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
B
2
C
3
D
4

Solution

## The correct option is B 2As for an ideal gas, CP−CV=R and γ=(CPCV) So, CV=R(γ−1) ∴(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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