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Question

A gaseous mixture enclosed in a vessel consists of 1g mole of gas A with (γ1=53) and another gas B with (γ2=75) at a temperature T. The gases A and B do not react with each other and assume to be ideal. Find the number of gram moles of the gas B, if γ for the gaseous mixture is (1913).

A
μ2=2gmol
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B
μ2=2.2gmol
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C
μ2=2.1gmol
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D
μ2=2.6gmol
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Solution

The correct option is A μ2=2gmol
As for ideal gas CpCv=R and γ=(CpCv),
γ1=RCv or Cv=R(γ1)
(Cv)1=R(53)1=32R
(Cv)2=R(75)1=5R2
(Cv)mix=R(1913)1=136R
Now from the conservation of energy, i.e., ΔU=ΔU1+ΔU2, we get
(μ1+μ2)(Cv)mixΔT=[μ1(Cv)1+μ2(Cv)2]ΔT
(Cv)mix=μ1(Cv)1+μ2(Cv)2μ1+μ2
136R=1×32R+μ2×52R1+μ2=(3+5μ2)R2(1+μ2)
Or 13+13μ2=9+15μ2, i.e., μ2=2gmol

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