Mixture of Gases Questions in English

(C) The ideal gas equation is given by $pV = nRT$.
For oxygen: $p_1 = 1 \,atm$,$V_1 = 1 \,L$. The number of moles is $n_{O_2} = \frac{p_1 V_1}{RT} = \frac{1 \times 1}{RT} = \frac{1}{RT}$.
For nitrogen: $p_2 = 0.5 \,atm$,$V_2 = 2 \,L$. The number of moles is $n_{N_2} = \frac{p_2 V_2}{RT} = \frac{0.5 \times 2}{RT} = \frac{1}{RT}$.
When these gases are introduced into a vessel of volume $V_{mix} = 1 \,L$ at the same temperature $T$,the total number of moles is $n_{mix} = n_{O_2} + n_{N_2} = \frac{1}{RT} + \frac{1}{RT} = \frac{2}{RT}$.
The final pressure $p_{mix}$ is given by $p_{mix} = \frac{n_{mix} RT}{V_{mix}} = \frac{(2/RT) \times RT}{1} = 2 \,atm$.

(D) For an ideal gas mixture, the final temperature $T$ is given by the conservation of internal energy. Since both gases are monoatomic, the molar heat capacity at constant volume $C_V$ is the same for both.
The formula for the final temperature is:
$T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$
Given:
$n_1 = 4 \,mol$, $T_1 = 400 \,K$
$n_2 = 2 \,mol$, $T_2 = 700 \,K$
Substituting the values:
$T = \frac{4(400) + 2(700)}{4 + 2}$
$T = \frac{1600 + 1400}{6}$
$T = \frac{3000}{6}$
$T = 500 \,K$

(B) For a mono-atomic gas,the degree of freedom $f_1 = 3$. For a diatomic gas,the degree of freedom $f_2 = 5$.
The number of moles are $n_1 = 3$ and $n_2 = 1$.
The degree of freedom for the mixture is given by:
$f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{3 \times 3 + 1 \times 5}{3 + 1} = \frac{9 + 5}{4} = \frac{14}{4} = 3.5$.
The adiabatic index $\gamma$ for the mixture is given by:
$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{3.5} = 1 + \frac{2}{7/2} = 1 + \frac{4}{7} = \frac{11}{7}$.

(C) Let $n_1$ be the number of moles of $CO_2$ and $n_2$ be the number of moles of $O_2$.
From the ideal gas equation,$PV = nRT$,where $n = n_1 + n_2$.
Given: $P = 100 \text{ kPa} = 10^5 \text{ Pa}$,$V = 8310 \text{ cm}^3 = 8.31 \times 10^{-3} \text{ m}^3$,$T = 300 \text{ K}$,$R = 8.31 \text{ J/mol.K}$.
Calculating total moles $n = n_1 + n_2 = \frac{PV}{RT} = \frac{10^5 \times 8.31 \times 10^{-3}}{8.31 \times 300} = \frac{100}{300} = \frac{1}{3} \approx 0.333 \text{ mol}$.
The total mass is $m = M_1 n_1 + M_2 n_2 = 44 n_1 + 32 n_2 = 13.2 \text{ g}$.
We have the system of equations:
$1) n_1 + n_2 = 0.333$
$2) 44 n_1 + 32 n_2 = 13.2$
From $(1)$,$n_2 = 0.333 - n_1$. Substituting into $(2)$:
$44 n_1 + 32(0.333 - n_1) = 13.2$
$44 n_1 + 10.656 - 32 n_1 = 13.2$
$12 n_1 = 2.544$
$n_1 = 0.212 \approx 0.21 \text{ mol}$.
$n_2 = 0.333 - 0.212 = 0.121 \approx 0.12 \text{ mol}$.
Thus,the number of moles of $CO_2$ and $O_2$ are $0.21$ and $0.12$ respectively. The correct option is $C$.

(D) The temperature of the mixture $T_{mix}$ is given by the formula $T_{mix} = \frac{n_1 C_{v1} T_1 + n_2 C_{v2} T_2}{n_1 C_{v1} + n_2 C_{v2}}$.
Since both gases are monoatomic,their molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.
Substituting the values into the formula: $T_{mix} = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.
Given $n_1 = 2, T_1 = T$ and $n_2 = 6, T_2 = 2T$.
$T_{mix} = \frac{2 \cdot T + 6 \cdot 2T}{2 + 6}$.
$T_{mix} = \frac{2T + 12T}{8} = \frac{14T}{8}$.
$T_{mix} = \frac{7}{4}T$.

(B) Using the ideal gas equation,the number of moles are $n_1 = \frac{P_1 V_1}{R T_1}$ and $n_2 = \frac{P_2 V_2}{R T_2}$.
For the mixture,the total number of moles is $n_{mix} = n_1 + n_2 = \frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2} = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{R T_1 T_2}$.
From the ideal gas law for the mixture,$P V = n_{mix} R T_{mix}$,so $T_{mix} = \frac{P V}{n_{mix} R}$.
Substituting $n_{mix}$,we get $T_{mix} = \frac{P V R T_1 T_2}{R (P_1 V_1 T_2 + P_2 V_2 T_1)} = \frac{P V T_1 T_2}{P_1 V_1 T_2 + P_2 V_2 T_1}$.