Mixture of Gases Questions in English

(D) The molar specific heat at constant volume for a mixture is given by the formula: $C_{V_{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$.
For a monoatomic gas,$C_{V_1} = \frac{3}{2} R$ and $n_1 = 1$.
For a diatomic gas without vibrational modes,$C_{V_2} = \frac{5}{2} R$ and $n_2 = 3$.
Substituting these values into the formula:
$C_{V_{mix}} = \frac{1 \cdot (\frac{3}{2} R) + 3 \cdot (\frac{5}{2} R)}{1 + 3} = \frac{\frac{3}{2} R + \frac{15}{2} R}{4} = \frac{\frac{18}{2} R}{4} = \frac{9 R}{4}$.
Given that $C_{V_{mix}} = \frac{\alpha^2}{4} R$,we equate the two expressions:
$\frac{9 R}{4} = \frac{\alpha^2}{4} R$.
This implies $\alpha^2 = 9$,so $\alpha = 3$ (taking the positive value).

(B) For a mixture of gases,the adiabatic index $\gamma_{\text{mix}}$ is given by the formula:
$\gamma_{\text{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}}$
Given $n_1 = 1$ mole,$\gamma_1 = 5/3$ and $n_2 = n$ moles,$\gamma_2 = 7/5$.
We know $C_v = \frac{R}{\gamma - 1}$ and $C_p = \frac{\gamma R}{\gamma - 1}$.
Substituting these into the formula:
$\gamma_{\text{mix}} = \frac{n_1 \frac{\gamma_1 R}{\gamma_1 - 1} + n_2 \frac{\gamma_2 R}{\gamma_2 - 1}}{n_1 \frac{R}{\gamma_1 - 1} + n_2 \frac{R}{\gamma_2 - 1}} = \frac{n_1 \frac{\gamma_1}{\gamma_1 - 1} + n_2 \frac{\gamma_2}{\gamma_2 - 1}}{n_1 \frac{1}{\gamma_1 - 1} + n_2 \frac{1}{\gamma_2 - 1}}$
For gas $A$: $\frac{\gamma_1}{\gamma_1 - 1} = \frac{5/3}{2/3} = 2.5$ and $\frac{1}{\gamma_1 - 1} = \frac{1}{2/3} = 1.5$.
For gas $B$: $\frac{\gamma_2}{\gamma_2 - 1} = \frac{7/5}{2/5} = 3.5$ and $\frac{1}{\gamma_2 - 1} = \frac{1}{2/5} = 2.5$.
Substituting values: $\frac{19}{13} = \frac{1(2.5) + n(3.5)}{1(1.5) + n(2.5)}$.
$19(1.5 + 2.5n) = 13(2.5 + 3.5n)$.
$28.5 + 47.5n = 32.5 + 45.5n$.
$2n = 4 \Rightarrow n = 2$.

(B) For a thermally isolated system,the total internal energy remains constant because $\Delta Q = 0$ and $\Delta W = 0$.
Since the initial temperatures of both boxes are the same $(T)$,the final equilibrium temperature $T_f$ will also be $T$.
Using the ideal gas law $PV = nRT$,we can find the number of moles in each box:
$n_1 = \frac{P_1 V_1}{RT} = \frac{1 \times V}{RT} = \frac{V}{RT}$
$n_2 = \frac{P_2 V_2}{RT} = \frac{0.5 \times 4V}{RT} = \frac{2V}{RT}$
When the valve is opened,the total number of moles $n_{total} = n_1 + n_2$ occupies the total volume $V_{total} = V_1 + V_2 = V + 4V = 5V$ at temperature $T$.
$n_{total} = \frac{V}{RT} + \frac{2V}{RT} = \frac{3V}{RT}$
Using the ideal gas law for the final state:
$P_f V_{total} = n_{total} R T$
$P_f (5V) = \left(\frac{3V}{RT}\right) RT$
$P_f (5V) = 3V$
$P_f = \frac{3}{5} \, atm = 0.6 \, atm$.

