*T₁/V₁...............(4)

same for second component

P₂=n₂*R*T₂/V₂..............(5)

total pressure is sum of partial pressures

P_t=P₁+P₂+...........=n₁*R*T₁/V₁+n₂*R*T₂/V₂+............(6)

now consider the each component is divided by total pressure

P₁/P_t=n₁*R*T₁/n_t*R*T_t*V_t/V₁..................(7)

P₁/P_t=n₁/n_t=X1.............(8)

P₁/P_t=X1

hence the given condition is satisfied

4. three way catalytic converter :

the exhaust system of si engines contain harmful gases like hydrocarbons,nitrogen,cabon monoxide etc.

this three way converter is used to harmful gases into harmless gases.

the three way caalytc converter uses two methods

1.reduction catalyst

2.oxidation catalyst

the reduction catalyst uses platinum and rhodium and oxidation catalyst uses platinum and palladium to convert gases.

stage -1: the exhaust gases are first sent to reduction catalyst then it converts oxides of nitrogen into nitrogen and oxygen

the reactions are :

no->n2+2o2

2no2->n2+2o2

stage -2: now the oxidation catalyst converts hydrocarbons and carbon monoxide in carbon dioxide and water

2co+o2->2co2

HC+o2->co2+h2o

in this way it converts the all three harmful gases into harmless gases.

2. in general the air fuel ratio is the ratio of air present in fuel in the combustion.

A/F=mass of air/mass of fuel

A/F=number of moles of air*moleculaar weight of air/number of moles of fuel*molecular weight of fuel

c₈h₁₅+(n+m/4)(o₂+3.78 N₂)=8co₂+m/2h₂o

n=8

m=15

A/F=11.75(2*16+3.78*28)/mass of fuel

A/F=15

mass of fuel per 100 gm burnt=107.97=108gm

3.co2 emission per km

first we need to calculate tco2 for that we have to calculate kg of co2 per 100 gm

for this we can just calculate the kgco2 from the formula

litre travelled per 100 km distance with kgco2 value of petrol taken from standard table

kgco2 of petrol=5.6*0.2=1.12 kg

now we need to calculate the co2 emission factor

density of the fuel *44/12=coef

coef=740*44/12=2713.3

now co2 emission=

co2 emission factor=1.12*2713.3=3038.8

3038.8 is the co2 emission factor per 100 km distance travelled