Two moles of an ideal gas with $$\dfrac{C_P}{C_V}=\dfrac{5}{3}$$ are mixed with $$3$$ moles of another ideal gas with $$\dfrac{C_P}{C_V}=\dfrac{4}{3}$$. The value of $$\dfrac{C_P}{C_V}$$ for the mixture is :
Correct option is C. $$1.42$$
Let $$n_1,n_2$$ be the number of moles of gases $$1$$ and $$2$$.
We have $$\dfrac{C_{p_1}}{C_{v_1}}=\dfrac{5}{3}$$ and We know $$C_{p_1}=C_{v_1}+R$$
$$\therefore \dfrac{5C_{v_1}}{3}=C_{v_1}+R,\dfrac{2C_{v_1}}{3}R_1,C_{v_1}=\dfrac{3R}{2}$$
$$A$$ and $$C_{p_1}=C_{ v_ 1 }+R=\dfrac{5R}{2}$$
Also, $$\dfrac{C_{ p_ 2}}{C_{ v_ 2 }}=\dfrac{4}{3}$$ and
$$C_{p_2}=C_{v_2}+R$$
$$\therefore \dfrac{4C_{ v_ 2 }}{3}=C_{ v_ 2 }+R, \dfrac{C_{ v_ 2 }}{3}=R,C_{ v_ 2 }=3R$$
$$C_{ p_ 2}=C_{ v_ 2 }+ R=4R$$
Now $$C_{ v_ {mix} }=\dfrac{n_1C_{ v_ 1 }+n_2C_{ v_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{3R}{2}+3\times 3R}{2+3}$$
$$=\dfrac{3R+9R}{5}=\dfrac{12R}{5}$$
$$C_{ p_ {mix} }=\dfrac{n_1C_{ p_ 1 }+n_2C_{ p_ 2 }}{n_1+n_2}=\dfrac{2\times \dfrac{5R}{2}+3\times 4R}{2+3}$$
$$=\dfrac{17R}{5}$$
$$l_{mix}=\dfrac{C_{ p_ {mix} }}{C_{ v_ {mix} }}=\dfrac{17}{12}=1.4166=1.42$$
Option (C) is correct.