Mixture of Gases Questions in English

(C) According to the ideal gas law,$PV = nRT$. Since the volume $V$,temperature $T$,and gas constant $R$ are constant,we have $P \propto n$.
Initial moles of nitrogen $(n_{N_2})$: $n_{N_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{35 \, kg}{28 \, g/mol} = \frac{35000 \, g}{28 \, g/mol} = 1250 \, mol$.
Using the proportionality $P_1/P_2 = n_1/n_2$:
$6/9 = 1250 / n_{total}$
$n_{total} = 1250 \times (9/6) = 1250 \times 1.5 = 1875 \, mol$.
The total moles $n_{total} = n_{N_2} + n_{O_2}$,so $n_{O_2} = 1875 - 1250 = 625 \, mol$.
Mass of oxygen supplied = $n_{O_2} \times \text{molar mass of } O_2 = 625 \, mol \times 32 \, g/mol = 20000 \, g = 20 \, kg$.

(A) Given: Volume $V = 0.02 \, m^3$,Temperature $T = 20 + 273 = 293 \, K$,Pressure $P = 2 \, atm = 2 \times 1.013 \times 10^5 \, Pa \approx 2.026 \times 10^5 \, Pa$. Total mass $M = 5 \, g$.
Let the mass of hydrogen be $m_H$ and the mass of helium be $m_{He}$. Then $m_H + m_{He} = 5 \, g$.
Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M_{mol}}$:
$PV = \left( \frac{m_H}{2} + \frac{m_{He}}{4} \right) RT$ (using molar masses $2 \, g/mol$ for $H_2$ and $4 \, g/mol$ for $He$ in $g/mol$ units).
$2.026 \times 10^5 \times 0.02 = \left( \frac{m_H}{2} + \frac{5 - m_H}{4} \right) \times 8.314 \times 293$.
$4052 = \left( \frac{2m_H + 5 - m_H}{4} \right) \times 2436$.
$4052 = \left( \frac{m_H + 5}{4} \right) \times 2436$.
$16208 = 2436m_H + 12180$.
$4028 = 2436m_H \implies m_H \approx 1.65 \, g$.
$m_{He} = 5 - 1.65 = 3.35 \, g$.
Ratio $m_H : m_{He} = 1.65 : 3.35 \approx 1 : 2$.

(C) The ideal gas law for a mixture is given by $PV = n_{total}RT$,where $n_{total} = \sum \frac{m_i}{M_i}$.
First,calculate the number of moles for each gas:
$n_{N_2} = \frac{28 \, g}{28 \, g/mol} = 1 \, mol$
$n_{H_2} = \frac{4 \, g}{2 \, g/mol} = 2 \, mol$
$n_{He} = \frac{8 \, g}{4 \, g/mol} = 2 \, mol$
Total moles $n_{total} = 1 + 2 + 2 = 5 \, mol$.
Total mass $m_{total} = 28 + 4 + 8 = 40 \, g = 0.04 \, kg$.
Using the ideal gas equation $PV = n_{total}RT$,we find the volume $V = \frac{n_{total}RT}{P}$.
$V = \frac{5 \times 8.3 \times 400}{8.3 \times 10^5} = \frac{2000}{10^5} = 0.02 \, m^3$.
Density $\rho = \frac{m_{total}}{V} = \frac{0.04 \, kg}{0.02 \, m^3} = 2 \, kg/m^3$.
Since $1 \, kg/m^3 = 1 \, g/litre$,the density is $2 \, g/litre$.

(B) The equivalent molar mass $(M_{mix})$ of a mixture of gases is given by the formula:
$M_{mix} = \frac{n_1 M_1 + n_2 M_2}{n_1 + n_2}$
Given:
$n_1 = 5 \ mol$ (Helium),$M_1 = 4 \ g/mol$
$n_2 = 2 \ mol$ (Hydrogen),$M_2 = 2 \ g/mol$
Substituting the values:
$M_{mix} = \frac{(5 \times 4) + (2 \times 2)}{5 + 2}$
$M_{mix} = \frac{20 + 4}{7}$
$M_{mix} = \frac{24}{7} \ g/mol$

(A) Helium $(He)$ is a monoatomic gas,so its degree of freedom $f_1 = 3$. The number of moles $n_1 = 5$.
Hydrogen $(H_2)$ is a diatomic gas,so its degree of freedom $f_2 = 5$. The number of moles $n_2 = 2$.
The equivalent degree of freedom $f_{mix}$ for a mixture of gases is given by the formula:
$f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$
Substituting the values:
$f_{mix} = \frac{(5 \times 3) + (2 \times 5)}{5 + 2}$
$f_{mix} = \frac{15 + 10}{7}$
$f_{mix} = \frac{25}{7} \approx 3.57$
Thus,the correct option is $A$.

(C) For a mixture of gases,the equivalent molar heat capacity at constant volume is given by $C_{V,mix} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$.
Helium is a monoatomic gas,so $C_{V1} = \frac{3}{2}R$. Hydrogen is a diatomic gas,so $C_{V2} = \frac{5}{2}R$.
Given $n_1 = 5 \ mol$ and $n_2 = 2 \ mol$,we have $C_{V,mix} = \frac{5(\frac{3}{2}R) + 2(\frac{5}{2}R)}{5 + 2} = \frac{7.5R + 5R}{7} = \frac{12.5R}{7}$.
The equivalent molar heat capacity at constant pressure is $C_{P,mix} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 + n_2}$.
For helium,$C_{P1} = \frac{5}{2}R$. For hydrogen,$C_{P2} = \frac{7}{2}R$.
$C_{P,mix} = \frac{5(\frac{5}{2}R) + 2(\frac{7}{2}R)}{5 + 2} = \frac{12.5R + 7R}{7} = \frac{19.5R}{7}$.
The adiabatic index $\gamma_{mix} = \frac{C_{P,mix}}{C_{V,mix}} = \frac{19.5R/7}{12.5R/7} = \frac{19.5}{12.5} = 1.56$.