(C) The internal energy $U$ of a gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.
For $He$ (monatomic gas),degrees of freedom $f_{He} = 3$.
For $O_2$ (diatomic gas),degrees of freedom $f_{O_2} = 5$.
For $O_3$ (non-linear polyatomic gas),degrees of freedom $f_{O_3} = 6$.
Given:
$n_{He} = 2$
$n_{O_2} = 4$
$n_{O_3} = 1$
The total internal energy of the mixture is the sum of the internal energies of individual components:
$U_{total} = U_{He} + U_{O_2} + U_{O_3}$
$U_{total} = \left( n_{He} \cdot \frac{f_{He}}{2} + n_{O_2} \cdot \frac{f_{O_2}}{2} + n_{O_3} \cdot \frac{f_{O_3}}{2} \right) RT$
$U_{total} = \left( 2 \cdot \frac{3}{2} + 4 \cdot \frac{5}{2} + 1 \cdot \frac{6}{2} \right) RT$
$U_{total} = (3 + 10 + 3) RT$
$U_{total} = 16 RT$.
Thus,the internal energy is $16 RT$.

(C) The molar mass of $CO$ is $28 \, g/mol$. Number of moles of $CO$ $(n_1)$ $= 14/28 = 0.5 \, mol$.
The molar mass of $O_2$ is $32 \, g/mol$. Number of moles of $O_2$ $(n_2)$ $= 16/32 = 0.5 \, mol$.
Both $CO$ and $O_2$ are diatomic gases. For diatomic gases,the degree of freedom $(f)$ is $5$ (vibration mode neglected).
The final temperature $T_f$ is given by the conservation of internal energy: $n_1 C_{v1} T_1 + n_2 C_{v2} T_2 = (n_1 C_{v1} + n_2 C_{v2}) T_f$.
Since $C_v = \frac{f}{2}R$ and $f$ is the same for both,the equation simplifies to: $n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_f$.
Convert temperatures to Kelvin: $T_1 = 27 + 273 = 300 \, K$ and $T_2 = 47 + 273 = 320 \, K$.
Substituting the values: $(0.5 \times 300) + (0.5 \times 320) = (0.5 + 0.5) T_f$.
$150 + 160 = 1 \times T_f$.
$T_f = 310 \, K$.
Converting back to Celsius: $T_f = 310 - 273 = 37^{\circ} C$.

(C) For a monatomic gas,the adiabatic index $\gamma_1 = 5/3$. The number of moles $n_1 = 1$.
For a diatomic gas (neglecting vibration),the adiabatic index $\gamma_2 = 7/5$. The number of moles $n_2 = 1$.
The equivalent adiabatic index $\gamma$ for a mixture is given by the formula:
$\frac{n_{mix}}{\gamma - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}$
Here,$n_{mix} = n_1 + n_2 = 1 + 1 = 2$.
Substituting the values:
$\frac{2}{\gamma - 1} = \frac{1}{5/3 - 1} + \frac{1}{7/5 - 1}$
$\frac{2}{\gamma - 1} = \frac{1}{2/3} + \frac{1}{2/5} = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4$
$\frac{2}{\gamma - 1} = 4 \implies \gamma - 1 = 0.5 \implies \gamma = 1.5$.

(A) For a monatomic gas,the degrees of freedom $f_1 = 3$. The molar specific heat at constant volume is $C_{v_1} = \frac{f_1}{2} R = \frac{3}{2} R$.
For a diatomic gas,the degrees of freedom $f_2 = 5$. The molar specific heat at constant volume is $C_{v_2} = \frac{f_2}{2} R = \frac{5}{2} R$.
Given $n_1 = 1$ mole and $n_2 = 3$ moles.
The molar specific heat of the mixture at constant volume $C_{v_{mix}}$ is given by the weighted average:
$C_{v_{mix}} = \frac{n_1 C_{v_1} + n_2 C_{v_2}}{n_1 + n_2}$
Substituting the values:
$C_{v_{mix}} = \frac{1 \times (\frac{3}{2} R) + 3 \times (\frac{5}{2} R)}{1 + 3}$
$C_{v_{mix}} = \frac{\frac{3}{2} R + \frac{15}{2} R}{4} = \frac{\frac{18}{2} R}{4} = \frac{9R}{4}$
Given $R = 8 \, J K^{-1} mol^{-1}$:
$C_{v_{mix}} = \frac{9 \times 8}{4} = 9 \times 2 = 18 \, J K^{-1} mol^{-1}$.

