A mixture of n1 moles of monoatomic gas and n2 moles of diatomic gas has (Cp/CV)=γ=1.5 . Then

Question

A mixture of n1 moles of monoatomic gas and n2 moles of diatomic gas has (Cp/CV)=&gamma;=1.5 . Then (A)  n1=n2  (B)  2n1=n2  (C)  n1=2n2  (D)  2n1=3n2

Tardigrade Admin · Accepted Answer

fa v=( text total number of degrees of freedom / text total number of molecules ) =(n1 NA f1+n2 NA f2/n1 NA+n2 NA) =(n1 f1+n2 f2/n1+n2) =(3 n1+5 n2/n1+n2) Also &gamma;=1+(2/fa v)=1.5 Or fa v=4 ∴ (3 n1+5 n2/n1+n2)=4 Or n1=n2

Q. A mixture of $n_{1}$ moles of monoatomic gas and $n_{2}$ moles of diatomic gas has $\frac{C _{p}}{C _{V}} = γ = 1.5.$ Then

$n_{1} = n_{2}$

$2 n_{1} = n_{2}$

$n_{1} = 2 n_{2}$

$2 n_{1} = 3 n_{2}$

Q. A mixture of n1​ moles of monoatomic gas and n2​ moles of diatomic gas has CV​Cp​​=γ=1.5. Then

n1​=n2​

2n1​=n2​

n1​=2n2​

2n1​=3n2​

fav​= total number of molecules total number of degrees of freedom ​ =n1​NA​+n2​NA​n1​NA​f1​+n2​NA​f2​​ =n1​+n2​n1​f1​+n2​f2​​ =n1​+n2​3n1​+5n2​​ Also γ=1+fav​2​=1.5 Or fav​=4 ∴n1​+n2​3n1​+5n2​​=4 Or n1​=n2​

Q. A mixture of $n_{1}$ moles of monoatomic gas and $n_{2}$ moles of diatomic gas has $\frac{C _{p}}{C _{V}} = γ = 1.5.$ Then

$n_{1} = n_{2}$

$2 n_{1} = n_{2}$

$n_{1} = 2 n_{2}$

$2 n_{1} = 3 n_{2}$