(D) The internal energy of a gas is an extensive property,meaning it depends on the total amount of matter present.
When two non-reacting gases are mixed,the total internal energy of the resulting mixture is the sum of the internal energies of the individual components.
Given:
Internal energy of $He$ sample $(U_{He})$ = $100\ J$
Internal energy of $H_2$ sample $(U_{H_2})$ = $200\ J$
Since the gases do not interact chemically,the total internal energy of the mixture $(U_{mix})$ is:
$U_{mix} = U_{He} + U_{H_2}$
$U_{mix} = 100\ J + 200\ J = 300\ J$

(D) Since there is no loss of energy and no external work is done,the total internal energy of the system is conserved.
The internal energy of a monoatomic gas is given by $U = \frac{3}{2} n R T$.
Let $T$ be the final temperature of the mixture.
The total initial internal energy is $U_i = \frac{3}{2} n_1 R T_1 + \frac{3}{2} n_2 R T_2$.
The total final internal energy is $U_f = \frac{3}{2} (n_1 + n_2) R T$.
By the law of conservation of energy,$U_i = U_f$.
$\frac{3}{2} n_1 R T_1 + \frac{3}{2} n_2 R T_2 = \frac{3}{2} (n_1 + n_2) R T$.
Dividing both sides by $\frac{3}{2} R$,we get $n_1 T_1 + n_2 T_2 = (n_1 + n_2) T$.
Therefore,the final temperature is $T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.

(D) Number of moles of $N_2$: $n_{N_2} = \frac{7 \, kg}{28 \, kg/kmol} = 0.25 \, kmol$.
Degrees of freedom for $N_2$: $f_{N_2} = \frac{2}{\gamma_{N_2}-1} = \frac{2}{1.4-1} = 5$.
Number of moles of $CO_2$: $n_{CO_2} = \frac{11 \, kg}{44 \, kg/kmol} = 0.25 \, kmol$.
Degrees of freedom for $CO_2$: $f_{CO_2} = \frac{2}{\gamma_{CO_2}-1} = \frac{2}{1.3-1} = \frac{20}{3}$.
Equivalent molecular weight $M_{mix} = \frac{m_1 + m_2}{n_1 + n_2} = \frac{7 + 11}{0.25 + 0.25} = \frac{18}{0.5} = 36 \, kg/kmol$.
Equivalent degrees of freedom $f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{0.25(5) + 0.25(20/3)}{0.25 + 0.25} = \frac{5 + 20/3}{2} = \frac{35/3}{2} = \frac{35}{6}$.
Equivalent adiabatic index $\gamma_{mix} = 1 + \frac{2}{f_{mix}} = 1 + \frac{2}{35/6} = 1 + \frac{12}{35} = \frac{47}{35}$.
Thus,both $(A)$ and $(B)$ are correct.

(B) Since the container is insulated,$Q = 0$. Since the partition is removed without doing any work,$W = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W = 0$.
This implies that the total internal energy remains constant: $U_{initial} = U_{final}$.
For an ideal gas,the internal energy is $U = n C_v T$.
Thus,$n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T$.
Canceling $C_v$ from both sides,we get $T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.
Using the ideal gas law $PV = nRT$,we have $n_1 = \frac{P_1 V_1}{R T_1}$ and $n_2 = \frac{P_2 V_2}{R T_2}$.
Substituting these into the expression for $T$:
$T = \frac{(\frac{P_1 V_1}{R T_1}) T_1 + (\frac{P_2 V_2}{R T_2}) T_2}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}} = \frac{P_1 V_1 + P_2 V_2}{\frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}}$.
Simplifying the denominator:
$T = \frac{R(P_1 V_1 + P_2 V_2)}{\frac{P_1 V_1 T_2 + P_2 V_2 T_1}{T_1 T_2}} = \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1}$.

(C) The total internal energy of the mixture is conserved since there is no loss of energy.
The internal energy of a gas with $n$ molecules at temperature $T$ is given by $U = \frac{f}{2} n k_B T$,where $f$ is the degrees of freedom and $k_B$ is the Boltzmann constant.
For the mixture,the total energy is the sum of the energies of the individual gases:
$U_{total} = U_1 + U_2 + U_3$
$\frac{f}{2} (n_1 + n_2 + n_3) k_B T_{mix} = \frac{f}{2} n_1 k_B T_1 + \frac{f}{2} n_2 k_B T_2 + \frac{f}{2} n_3 k_B T_3$
Canceling the common terms $\frac{f}{2} k_B$ from both sides,we get:
$(n_1 + n_2 + n_3) T_{mix} = n_1 T_1 + n_2 T_2 + n_3 T_3$
Therefore,the final temperature of the mixture is:
$T_{mix} = \frac{n_1 T_1 + n_2 T_2 + n_3 T_3}{n_1 + n_2 + n_3}$

(B) According to the kinetic theory of gases,the average kinetic energy of a gas molecule depends only on the absolute temperature $T$ and is given by $K_{av} = \frac{3}{2} kT$. Since both gases are at the same room temperature,the average kinetic energy per molecule is the same for both $H_2$ and $Ar$. Thus,statement $(i)$ is correct.
For an ideal gas mixture in a closed container,the partial pressure $p_i$ of a component is given by $p_i = \frac{n_i}{n_{total}} P_{total}$. Since the number of moles of hydrogen $(n_{H_2} = 2)$ is greater than the number of moles of argon $(n_{Ar} = 1)$,the partial pressure of hydrogen will be greater than the partial pressure of argon. Thus,statement $(ii)$ is incorrect.

(A) The number of moles of $H_2$ is $n_1 = \frac{1 \, g}{2 \, g/mol} = 0.5 \, mol$.
The number of moles of $O_2$ is $n_2 = \frac{16 \, g}{32 \, g/mol} = 0.5 \, mol$.
Both $H_2$ and $O_2$ are diatomic gases,so their molar heat capacities at constant volume are equal $(C_{v1} = C_{v2} = \frac{5}{2}R)$.
The final temperature $T$ of the mixture is given by the principle of conservation of energy:
$n_1 C_v (T - T_1) + n_2 C_v (T - T_2) = 0$
$T = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$
Convert temperatures to Kelvin: $T_1 = 27 + 273 = 300 \, K$ and $T_2 = 37 + 273 = 310 \, K$.
$T = \frac{0.5 \times 300 + 0.5 \times 310}{0.5 + 0.5} = \frac{150 + 155}{1} = 305 \, K$.
Converting back to Celsius: $T = 305 - 273 = 32 \, ^oC$.

(C) For Helium $(He)$: Molar mass $M_1 = 4 \ g/mol$,degrees of freedom $f_1 = 3$,mass $m_1 = 8 \ g$. Number of moles $n_1 = 8/4 = 2 \ mol$.
For Oxygen $(O_2)$: Molar mass $M_2 = 32 \ g/mol$,degrees of freedom $f_2 = 5$,mass $m_2 = 16 \ g$. Number of moles $n_2 = 16/32 = 0.5 \ mol$.
The average degrees of freedom $f_{av}$ is given by:
$f_{av} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{(2 \times 3) + (0.5 \times 5)}{2 + 0.5} = \frac{6 + 2.5}{2.5} = \frac{8.5}{2.5} = \frac{17}{5} = 3.4$.
The ratio of specific heats $\gamma_{av} = 1 + \frac{2}{f_{av}} = 1 + \frac{2}{3.4} = 1 + \frac{20}{34} = 1 + \frac{10}{17} = \frac{27}{17}$.