(A) For a monoatomic gas,the degree of freedom $f_1 = 3$.
For a polyatomic gas,neglecting the vibrational mode of freedom,the degree of freedom $f_2 = 6$ ($3$ translational + $3$ rotational).
The mixture behaves like a diatomic gas,so the degree of freedom of the mixture $f_{\text{mix}} = 5$ ($3$ translational + $2$ rotational).
The formula for the degree of freedom of a mixture is given by:
$f_{\text{mix}} = \frac{f_1 n_1 + f_2 n_2}{n_1 + n_2}$
Substituting the given values:
$5 = \frac{3 \alpha + 6 \beta}{\alpha + \beta}$
$5(\alpha + \beta) = 3 \alpha + 6 \beta$
$5 \alpha + 5 \beta = 3 \alpha + 6 \beta$
$2 \alpha = \beta$

(D) Let $n_1$ be the number of moles of hydrogen $(H_2)$ and $n_2$ be the number of moles of helium $(He)$.
Using the ideal gas equation $PV = nRT$,where $n = n_1 + n_2$ is the total number of moles.
Given: $P = 2 \ atm = 2 \times 1.013 \times 10^5 \ Pa$,$V = 20 \ L = 20 \times 10^{-3} \ m^3$,$T = 27^{\circ} C = 300 \ K$,$R = 8.314 \ J/mol \cdot K$.
Total moles $n = n_1 + n_2 = \frac{PV}{RT} = \frac{2 \times 1.013 \times 10^5 \times 20 \times 10^{-3}}{8.314 \times 300} \approx 1.62 \ mol$.
Mass of mixture: $m_{H_2} + m_{He} = 5 \ g$.
Since $m = n \times M$,where $M$ is molar mass: $n_1(2) + n_2(4) = 5 \implies n_1 + 2n_2 = 2.5$.
Subtracting the first equation $(n_1 + n_2 = 1.62)$ from the second $(n_1 + 2n_2 = 2.5)$: $n_2 = 0.88 \ mol$.
Then $n_1 = 1.62 - 0.88 = 0.74 \ mol$.
Mass ratio $\frac{m_{H_2}}{m_{He}} = \frac{n_1 \times 2}{n_2 \times 4} = \frac{0.74 \times 2}{0.88 \times 4} = \frac{1.48}{3.52} = \frac{2}{5}$.

(B) According to the ideal gas law,$PV = nRT$. Since the temperature $T$ is constant,the total number of moles $n_{total} = n_A + n_B$ remains constant.
For container $A$: $n_A = \frac{P_A V_A}{RT}$
For container $B$: $n_B = \frac{P_B V_B}{RT}$
When the containers are connected,the final pressure $P_f$ is given by:
$P_f(V_A + V_B) = (n_A + n_B)RT$
$P_f(V_A + V_B) = P_A V_A + P_B V_B$
Substituting the given values:
$P_f = \frac{P_A V_A + P_B V_B}{V_A + V_B}$
$P_f = \frac{(1.40 \, MPa \times 0.10 \, m^3) + (0.7 \, MPa \times 0.15 \, m^3)}{0.10 \, m^3 + 0.15 \, m^3}$
$P_f = \frac{0.14 + 0.105}{0.25} \, MPa$
$P_f = \frac{0.245}{0.25} \, MPa = 0.98 \, MPa$.

(B) The internal energy of an ideal gas is given by $U = n \left( \frac{f}{2} RT \right)$. Assuming both gases are monatomic $(f=3)$, the total energy is conserved.
Energy of the first gas at $T_1 = 27^{\circ}\, \text{C} = 300\, \text{K}$ is $E_1 = 3 \left( \frac{3}{2} R \times 300 \right) = 1350\, R$.
Energy of the second gas at $T_2 = 227^{\circ}\, \text{C} = 500\, \text{K}$ is $E_2 = 2 \left( \frac{3}{2} R \times 500 \right) = 1500\, R$.
Let $T$ be the equilibrium temperature in Kelvin. The total energy of the mixture is $E_m = (3+2) \left( \frac{3}{2} RT \right) = 7.5\, RT$.
By conservation of energy, $E_m = E_1 + E_2$.
$7.5\, RT = 1350\, R + 1500\, R$.
$7.5\, T = 2850$.
$T = \frac{2850}{7.5} = 380\, \text{K}$.
Converting to Celsius: $T(^{\circ}\, \text{C}) = 380 - 273 = 107^{\circ}\, \text{C}$.

