Question

Two moles of at temperature Y=7/5 T0 and 3 moles of CO2 Y=4/3 at temperature 2T0 are allowed to mix together in a closed adiabatic vessel. The resulting mixture finally comes in thermal equilibrium. Then

Answer 1

Given :

Number of moles of O₂ = 2

Number of moles of CO₂ =3

Temperature of O₂ = T₀

Temperature of CO₂ = 2T₀

To Find :

a) The final temperature of the mixture

b) The adiabatic exponent of the mixture formed

Solution :

• From the relation

    $n_{1} C_{v_{1} }T_{1} + n_{2} C_{v_{2}} T_{2} = (n_{1} +n_{2} )C_{v}T$

• By substituting the values in the relation

        $n_{1} C_{v_{1} }T_{1} + n_{2} C_{v_{2}} T_{2} = (n_{1} +n_{2} )C_{v}T$

        $2(\frac{5R}{2} )T_{0}+3(3R)2 T_{0}=5C_{v}T$

        $\frac{23}{5} RT_{0}= C_{v}T\rightarrow Equation(1)$

• From the  relation

       $C_{v} =\frac{n_{1}C_{v1} + n_{2}C_{v2}}{n_{1}+n_{2} }$

       $C_{v}=\frac{2\times \frac{5}{2}R + 3\times 2R }{3+2}$

       $C_{v}=\frac{14}{5} R$

       $C_{p} =\frac{19}{5} R$

       $r = \frac{C_{p}}{C_{v}}$

       $r = \frac{19}{14}$

The adiabatic exponent of the mixture is 19/14

• By substituting the value in equation(1)

        $\frac{23}{5} RT_{0}= C_{v}T$    

        $T = \frac{23}{14} T_{0}$

The final temperature of the mixture is (23/14)T₀