(B) The total number of moles of the gas remains conserved when the two containers are joined.
The number of moles in the first container is $n_1 = \frac{P_1 V}{R T_1}$.
The number of moles in the second container is $n_2 = \frac{P_2 V}{R T_2}$.
After joining,the total volume becomes $2V$,and the gas reaches a common pressure $P$ and temperature $T$. The total number of moles is $n = n_1 + n_2 = \frac{P(2V)}{RT}$.
Equating the total moles:
$\frac{P_1 V}{R T_1} + \frac{P_2 V}{R T_2} = \frac{P(2V)}{RT}$
Dividing both sides by $V/R$:
$\frac{P_1}{T_1} + \frac{P_2}{T_2} = \frac{2P}{T}$
Rearranging for $P$:
$P = \frac{1}{2} \left( \frac{P_1}{T_1} + \frac{P_2}{T_2} \right) T$

(C) For a gas with $\gamma_1 = \frac{7}{5}$,the molar heat capacity at constant volume is $C_{v1} = \frac{R}{\gamma_1 - 1} = \frac{R}{\frac{7}{5} - 1} = \frac{5}{2} R$.
For a gas with $\gamma_2 = \frac{4}{3}$,the molar heat capacity at constant volume is $C_{v2} = \frac{R}{\gamma_2 - 1} = \frac{R}{\frac{4}{3} - 1} = 3R$.
For the mixture of $n_1 = 1 \, \text{mole}$ and $n_2 = 1 \, \text{mole}$,the equivalent molar heat capacity at constant volume is given by:
$C_{v, \text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{1 \cdot \frac{5}{2} R + 1 \cdot 3R}{1 + 1} = \frac{\frac{11}{2} R}{2} = \frac{11}{4} R$.
The molar heat capacity at constant pressure for the mixture is $C_{p, \text{mix}} = C_{v, \text{mix}} + R = \frac{11}{4} R + R = \frac{15}{4} R$.
Therefore,the adiabatic index for the mixture is $\gamma_{\text{mix}} = \frac{C_{p, \text{mix}}}{C_{v, \text{mix}}} = \frac{\frac{15}{4} R}{\frac{11}{4} R} = \frac{15}{11}$.

(B) For a process at constant pressure,the heat supplied is given by $Q = n C_p \Delta T$.
From the graph,the slope of the line is $C_p = \frac{Q}{n \Delta T} = \frac{2500 \text{ J}}{1 \text{ mol} \times 100 \text{ K}} = 25 \text{ J/(mol K)}$.
We know that $C_p = \frac{f+2}{2} R$,where $f$ is the degree of freedom.
Taking $R \approx 8.314 \text{ J/(mol K)}$,we have $25 = \frac{f+2}{2} \times 8.314$.
$\frac{f+2}{2} \approx 3.007 \Rightarrow f+2 \approx 6.014 \Rightarrow f \approx 4.014 \approx 4$.
The adiabatic index $\gamma$ is given by $\gamma = 1 + \frac{2}{f}$.
Substituting $f = 4$,we get $\gamma = 1 + \frac{2}{4} = 1 + 0.5 = 1.5$.

(B) The speed of sound in a gas is given by $v = \sqrt{\frac{\gamma P}{\rho}}$. Assuming $\gamma$ remains approximately constant for the mixture,$v \propto \frac{1}{\sqrt{\rho}}$.
Let the volume of oxygen be $V$. Then the volume of hydrogen is $4V$. The total volume is $5V$.
The molar mass of hydrogen $(H_2)$ is $2 \, g/mol$ and oxygen $(O_2)$ is $32 \, g/mol$.
The density $\rho$ is proportional to the molar mass $M$. The effective molar mass of the mixture $M_{mix}$ is given by:
$M_{mix} = \frac{n_1 M_1 + n_2 M_2}{n_1 + n_2} = \frac{4(2) + 1(32)}{4 + 1} = \frac{8 + 32}{5} = \frac{40}{5} = 8 \, g/mol$.
The density of hydrogen is proportional to its molar mass $M_{H_2} = 2 \, g/mol$.
Thus,the ratio of densities is $\frac{\rho_{mix}}{\rho_{H_2}} = \frac{8}{2} = 4$.
Using the relation $v_{mix} = v_{H_2} \times \sqrt{\frac{\rho_{H_2}}{\rho_{mix}}}$:
$v_{mix} = 1270 \times \sqrt{\frac{1}{4}} = 1270 \times \frac{1}{2} = 635 \, m/s$.

(B) For a mixture of gases,the adiabatic index $\gamma_{mix}$ is given by the formula:
$\gamma_{mix} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 C_{V1} + n_2 C_{V2}}$
Alternatively,using the degrees of freedom $f$:
$C_V = \frac{f}{2}R$ and $C_P = (\frac{f}{2} + 1)R$
For Helium (monoatomic),$f_1 = 3$,so $C_{V1} = \frac{3}{2}R$ and $C_{P1} = \frac{5}{2}R$.
For Hydrogen (diatomic),$f_2 = 5$,so $C_{V2} = \frac{5}{2}R$ and $C_{P2} = \frac{7}{2}R$.
Given $n_1 = 2$ and $n_2 = n$,the mixture ratio is $\frac{C_P}{C_V} = \frac{3}{2}$.
Using the relation $\frac{C_{P,mix}}{C_{V,mix}} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 C_{V1} + n_2 C_{V2}} = \frac{3}{2}$:
$\frac{2(\frac{5}{2}R) + n(\frac{7}{2}R)}{2(\frac{3}{2}R) + n(\frac{5}{2}R)} = \frac{3}{2}$
$\frac{5 + 3.5n}{3 + 2.5n} = \frac{3}{2}$
$10 + 7n = 9 + 7.5n$
$0.5n = 1$
$n = 2$.

(A) The total internal energy $U_{\text{total}}$ of the system is the sum of the internal energies of the individual gases.
For oxygen $(O_2)$,which is a diatomic gas,the degrees of freedom $f_1 = 5$ ($3$ translational + $2$ rotational).
The internal energy of oxygen is $U_{O_2} = n_1 \frac{f_1}{2} RT = 3 \times \frac{5}{2} RT = 7.5 \, RT$.
For argon $(Ar)$,which is a monatomic gas,the degrees of freedom $f_2 = 3$ ($3$ translational).
The internal energy of argon is $U_{Ar} = n_2 \frac{f_2}{2} RT = 5 \times \frac{3}{2} RT = 7.5 \, RT$.
The total internal energy is $U_{\text{total}} = U_{O_2} + U_{Ar} = 7.5 \, RT + 7.5 \, RT = 15 \, RT$.

