
Concept explainers
Interpretation: -Volume of a gas and partial pressure of each gas in a mixture needs to be calculated using variable values given.
Concept Introduction: - For volume calculation first of all one have to calculate number of moles (n ) of each gas.
n = massmolar mass
Then, volume is calculated by Ideal gas equation
PV=nRTV=nPRT
For calculating partial pressure, Dalton’s law of partial pressure is used
PT=P1+P2+P3+P4

Answer to Problem 45A
Volume of the mixture is 15.01mL
Partial pressure of gases:
PO2=0.22atmPN2=0.25atmPCO2=0.16atmPNe=0.35atm
Explanation of Solution
There is a relation between volume of gas and molecules of gas present in mixture which is expressed by
V∝N
Now,
V=nNA { NA = Avogadro Number}
n∝N
After calculating number of moles, volume is calculated.
No. of moles (n) = massmolar mass
Putting the values,
No.ofmolesofO2=532=0.15moln1=0.15mol.
Also,
molesofN2=528.014=0.17moln2=0.17mol.
Or,
molesofCO2=544.01=0.11moln3=0.11mol.
Now,
molesofNe=520.17=0.24moln4=0.24mol.
Therefore,
nT=n1+n2+n3+n4nT=0.15+0.17+0.11+0.24nT=0.67mol.
Volume of mixture of gases will be:
PV=nTRTV=nTRTP
Putting the values,
V=0.67mol1atm×0.0821atmLKmol×273KV=0.67×0.0821×273V=15.01L
Volume of the mixture of gas is 15.01 L
For calculating Partial pressure of gases present in mixture, Dalton law is applied
PT=P1+P2+P3+P4
For calculating partial pressure of individual gases.
PPT=nnT
Where P = Partial Pressure of gas
PT = Total pressure
n = no. of moles of gas
nT = Total moles.
1) Partial Pressure of Oxygen
PO2PT=nO2nTPO2=no2nT×PTPO2=0.150.67×1atm=0.22atm
2) Partial Pressure of H2
PH2PT=nH2nTPH2=nH2nT×PTPH2=0.170.67×1atm=0.25atm
1) Partial Pressure of CO2
PCO2PT=nCO2nTPCO2=nCO2nT×PTPCO2=0.110.67×1atm=0.16atm
4) Partial Pressure of Ne
PNePT=nNenTPNe=nNenT×PTPO2=0.240.67×1atm=0.35atm
Partial pressure of gases:
a)PO2=0.22atmb)PN2=0.25atmc)PCO2=0.16atmd)PNe=0.35atm
Volume of mixture = 15.01L
Chapter 13 Solutions
World of Chemistry, 3rd edition
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