(D) The number of moles for each gas are: $n_1 = 4$ (hydrogen,diatomic),$n_2 = 2$ (helium,monatomic),$n_3 = 1$ (water vapour,polyatomic).
The molar specific heat at constant volume $(C_v)$ for each gas is:
For hydrogen (diatomic): $(C_v)_1 = \frac{5}{2} R$
For helium (monatomic): $(C_v)_2 = \frac{3}{2} R$
For water vapour (polyatomic): $(C_v)_3 = 3 R$
The equivalent molar specific heat at constant volume for the mixture is given by:
$(C_v)_{\text{mix}} = \frac{n_1(C_v)_1 + n_2(C_v)_2 + n_3(C_v)_3}{n_1 + n_2 + n_3}$
$(C_v)_{\text{mix}} = \frac{4 \times \frac{5}{2} R + 2 \times \frac{3}{2} R + 1 \times 3 R}{4 + 2 + 1} = \frac{10 R + 3 R + 3 R}{7} = \frac{16}{7} R$
Using the relation $C_p = C_v + R$ for the mixture:
$(C_p)_{\text{mix}} = (C_v)_{\text{mix}} + R = \frac{16}{7} R + R = \frac{23}{7} R$.

(C) The molar heat capacity at constant volume for a mixture is given by $C_{V, \text{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2} = \frac{13R}{6}$.
Given the ratio $n_1 : n_2 = 1 : 2$,let $n_1 = n$ and $n_2 = 2n$.
Substituting these values: $\frac{n C_{V_1} + 2n C_{V_2}}{n + 2n} = \frac{13R}{6}$.
$\frac{C_{V_1} + 2C_{V_2}}{3} = \frac{13R}{6} \implies C_{V_1} + 2C_{V_2} = \frac{13R}{2}$.
For monatomic gases,$C_V = 3R/2$. For diatomic gases,$C_V = 5R/2$.
If gas $1$ is monatomic $(C_{V_1} = 3R/2)$ and gas $2$ is diatomic $(C_{V_2} = 5R/2)$:
$3R/2 + 2(5R/2) = 3R/2 + 5R = 13R/2$.
This matches the condition. Thus,the first gas is monatomic $(He)$ and the second is diatomic $(N_2)$.

(D) The total internal energy $U$ of a system is the sum of the internal energies of its components.
For oxygen $(O_2)$,which is a diatomic gas,the degrees of freedom $f_1 = 5$ (neglecting vibrational modes).
For neon $(Ne)$,which is a monatomic gas,the degrees of freedom $f_2 = 3$.
The internal energy of $n_1$ moles of gas $1$ is $U_1 = n_1 \frac{f_1}{2} RT$.
The internal energy of $n_2$ moles of gas $2$ is $U_2 = n_2 \frac{f_2}{2} RT$.
Total internal energy $U = U_1 + U_2 = (n_1 \frac{f_1}{2} + n_2 \frac{f_2}{2}) RT$.
Given $n_1 = 2$ (oxygen) and $n_2 = 4$ (neon).
$U = (2 \times \frac{5}{2} + 4 \times \frac{3}{2}) RT$.
$U = (5 + 6) RT = 11 RT$.
Thus,the total internal energy is $11 RT$.

(C) The equivalent degree of freedom $f_{eq}$ for a mixture of gases is given by the formula:
$f_{eq} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$
For a diatomic gas,the degree of freedom is $f_{eq} = 5$.
Given:
$n_1 = N$,$f_1 = 6$ (polyatomic gas)
$n_2 = 2$,$f_2 = 3$ (monoatomic gas)
Substituting these values into the formula:
$5 = \frac{(N)(6) + (2)(3)}{N + 2}$
Multiplying both sides by $(N + 2)$:
$5(N + 2) = 6N + 6$
$5N + 10 = 6N + 6$
Rearranging the terms to solve for $N$:
$10 - 6 = 6N - 5N$
$N = 4$

(C) For a monoatomic gas,the degrees of freedom $f_1 = 3$. For a diatomic gas,the degrees of freedom $f_2 = 5$.
Given $n_1 = 3$ moles and $n_2 = 2$ moles.
The 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: $f_{\text{mix}} = \frac{3(3) + 2(5)}{3 + 2} = \frac{9 + 10}{5} = \frac{19}{5} = 3.8$.
The adiabatic exponent $\gamma$ is related to the degrees of freedom $f$ by the formula $\gamma = 1 + \frac{2}{f}$.
Therefore,$\gamma_{\text{mix}} = 1 + \frac{2}{f_{\text{mix}}} = 1 + \frac{2}{3.8} = 1 + \frac{20}{38} = 1 + \frac{10}{19} = \frac{29}{19} \approx 1.526$.
Rounding to two decimal places,we get $\gamma_{\text{mix}} = 1.52$.