(A) For helium (monoatomic gas),the degrees of freedom $f_1 = 3$.
For hydrogen (rigid diatomic gas),the degrees of freedom $f_2 = 5$.
The number of moles are $n_1 = 2$ and $n_2 = 3$.
The effective degrees of freedom for the mixture is given by $f_{\text{mix}} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$.
Substituting the values: $f_{\text{mix}} = \frac{2 \times 3 + 3 \times 5}{2 + 3} = \frac{6 + 15}{5} = \frac{21}{5} = 4.2$.
The molar specific heat at constant volume is $C_{v, \text{mix}} = \frac{f_{\text{mix}} R}{2}$.
Substituting the values: $C_{v, \text{mix}} = \frac{4.2 \times 8.3}{2} = 2.1 \times 8.3 = 17.43\, J/mol\, K$.
Rounding to the nearest option,the answer is $17.4\, J/mol\, K$.

(C) The number of moles of oxygen $(n_1 = 3)$ is diatomic,so its degrees of freedom $f_1 = 5$. The number of moles of helium $(n_2 = 2)$ is monatomic,so its degrees of freedom $f_2 = 3$.
The equivalent degrees of freedom for the mixture is given by $f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$.
Substituting the values: $f_{mix} = \frac{3 \times 5 + 2 \times 3}{3 + 2} = \frac{15 + 6}{5} = \frac{21}{5} = 4.2$.
The ratio of specific heats for the mixture is $\gamma_{mix} = 1 + \frac{2}{f_{mix}}$.
$\gamma_{mix} = 1 + \frac{2}{4.2} = 1 + \frac{20}{42} = 1 + \frac{10}{21} = \frac{31}{21} \approx 1.476$.
Rounding to the nearest provided option,the value is $1.5$.

(B) The final temperature of a mixture of non-reacting gases is given by the formula: $T_{\text{mix}} = \frac{n_1 C_{v1} T_1 + n_2 C_{v2} T_2}{n_1 C_{v1} + n_2 C_{v2}}$.
For $N_2$ (diatomic gas),$n_1 = 2 \text{ mole}$ and $C_{v1} = \frac{5}{2}R$. The temperature $T_1 = 27 + 273 = 300 \, K$.
For $He$ (monatomic gas),$n_2 = 1 \text{ mole}$ and $C_{v2} = \frac{3}{2}R$. The temperature $T_2 = -73 + 273 = 200 \, K$.
Substituting the values:
$T_{\text{mix}} = \frac{2 \times (\frac{5}{2}R) \times 300 + 1 \times (\frac{3}{2}R) \times 200}{2 \times (\frac{5}{2}R) + 1 \times (\frac{3}{2}R)}$
$T_{\text{mix}} = \frac{1500R + 300R}{5R + 1.5R} = \frac{1800R}{6.5R} = \frac{1800}{6.5} \approx 276.92 \, K \approx 277 \, K$.

(A) For a monoatomic gas,$C_{v1} = \frac{R}{\gamma_1 - 1} = \frac{R}{5/3 - 1} = \frac{3R}{2}$ and $C_{p1} = \gamma_1 C_{v1} = \frac{5}{3} \times \frac{3R}{2} = \frac{5R}{2}$.
For a diatomic gas,$C_{v2} = \frac{R}{\gamma_2 - 1} = \frac{R}{7/5 - 1} = \frac{5R}{2}$ and $C_{p2} = \gamma_2 C_{v2} = \frac{7}{5} \times \frac{5R}{2} = \frac{7R}{2}$.
For the mixture,the total heat capacity at constant volume is $C_{v, \text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{1 \times \frac{3R}{2} + 1 \times \frac{5R}{2}}{1 + 1} = \frac{4R}{2} = 2R$.
The total heat capacity at constant pressure is $C_{p, \text{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} = \frac{1 \times \frac{5R}{2} + 1 \times \frac{7R}{2}}{1 + 1} = \frac{6R}{2} = 3R$.
Thus,$\gamma_{\text{mix}} = \frac{C_{p, \text{mix}}}{C_{v, \text{mix}}} = \frac{3R}{2R} = 3/2$.

(B) For a mixture of gases,the molar specific heat at constant volume $(C_V)_{mix}$ is given by the formula:
$(C_V)_{mix} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$
Here,$n_1 = 1$ mole of $He$ (monatomic gas),so $C_{V1} = \frac{3}{2} R$.
$n_2 = 3$ moles of $O_2$ (diatomic gas),so $C_{V2} = \frac{5}{2} R$.
Substituting the values:
$(C_V)_{mix} = \frac{1 \times (\frac{3}{2} R) + 3 \times (\frac{5}{2} R)}{1 + 3}$
$(C_V)_{mix} = \frac{1.5 R + 7.5 R}{4}$
$(C_V)_{mix} = \frac{9 R}{4} = 2.25 R$.

(D) For an ideal gas,the internal energy is given by $U = U_0 + nC_vT$.
Given the relation $U = 2 + 2PV$.
Using the ideal gas equation $PV = nRT$,we substitute this into the given relation:
$U = 2 + 2nRT$.
Differentiating with respect to temperature $T$,we get:
$\frac{dU}{dT} = 2nR$.
Since $C_v = \frac{1}{n} \frac{dU}{dT}$,we have $C_v = 2R$.
For a monoatomic gas,$C_v = \frac{3}{2}R = 1.5R$.
For a diatomic gas,$C_v = \frac{5}{2}R = 2.5R$.
Since the calculated $C_v = 2R$ lies between $1.5R$ and $2.5R$,the gas must be a mixture of monoatomic and diatomic gases.

(A) For a monoatomic gas,$C_{v1} = \frac{3}{2}R$ and $C_{p1} = \frac{5}{2}R$. For a diatomic gas,$C_{v2} = \frac{5}{2}R$ and $C_{p2} = \frac{7}{2}R$.
Given $n_1 = 1$ mole and $n_2 = 1$ mole.
The equivalent molar heat capacity at constant volume for the mixture is $(C_v)_{\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{1(\frac{3}{2}R) + 1(\frac{5}{2}R)}{1 + 1} = \frac{4R}{2} = 2R$.
The equivalent molar heat capacity at constant pressure for the mixture is $(C_p)_{\text{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} = \frac{1(\frac{5}{2}R) + 1(\frac{7}{2}R)}{1 + 1} = \frac{6R}{2} = 3R$.
The adiabatic index for the mixture is $\gamma_{\text{mix}} = \frac{(C_p)_{\text{mix}}}{(C_v)_{\text{mix}}} = \frac{3R}{2R} = 1.5 = 3/2$.

