Kinetic Theory

JEE Main level, single correct, medium

111 of 168

Question 111 of 168

Two moles of helium are mixed with n moles of hydrogen. If CpCv=32\dfrac{{C}_{p}}{{C}_{v}}=\dfrac{3}{2} for the mixture then the value of nn is,

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Answer: (C) 2

γmix=1+2fmix{\gamma }_{mix}=1+\dfrac{2}{{f}_{mix}}

CpCv=fmix+2fmix=32\dfrac{{C}_{p}}{{C}_{v}}=\dfrac{{f}_{mix}+2}{{f}_{mix}}=\dfrac{3}{2}

⇒fmix=4\Rightarrow {f}_{mix}=4

fmix=n1f1+nf2n1+n2{f}_{mix}=\dfrac{{n}_{1}{f}_{1}+{n}_{}{f}_{2}}{{n}_{1}+{n}_{2}}

⇒4=2×3+n×52+n\Rightarrow 4=\dfrac{2\times 3+n\times 5}{2+n}

⇒n=2mole\Rightarrow n=2mole

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