When 1 mole of a monatomic gas is mixed with 3 moles of a diatomic gas, the value of adiabatic exponent γ for the mixture is:

Question

When 1 mole of a monatomic gas is mixed with 3 moles of a diatomic gas, the value of adiabatic exponent &gamma; for the mixture is: (A) 5/3 (B) 1.5 (C) 1.4 (D) 13/9

Tardigrade Admin · Accepted Answer

(n1+n2/&gamma;-1) =(n1/&gamma;1-1)+(n2/&gamma;2-1) (1+3/&gamma;-1) =(1/(5)3-1)+(3/(7)5-1) (4/&gamma;-1) =(3/2)+(15/2)=9 or &gamma;-1 =(4/9) ⇒ &gamma;=1+(4/9)=(13/9)

Q. When 1 mole of a monatomic gas is mixed with 3 moles of a diatomic gas, the value of adiabatic exponent $γ$ for the mixture is:

5/3

1.5

1.4

13/9

$\frac{n _{1} + n _{2}}{γ - 1} = \frac{n _{1}}{γ _{1} - 1} + \frac{n _{2}}{γ _{2} - 1}$
$\frac{1 + 3}{γ - 1} = \frac{1}{\frac{5}{3} - 1} + \frac{3}{\frac{7}{5} - 1}$
$\frac{4}{γ - 1} = \frac{3}{2} + \frac{15}{2} = 9$
or $γ - 1 = \frac{4}{9}$
$\Rightarrow γ = 1 + \frac{4}{9} = \frac{13}{9}$

Q. When 1 mole of a monatomic gas is mixed with 3 moles of a diatomic gas, the value of adiabatic exponent γ for the mixture is:

5/3

1.5

1.4

13/9

γ−1n1​+n2​​=γ1​−1n1​​+γ2​−1n2​​ γ−11+3​=35​−11​+57​−13​ γ−14​=23​+215​=9 or γ−1=94​ ⇒γ=1+94​=913​

Q. When 1 mole of a monatomic gas is mixed with 3 moles of a diatomic gas, the value of adiabatic exponent $γ$ for the mixture is:

$\frac{n _{1} + n _{2}}{γ - 1} = \frac{n _{1}}{γ _{1} - 1} + \frac{n _{2}}{γ _{2} - 1}$
$\frac{1 + 3}{γ - 1} = \frac{1}{\frac{5}{3} - 1} + \frac{3}{\frac{7}{5} - 1}$
$\frac{4}{γ - 1} = \frac{3}{2} + \frac{15}{2} = 9$
or $γ - 1 = \frac{4}{9}$
$\Rightarrow γ = 1 + \frac{4}{9} = \frac{13}{9}$