Considering the gases to be ideal,the value o | vedclass

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(B) For a mixture of gases,the equivalent adiabatic exponent $\gamma_{mix}$ is given by the relation:$\frac{
u_{mix}}{\gamma_{mix} - 1} = \frac{
u_1

Vedclass · Accepted Answer

$1.33$

Vedclass · Answer

(B) For a mixture of gases,the equivalent adiabatic exponent $\gamma_{mix}$ is given by the relation:$\frac{
u_{mix}}{\gamma_{mix} - 1} = \frac{
u_1}{\gamma_1 - 1} + \frac{
u_2}{\gamma_2 - 1}$Where $
u_{mix} = 
u_1 + 
u_2$.Given:$
u_1 = 2 \ moles$ $(O_2)$,$\gamma_1 = 1.4$$
u_2 = 3 \ moles$ $(CO_2)$,$\gamma_2 = 1.3$$
u_{mix} = 2 + 3 = 5 \ moles$Substituting the values:$\frac{5}{\gamma_{mix} - 1} = \frac{2}{1.4 - 1} + \frac{3}{1.3 - 1}$$\frac{5}{\gamma_{mix} - 1} = \frac{2}{0.4} + \frac{3}{0.3}$$\frac{5}{\gamma_{mix} - 1} = 5 + 10 = 15$$\gamma_{mix} - 1 = \frac{5}{15} = \frac{1}{3}$$\gamma_{mix} = 1 + \frac{1}{3} = 1.33$.

Considering the gases to be ideal,the value of $\gamma = \frac{C_p}{C_v}$ for a gaseous mixture consisting of $3 \ moles$ of carbon dioxide $(CO_2)$ and $2 \ moles$ of oxygen $(O_2)$ will be $(\gamma_{O_2} = 1.4, \gamma_{CO_2} = 1.3)$.

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