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Question

A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with γ=53 and some amount of gas B with γ=75 at a temperature T. The gases A and B do not react with each other and are assumed to be ideal. Find the number of moles of the gas B if γ for the gaseous mixture is (1913).

A
1
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B
2
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C
3
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D
4
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Solution

The correct option is B 2
As for an ideal gas, CPCV=R and γ=(CPCV)
So, CV=R(γ1)
(CV)1=R(53)1=32R;
& (CV)2=R(75)1=52R
Also, (CV)mix=R(1913)1=136R

Since both the gases do not react,
ΔUmix=ΔU1+ΔU2
i.e. (n1+n2)(CV)mixΔT=n1(CV)1+n2(CV)2ΔT
(CV)mix=n1(CV)1+n2(CV)2n1+n2
Substituting the data,
136R=1×32R+n252R1+n2=(3+5n2)R2(1+n2)
or 13+13n2=9+15n2
or n2=2 is the no. of moles of gas B.

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