(C) The specific heat at constant volume for a mixture is given by $C_{V, \text{mix}} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$.
Given $n_1 : n_2 = 1 : 2$,let $n_1 = 1$ and $n_2 = 2$.
For a gas with $f$ degrees of freedom,$C_V = \frac{f}{2} R$.
Substituting the values: $\frac{13}{6} R = \frac{1 \cdot (\frac{f_1}{2} R) + 2 \cdot (\frac{f_2}{2} R)}{1 + 2}$.
$\frac{13}{6} R = \frac{\frac{f_1}{2} R + f_2 R}{3} = \frac{f_1 + 2f_2}{6} R$.
Thus,$f_1 + 2f_2 = 13$.
For $He$ (monatomic),$f = 3$. For $N_2$ (diatomic),$f = 5$.
If $f_1 = 3$ $(He)$ and $f_2 = 5$ $(N_2)$,then $3 + 2(5) = 13$,which satisfies the equation.
Therefore,the gases are $He$ and $N_2$.

(A) The number of moles of $H_2$ is $n_1 = \frac{1\,g}{2\,g/mol} = 0.5\,mol$.
The number of moles of $O_2$ is $n_2 = \frac{16\,g}{32\,g/mol} = 0.5\,mol$.
The temperatures in Kelvin are $T_1 = 27 + 273 = 300\,K$ and $T_2 = 37 + 273 = 310\,K$.
Assuming the gases are ideal and the system is adiabatic,the final temperature $T$ is given by the weighted average based on the degrees of freedom. Since both $H_2$ and $O_2$ are diatomic gases,their degrees of freedom $f = 5$ are the same.
$T = \frac{n_1 f_1 T_1 + n_2 f_2 T_2}{n_1 f_1 + n_2 f_2} = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$.
$T = \frac{0.5 \times 300 + 0.5 \times 310}{0.5 + 0.5} = \frac{150 + 155}{1} = 305\,K$.
Converting back to Celsius: $T = 305 - 273 = 32\,^{\circ}C$.

(C) The total internal energy of the mixture is the sum of the internal energies of the individual gases.
For a gas,the internal energy is given by $U = \frac{3}{2} N k T$.
Total internal energy $U_{total} = U_1 + U_2 = \frac{3}{2} N_1 k T_1 + \frac{3}{2} N_2 k T_2$.
Let the final temperature of the mixture be $T$. The total internal energy of the mixture is $U_{final} = \frac{3}{2} (N_1 + N_2) k T$.
By the law of conservation of energy,$U_{total} = U_{final}$.
$\frac{3}{2} N_1 k T_1 + \frac{3}{2} N_2 k T_2 = \frac{3}{2} (N_1 + N_2) k T$.
Canceling the common factor $\frac{3}{2} k$ from both sides:
$N_1 T_1 + N_2 T_2 = (N_1 + N_2) T$.
Therefore,the resulting temperature is $T = \frac{N_1 T_1 + N_2 T_2}{N_1 + N_2}$.

(A) The $r.m.s.$ speed of a gas molecule is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since the temperature $T$ is the same for all gases in the mixture,$v_{rms} \propto \frac{1}{\sqrt{M}}$.
Given $m_1 > m_2 > m_3$,it follows that $(v_{rms})_1 < (v_{rms})_2 < (v_{rms})_3$.
The average kinetic energy of a gas molecule is given by $\bar K = \frac{3}{2} k_B T$.
Since the temperature $T$ is the same for all gases in the mixture,the average kinetic energy $\bar K$ is the same for all types of molecules,i.e.,$(\bar K)_1 = (\bar K)_2 = (\bar K)_3$.

(B) For gas $1$: $n_1 = 2$,$\gamma_1 = \frac{5}{3}$. Since $\gamma = 1 + \frac{2}{f}$,$f_1 = 3$. Thus,$C_{V_1} = \frac{3}{2}R$ and $C_{P_1} = \frac{5}{2}R$.
For gas $2$: $n_2 = 3$,$\gamma_2 = \frac{4}{3}$. Since $\gamma = 1 + \frac{2}{f}$,$f_2 = 6$. Thus,$C_{V_2} = \frac{6}{2}R = 3R$ and $C_{P_2} = 4R$.
The molar heat capacity at constant volume for the mixture is $C_{V_{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2} = \frac{2(\frac{3}{2}R) + 3(3R)}{2+3} = \frac{3R + 9R}{5} = \frac{12R}{5} = 2.4R$.
The molar heat capacity at constant pressure for the mixture is $C_{P_{mix}} = \frac{n_1 C_{P_1} + n_2 C_{P_2}}{n_1 + n_2} = \frac{2(\frac{5}{2}R) + 3(4R)}{2+3} = \frac{5R + 12R}{5} = \frac{17R}{5} = 3.4R$.
The ratio $\gamma_{mix} = \frac{C_{P_{mix}}}{C_{V_{mix}}} = \frac{17R/5}{12R/5} = \frac{17}{12} \approx 1.4167 \approx 1.42$.

(B) For a mixture of gases,the ratio of specific heats is given by $\frac{C_{P, \text{mix}}}{C_{V, \text{mix}}} = \frac{n_1 C_{P1} + n_2 C_{P2}}{n_1 C_{V1} + n_2 C_{V2}}$.
Helium is a monoatomic gas,so $C_{V1} = \frac{3R}{2}$ and $C_{P1} = \frac{5R}{2}$.
Oxygen is a diatomic rigid gas,so $C_{V2} = \frac{5R}{2}$ and $C_{P2} = \frac{7R}{2}$.
Given $n_1 = n$ and $n_2 = 2n$.
Substituting these values:
$\frac{C_{P, \text{mix}}}{C_{V, \text{mix}}} = \frac{n(\frac{5R}{2}) + 2n(\frac{7R}{2})}{n(\frac{3R}{2}) + 2n(\frac{5R}{2})}$
$= \frac{\frac{5nR}{2} + \frac{14nR}{2}}{\frac{3nR}{2} + \frac{10nR}{2}} = \frac{19nR/2}{13nR/2} = \frac{19}{13}$.

