Sapting Learting A mixture of methane gas, CH4(g), and pentane gas, CsHg), has a

Question

Sapting Learting A mixture of methane gas, CH4(g), and pentane gas, CsHg), has a pressure of 0.5285 bar when placed in a sealed container. The complete combustion of the mixture to carbon dioxide gas, CO-kg), and water vapor, H20(9), was achieved by adding exactly enough oxygen gas, O-4g), to the container. The pressure of the product mixture in the sealed container is 2.486 bar. Calculate the mole fraction of methane in the initial mixture assuming the temperature and volume remain constant. Number Xcu

Explanation / Answer

given:

Before combustion of methane total pressure in the container, Pi=p(CH4)+p(C5H12)=0.5285 bar..........(1)

After combustion ,Pf =p(CO2)+p(H2O)+p(C5H12)=2.486 bar.............(2)

Substracting Equation (2) from (1) ,

p(CO2)+p(H2O)-p(CH4)=1.9575 bar.............(3)

CH4+2O2 --->CO2+2H2O

mole fraction of CH4=XCH4=p(CH4)/Pi

or, p(CH4)=XCH4*Pi [dalton's law of partial pressures]

mol fraction of CO2=xCO2=p(CO2)/Pf

mol fraction of H2O=XH2O=p(H2O)/Pf

But ,1 mol CH4 produced ,2mol of H2O and 1 mol CO2 are produced

So, XCH4=2(XH2O)=X(CO2)

from eqn (3),p(CO2)+p(H2O)-p(CH4)=1.9575 bar

or,  XCO2*Pf+XH2O*Pf- XCH4*Pi=1.9575 bar

or, XCH4*Pf+2*XCH4*Pf-XCH4*Pi=1.9575 bar

or,3XCH4*(2.486 bar)-XCH4*(0.5285 bar)=1.9575 bar

or, 6.9295 bar*XCH4=1.9575 bar

XCH4=1.9575/6.9295=0.282

XCH4=0.3

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