Question

Two moles of ${O_2}(\gamma = \dfrac{7}{5})$ at temperature ${T_0}$and 3 moles of $C{O_2}(\gamma = \dfrac{4}{3})$ at temperature $2{T_0}$ are allowed to mix together in a rigid closed adiabatic vessel. The resulting mixture finally comes in thermal equilibrium. Then,
(A) Final temperature of the mixture is $\dfrac{{23{T_0}}}{{14}}$
(B) Final temperature of the mixture is $\dfrac{{31{T_0}}}{{19}}$
(C) Adiabatic exponent of the mixture formed is $\dfrac{{14}}{5}$
(D) Adiabatic exponent of the mixture formed is $\dfrac{{19}}{{14}}$

Answer 1

In the question, it is mentioned that the reaction takes place in adiabatic vessel. In adiabatic processes, there is no exchange of heat between the surrounding and the system. Hence, no heat will leave the vessel.

Formula used:
$Q = n{C_v}\Delta T$

Complete step by step answer:
We know in adiabatic processes, there is no heat exchange. In case some amount of heat is produced, it does not go into the surrounding instead it is used to increase the temperature of the system and incase some heat is absorbed, then the system does not take heat from the surrounding instead it is used in lowering the temperature.
We know that amount of heat is given by $Q = n{C_v}\Delta T$where n is the number of moles , ${C_v}$is the specific heat of the gas at constant volume and$\Delta T$is the temperature change
Keeping these in mind, at equilibrium, the sum of the amount of heat of oxygen and carbon dioxide will be equal to the amount of heat of the mixture produced. So,
${n_1}{C_p}_1\Delta {T_1} + {n_2}{C_p}_2\Delta {T_2} = ({n_1} + {n_2}){C_p}\Delta T$----- (1)
For 2 moles of ${O_2}(\gamma = \dfrac{7}{5})$, ${C_v} = \dfrac{1}{2}fR,f = \dfrac{2}{{\gamma - 1}} \Rightarrow f = 5,{C_v} = \dfrac{{5R}}{2}$
And 3 moles of $C{O_2}(\gamma = \dfrac{4}{3})$ , ${C_v} = \dfrac{1}{2}fR,f = \dfrac{2}{{\gamma - 1}} \Rightarrow f = 6,{C_v} = 3R$
Here R is the gas constant. Let the final temperature of mixture be T and the specific heat capacity be ${C_v}(mix)$
On substituting these values equation (1) becomes,
$2 \times \dfrac{{5R}}{2} \times {T_0} + 3 \times (3R) \times 2{T_0} = (5) \times {C_v}(mix) \times T$
We know that when two gases are mixed the ${C_v}(mix) = \dfrac{{{n_1}{C_{v1}} + {n_2}{C_{v2}}}}{{{n_1} + {n_2}}} = \dfrac{{2 \times \dfrac{{5R}}{2} + 3 \times 3R}}{{2 + 3}} = \dfrac{{14R}}{5}$
Putting this
$5R \times {T_0} + 18R{T_0} = 5 \times \dfrac{{14R}}{5} \times T \Rightarrow 23R{T_0} = 14RT \Rightarrow T = \dfrac{{23{T_0}}}{{14}}$
${C_P}(mix) = {C_V}(mix) + R = \dfrac{{14R}}{5} + R = \dfrac{{19R}}{5}$(This equation is known as Mayer’s equation)
Adiabatic exponent is nothing but the ratio of specific heats or we can say $\gamma$
Therefore $\gamma = \dfrac{{{C_p}}}{{{C_v}}} = \dfrac{{19}}{{14}}$

Hence, the correct options are A and D.

Note:
Degree of freedom tells us how many independent motions a particle can do.This includes motion along the three axes as well as the rotational motion done by the particle. $\gamma$ is inversely proportional to the degree of freedom.