Two ideal monoatomic and diatomic gases are mixed with one another to form an ideal gas mixture. The equation of the adiabatic process of the mixture is PVγ = constant, where γ =(11/7) If n1 and n2 are the number of moles of the monoatomic and diatomic gases in the mixture respectively, find the ratio (n1/n2)

Question

Two ideal monoatomic and diatomic gases are mixed with one another to form an ideal gas mixture. The equation of the adiabatic process of the mixture is PV&gamma; = constant, where &gamma; =(11/7) If n1 and n2 are the number of moles of the monoatomic and diatomic gases in the mixture respectively, find the ratio (n1/n2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

&gamma; m i x=(n1 C p1 + n2 C p2/n1 C v1 + n2 C v2) (11/7)=(n1 ((5/2) R) + n2 ((7/2) R)/n1 ((3)2 R) + n2 ((5)2 R) (n1/n2)=3

γmix​=n1​Cv1​+n2​Cv2​n1​Cp1​+n2​Cp2​​ 711​=n1​(23​R)+n2​(25​R)n1​(25​R)+n2​(27​R)​ n2​n1​​=3

$γ_{mi x} = \frac{n _{1} C p _{1} + n _{2} C p _{2}}{n _{1} C v _{1} + n _{2} C v _{2}}$
$\frac{11}{7} = \frac{n _{1} ( \frac{5}{2} R ) + n _{2} ( \frac{7}{2} R )}{n _{1} ( \frac{3}{2} R ) + n _{2} ( \frac{5}{2} R )}$
$\frac{n _{1}}{n _{2}} = 3$