(N/A) The partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the vessel at the same volume and temperature. Since both gases are assumed to be ideal and share the same volume $V$ and temperature $T$,we use the ideal gas law $PV = \mu RT$,where $\mu$ is the number of moles.
For neon $(1)$ and oxygen $(2)$:
$P_1 V = \mu_1 RT$ and $P_2 V = \mu_2 RT$
Thus,$\frac{P_1}{P_2} = \frac{\mu_1}{\mu_2}$. Given $\frac{P_1}{P_2} = \frac{3}{2}$,we have $\frac{\mu_1}{\mu_2} = \frac{3}{2}$.
$(i)$ Since $\mu = \frac{N}{N_A}$,where $N$ is the number of molecules and $N_A$ is Avogadro's number:
$\frac{N_1}{N_2} = \frac{\mu_1}{\mu_2} = \frac{3}{2} = 1.5$.
$(ii)$ Mass density $\rho = \frac{m}{V}$. Since $\mu = \frac{m}{M}$ (where $M$ is molar mass),we have $m = \mu M$.
$\frac{\rho_1}{\rho_2} = \frac{m_1/V}{m_2/V} = \frac{m_1}{m_2} = \frac{\mu_1 M_1}{\mu_2 M_2} = \left(\frac{\mu_1}{\mu_2}\right) \times \left(\frac{M_1}{M_2}\right)$.
Substituting the values: $\frac{\rho_1}{\rho_2} = \frac{3}{2} \times \frac{20.2}{32.0} = 1.5 \times 0.63125 = 0.947$.

The pressure of an ideal gas is given by $P = \frac{1}{3} n m \langle v^2 \rangle$.
Consider a unit volume containing a mixture of non-reactive gases. Let the number densities of molecules be $n_1, n_2, \ldots$,their masses be $m_1, m_2, \ldots$,and their mean square speeds be $\langle v_1^2 \rangle, \langle v_2^2 \rangle, \ldots$.
The total pressure $P$ of the mixture is the sum of the partial pressures of the individual gases:
$P = \frac{1}{3} n_1 m_1 \langle v_1^2 \rangle + \frac{1}{3} n_2 m_2 \langle v_2^2 \rangle + \ldots$ ....$(1)$
In thermal equilibrium,the average kinetic energy of the molecules of each gas is equal:
$\langle \frac{1}{2} m_1 v_1^2 \rangle = \langle \frac{1}{2} m_2 v_2^2 \rangle = \ldots = \frac{3}{2} k_B T$ ....$(2)$
From equation $(2)$,we have:
$m_1 \langle v_1^2 \rangle = m_2 \langle v_2^2 \rangle = \ldots = 3 k_B T$
Substituting this into equation $(1)$:
$P = \frac{1}{3} [n_1 (3 k_B T) + n_2 (3 k_B T) + \ldots]$
$P = (n_1 + n_2 + \ldots) k_B T$
This represents Dalton's law of partial pressure,where $n = n_1 + n_2 + \ldots$ is the total number density of molecules.

(A) The average kinetic energy per molecule of an ideal gas is given by the formula $K_{avg} = \frac{3}{2} k_{B} T$,where $k_{B}$ is the Boltzmann constant and $T$ is the absolute temperature of the gas.
Since both gases are in the same container,they are in thermal equilibrium and must be at the same temperature $T$.
Because the average kinetic energy depends only on the temperature $T$ and not on the mass or nature of the gas molecules,both gases will have the same average kinetic energy per molecule.

(A) Oxygen $(O_{2})$ is a diatomic gas. Neglecting vibrational modes, its degrees of freedom $(f_{O_2})$ are $3$ (translational) $+ 2$ (rotational) $= 5$.
The internal energy of $n$ moles of an ideal gas is given by $U = n \cdot \frac{f}{2} RT$.
For $2.0$ moles of $O_{2}$:
$U_{O_2} = 2.0 \times \frac{5}{2} RT = 5 RT$.
Neon $(Ne)$ is a monoatomic gas. Its degrees of freedom $(f_{Ne})$ are $3$ (translational).
For $4.0$ moles of $Ne$:
$U_{Ne} = 4.0 \times \frac{3}{2} RT = 6 RT$.
The total internal energy of the system is the sum of the internal energies of the individual components:
$U_{total} = U_{O_2} + U_{Ne} = 5 RT + 6 RT = 11 RT$.

(1.60 ATM) Given:
$V_1 = 2.0 \, L, V_2 = 3.0 \, L$
$\mu_1 = 4.0 \, mol, \mu_2 = 5.0 \, mol$
$P_1 = 1.00 \, atm, P_2 = 2.00 \, atm$
For an ideal gas,the internal energy $U$ is given by $U = \frac{f}{2} PV$,where $f$ is the degrees of freedom.
Since the gases mix and reach thermal equilibrium,the total internal energy is conserved.
$U_{total} = U_1 + U_2$
$\frac{f}{2} P(V_1 + V_2) = \frac{f}{2} P_1 V_1 + \frac{f}{2} P_2 V_2$
Canceling $\frac{f}{2}$ from both sides,we get:
$P(V_1 + V_2) = P_1 V_1 + P_2 V_2$
Solving for the final pressure $P$:
$P = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2}$
Substituting the values:
$P = \frac{(1.00 \times 2.0) + (2.00 \times 3.0)}{2.0 + 3.0} \, atm$
$P = \frac{2.0 + 6.0}{5.0} \, atm$
$P = \frac{8.0}{5.0} \, atm = 1.60 \, atm$
The final pressure of the mixture is $1.60 \, atm$.

(N/A) In a gas mixture at thermal equilibrium,all molecules are at the same temperature. The average kinetic energy of a gas molecule is given by $K.E._{avg} = \frac{3}{2} k_B T$. Since $k_B$ is the Boltzmann constant and $T$ is the same for all molecules in the mixture,the average kinetic energy is the same for all: $(K.E.)_A = (K.E.)_B = (K.E.)_C$.
$(b)$ The $rms$ speed of a gas molecule is given by $v_{rms} = \sqrt{\frac{3 k_B T}{m}}$,where $m$ is the mass of the molecule. Since $v_{rms} \propto \frac{1}{\sqrt{m}}$,the molecule with the smallest mass will have the highest $rms$ speed. Given $m_A > m_B > m_C$,the decreasing order of $rms$ speeds is $(v_{rms})_C > (v_{rms})_B > (v_{rms})_A$.

(B) The internal energy $U$ of a gas mixture is given by the sum of the internal energies of its components: $U = U_1 + U_2$.
For an ideal gas,the internal energy is $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom.
Oxygen $(O_2)$ is a diatomic gas. Since the bond is rigid,its degrees of freedom $f_1 = 5$ ($3$ translational + $2$ rotational).
Argon $(Ar)$ is a monatomic gas,so its degrees of freedom $f_2 = 3$.
Given $n_1 = 3$ moles of oxygen and $n_2 = 5$ moles of argon.
Total internal energy $U = \frac{f_1}{2} n_1 RT + \frac{f_2}{2} n_2 RT$.
$U = \left( \frac{5}{2} \times 3 \times RT \right) + \left( \frac{3}{2} \times 5 \times RT \right)$.
$U = \frac{15}{2} RT + \frac{15}{2} RT = \frac{30}{2} RT = 15 RT$.
Thus,the total internal energy in units of $RT$ is $15$.

(D) Since the vessel is closed and insulated,no heat is exchanged with the surroundings and no work is done by or on the gas. Therefore,the total internal energy of the system remains conserved.
The internal energy of a monoatomic ideal gas is given by $U = n C_v T$,where $C_v = \frac{3}{2}R$.
Let $n_1 = 0.1$ mole at $T_1 = 200\, K$ and $n_2 = 0.05$ mole at $T_2 = 400\, K$.
By the principle of conservation of energy:
$n_1 C_v T_1 + n_2 C_v T_2 = (n_1 + n_2) C_v T_{final}$
Since $C_v$ is the same for both,it cancels out:
$n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_{final}$
$(0.1 \times 200) + (0.05 \times 400) = (0.1 + 0.05) T_{final}$
$20 + 20 = 0.15 \times T_{final}$
$40 = 0.15 \times T_{final}$
$T_{final} = \frac{40}{0.15} = \frac{4000}{15} = \frac{800}{3} \approx 266.67\, K$.

(A) $1$. Calculate the number of moles for each gas:
$n_{1} (CO_{2}) = \frac{11 \, g}{44 \, g/mol} = 0.25 \, mol$
$n_{2} (N_{2}) = \frac{14 \, g}{28 \, g/mol} = 0.50 \, mol$
$2$. Identify the degrees of freedom $(f)$ and adiabatic index $(\gamma)$ for each gas:
$CO_{2}$ is a polyatomic gas,$f_{1} = 6$,$\gamma_{1} = 1 + \frac{2}{f_{1}} = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.33$
$N_{2}$ is a diatomic gas,$f_{2} = 5$,$\gamma_{2} = 1 + \frac{2}{f_{2}} = 1 + \frac{2}{5} = 1.4$
$3$. Use the formula for the adiabatic index of a mixture:
$\frac{n_{1} + n_{2}}{\gamma_{\text{mix}} - 1} = \frac{n_{1}}{\gamma_{1} - 1} + \frac{n_{2}}{\gamma_{2} - 1}$
$4$. Substitute the values:
$\frac{0.25 + 0.50}{\gamma_{\text{mix}} - 1} = \frac{0.25}{\frac{4}{3} - 1} + \frac{0.50}{1.4 - 1}$
$\frac{0.75}{\gamma_{\text{mix}} - 1} = \frac{0.25}{1/3} + \frac{0.50}{0.4}$
$\frac{0.75}{\gamma_{\text{mix}} - 1} = 0.75 + 1.25 = 2.0$
$5$. Solve for $\gamma_{\text{mix}}$:
$\gamma_{\text{mix}} - 1 = \frac{0.75}{2.0} = 0.375$
$\gamma_{\text{mix}} = 1.375 = \frac{11}{8}$

(C) For $N_{2}$ (diatomic gas),the degree of freedom $f_{1} = 5$. Molar mass $M_{1} = 28 \, g/mol$. Number of moles $n_{1} = 7/28 = 1/4 \, mol$.
For $Ar$ (monatomic gas),the degree of freedom $f_{2} = 3$. Molar mass $M_{2} = 40 \, g/mol$. Number of moles $n_{2} = 20/40 = 1/2 \, mol$.
$C_{V,mix} = \frac{n_{1}C_{V,1} + n_{2}C_{V,2}}{n_{1} + n_{2}} = \frac{\frac{1}{4}(\frac{5}{2}R) + \frac{1}{2}(\frac{3}{2}R)}{\frac{1}{4} + \frac{1}{2}} = \frac{\frac{5}{8}R + \frac{3}{4}R}{\frac{3}{4}} = \frac{\frac{11}{8}R}{\frac{3}{4}} = \frac{11}{6}R$.
$C_{P,mix} = \frac{n_{1}C_{P,1} + n_{2}C_{P,2}}{n_{1} + n_{2}} = \frac{\frac{1}{4}(\frac{7}{2}R) + \frac{1}{2}(\frac{5}{2}R)}{\frac{1}{4} + \frac{1}{2}} = \frac{\frac{7}{8}R + \frac{5}{4}R}{\frac{3}{4}} = \frac{\frac{17}{8}R}{\frac{3}{4}} = \frac{17}{6}R$.
Ratio $\gamma_{mix} = \frac{C_{P,mix}}{C_{V,mix}} = \frac{17/6 R}{11/6 R} = \frac{17}{11}$.

(C) The total pressure $P$ of a mixture of ideal gases is given by the ideal gas equation $PV = n_{total} RT$,where $n_{total} = n_1 + n_2 + n_3$.
First,calculate the number of moles for each gas:
$n_{O_2} = \frac{16 \, g}{32 \, g/mol} = 0.5 \, mol$
$n_{N_2} = \frac{28 \, g}{28 \, g/mol} = 1.0 \, mol$
$n_{CO_2} = \frac{44 \, g}{44 \, g/mol} = 1.0 \, mol$
Total moles $n_{total} = 0.5 + 1.0 + 1.0 = 2.5 \, mol = \frac{5}{2} \, mol$.
Substituting into the ideal gas equation:
$PV = (2.5) RT$
$P = \frac{5}{2} \frac{RT}{V}$.

(B) Let the final temperature of the mixture be $T$. Since there is no loss of energy,the total internal energy of the system is conserved.
The internal energy of a gas is given by $U = \frac{F}{2} n R T$,where $n$ is the number of moles. Since $n$ is proportional to the number of molecules $N$,we can write $U = \frac{F}{2} N k_{B} T$.
For the mixture,the total energy before mixing equals the total energy after mixing:
$U_{1} + U_{2} = U_{mix}$
$\frac{F_{1}}{2} n_{1} k_{B} T_{1} + \frac{F_{2}}{2} n_{2} k_{B} T_{2} = \frac{F_{1}}{2} n_{1} k_{B} T + \frac{F_{2}}{2} n_{2} k_{B} T$
Canceling the common terms $\frac{1}{2} k_{B}$ from both sides:
$F_{1} n_{1} T_{1} + F_{2} n_{2} T_{2} = T (F_{1} n_{1} + F_{2} n_{2})$
Solving for $T$:
$T = \frac{F_{1} n_{1} T_{1} + F_{2} n_{2} T_{2}}{F_{1} n_{1} + F_{2} n_{2}}$

(C) Using the ideal gas law, the total number of moles $n_{total} = n_1 + n_2 = 3.0 + 4.0 = 7.0 \, \text{moles}$.
The total volume $V_{total} = V_1 + V_2 = 4.5 + 5.5 = 10.0 \, L$.
Assuming the temperature $T$ remains constant (as the problem implies a simple mixing process), we use the relation $P_{final} V_{total} = (n_1 + n_2) RT$.
From the initial states:
$P_1 V_1 = n_1 RT \Rightarrow (2.0)(4.5) = 3.0 RT \Rightarrow 9.0 = 3.0 RT \Rightarrow RT = 3.0$.
$P_2 V_2 = n_2 RT \Rightarrow (3.0)(5.5) = 4.0 RT \Rightarrow 16.5 = 4.0 RT \Rightarrow RT = 4.125$.
Since the temperatures are different, we use the conservation of internal energy: $U_1 + U_2 = U_{mix}$.
$U = \frac{f}{2} nRT = \frac{f}{2} PV$.
$\frac{f}{2} P_1 V_1 + \frac{f}{2} P_2 V_2 = \frac{f}{2} P_{final} (V_1 + V_2)$.
$(2.0)(4.5) + (3.0)(5.5) = P_{final} (4.5 + 5.5)$.
$9.0 + 16.5 = P_{final} (10.0)$.
$25.5 = 10.0 P_{final} \Rightarrow P_{final} = 2.55 \, \text{atm}$.
Given $P_{final} = x \times 10^{-1} \, \text{atm}$, we have $2.55 = x \times 10^{-1} \Rightarrow x = 25.5$.

(C) Given:
Volume $V = 4.0 \times 10^{-3} \, m^3$
Total number of moles $n = 1 \, mol \, (H_2) + 2 \, mol \, (CO_2) = 3 \, mol$
Temperature $T = 400 \, K$
Gas constant $R = 8.3 \, J \, mol^{-1} \, K^{-1}$
Using the ideal gas equation $PV = nRT$,we can find the pressure $P$:
$P = \frac{nRT}{V}$
Substituting the values:
$P = \frac{3 \times 8.3 \times 400}{4.0 \times 10^{-3}}$
$P = \frac{9960}{4.0 \times 10^{-3}}$
$P = 2490 \times 10^3 \, Pa = 24.9 \times 10^5 \, Pa$
Therefore,the correct option is $C$.

(C) Using the ideal gas equation $PV = nRT$,where $n$ is the total number of moles.
Given: $P = 400 \times 10^3 \, Pa$,$V = 500 \times 10^{-6} \, m^3$,$T = 300 \, K$,$R = 8.314 \, J/(mol \cdot K) \approx 25/3 \, J/(mol \cdot K)$.
$400 \times 10^3 \times 500 \times 10^{-6} = n \times (25/3) \times 300$
$200 = n \times 2500$
$n = 200/2500 = 2/25 = 0.08 \, mol$.
Let $m_H$ be the mass of hydrogen and $m_O$ be the mass of oxygen.
$m_H + m_O = 0.76 \, g$.
The total number of moles $n = n_H + n_O = m_H/2 + m_O/32 = 0.08$.
Multiplying by $32$: $16m_H + m_O = 0.08 \times 32 = 2.56$.
We have the system:
$1) \, m_H + m_O = 0.76$
$2) \, 16m_H + m_O = 2.56$
Subtracting $(1)$ from $(2)$: $15m_H = 1.80 \implies m_H = 1.80/15 = 0.12 \, g$.
Then $m_O = 0.76 - 0.12 = 0.64 \, g$.
The ratio of mass of oxygen to hydrogen is $m_O/m_H = 0.64/0.12 = 64/12 = 16/3$.

(B) Using the ideal gas equation $PV = nRT$,where $n$ is the total number of moles.
Given: $P = 100 \; kPa = 10^{5} \; Pa$,$V = 2000 \; cm^{3} = 2 \times 10^{-3} \; m^{3}$,$T = 300 \; K$,$R = 8.314 \; J/(mol \cdot K) \approx 25/3 \; J/(mol \cdot K)$.
Total moles $n = \frac{PV}{RT} = \frac{10^{5} \times 2 \times 10^{-3}}{(25/3) \times 300} = \frac{200}{2500} = 0.08 \; mol$.
Let $n_{1}$ be the moles of $H_{2}$ and $n_{2}$ be the moles of $O_{2}$.
$n_{1} + n_{2} = 0.08$ (Equation $1$).
The total mass is $m = n_{1}M_{1} + n_{2}M_{2} = 0.76 \; g$.
$2n_{1} + 32n_{2} = 0.76$ (Equation $2$).
From Equation $1$,$n_{1} = 0.08 - n_{2}$.
Substitute into Equation $2$: $2(0.08 - n_{2}) + 32n_{2} = 0.76$.
$0.16 - 2n_{2} + 32n_{2} = 0.76 \implies 30n_{2} = 0.60 \implies n_{2} = 0.02 \; mol$.
Then $n_{1} = 0.08 - 0.02 = 0.06 \; mol$.
The ratio $n_{1}/n_{2} = 0.06/0.02 = 3/1$.

(B) The speed of sound in a gas is given by $v_s = \sqrt{\frac{\gamma RT}{M}}$.
The root-mean-square speed of gas molecules is $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Given that $v_{rms} = \sqrt{2} v_s$,we have $\frac{v_s}{v_{rms}} = \frac{1}{\sqrt{2}}$.
Substituting the formulas: $\sqrt{\frac{\gamma}{3}} = \frac{1}{\sqrt{2}} \Rightarrow \frac{\gamma}{3} = \frac{1}{2} \Rightarrow \gamma = \frac{3}{2} = 1.5$.
For a mixture,the adiabatic index is $\gamma = 1 + \frac{2}{f_{mix}}$,where $f_{mix}$ is the effective degree of freedom.
For Helium (monoatomic),$f_1 = 3$. For Hydrogen (diatomic),$f_2 = 5$.
$f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{2 \times 3 + n \times 5}{n + 2} = \frac{6 + 5n}{n + 2}$.
Substituting $f_{mix}$ into the $\gamma$ formula: $\gamma = 1 + \frac{2(n + 2)}{6 + 5n} = \frac{6 + 5n + 2n + 4}{6 + 5n} = \frac{7n + 10}{5n + 6}$.
Equating to $1.5$: $\frac{7n + 10}{5n + 6} = \frac{3}{2} \Rightarrow 14n + 20 = 15n + 18$.
Solving for $n$: $n = 2$.