(C) The total internal energy $U$ of a gas mixture is the sum of the internal energies of its individual components.
For a gas with $n$ moles,the internal energy is given by $U = n C_V T$,where $C_V$ is the molar heat capacity at constant volume.
Argon is a monatomic gas,so its degrees of freedom $f_1 = 3$,and $C_{V1} = \frac{3}{2} R$.
Oxygen is a diatomic gas,so its degrees of freedom $f_2 = 5$ (neglecting vibrational modes),and $C_{V2} = \frac{5}{2} R$.
The total internal energy is $U = n_1 C_{V1} T + n_2 C_{V2} T$.
Substituting the given values: $U = (8 \times \frac{3}{2} R \times T) + (6 \times \frac{5}{2} R \times T)$.
$U = (12 RT) + (15 RT) = 27 RT$.

(A) The molar specific heat at constant volume for a mixture is given by the formula: $C_{V,mix} = \frac{n_1 C_{V,1} + n_2 C_{V,2}}{n_1 + n_2}$.
For a monoatomic gas,$C_{V,1} = \frac{3}{2} R$ and $n_1 = 2$.
For a diatomic gas,$C_{V,2} = \frac{5}{2} R$ and $n_2 = 6$.
Substituting these values into the formula:
$C_{V,mix} = \frac{2 \times (\frac{3}{2} R) + 6 \times (\frac{5}{2} R)}{2 + 6}$
$C_{V,mix} = \frac{3R + 15R}{8}$
$C_{V,mix} = \frac{18R}{8} = \frac{9}{4} R$.

(D) For a mixture of gases at constant pressure,the work done is given by $W = n_{mix} R \Delta T = 66 \ J$.
The molar heat capacity at constant volume for the mixture is given by $(C_V)_{mix} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$.
For a monatomic gas,$C_{V1} = \frac{3}{2} R$. For a diatomic gas,$C_{V2} = \frac{5}{2} R$.
Substituting the values: $(C_V)_{mix} = \frac{2 \times (3/2)R + 1 \times (5/2)R}{2 + 1} = \frac{3R + 2.5R}{3} = \frac{5.5R}{3} = \frac{11}{6} R$.
The change in internal energy is $\Delta U = n_{total} (C_V)_{mix} \Delta T$.
Since $n_{total} = n_1 + n_2 = 3$,we have $\Delta U = 3 \times (\frac{11}{6} R) \Delta T = \frac{11}{2} R \Delta T$.
From the work equation,$n_{total} R \Delta T = 66$,so $3 R \Delta T = 66$,which means $R \Delta T = 22$.
Substituting this into the internal energy equation: $\Delta U = \frac{11}{2} \times 22 = 11 \times 11 = 121 \ J$.

(D) For the mixture: $n_1 = 5$ (monatomic),$C_{v1} = 3R/2$; $n_2 = 1$ (diatomic),$C_{v2} = 5R/2$.
Total moles $n = 5 + 1 = 6$.
$(C_v)_m = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{5(3R/2) + 1(5R/2)}{6} = \frac{10R}{6} = \frac{5R}{3}$.
$(C_p)_m = (C_v)_m + R = \frac{5R}{3} + R = \frac{8R}{3}$.
Adiabatic constant $\gamma_m = \frac{(C_p)_m}{(C_v)_m} = \frac{8/3}{5/3} = 1.6$. Thus,$(4)$ is correct.
For adiabatic compression: $P_0 V_0^{\gamma} = P_f (V_0/4)^{\gamma}$.
$P_f = P_0 (4)^{1.6} = P_0 (2^2)^{1.6} = P_0 (2^{3.2}) = 9.2 P_0$. Thus,$(1)$ is correct.
Using $T_f V_f^{\gamma-1} = T_0 V_0^{\gamma-1}$,$T_f = T_0 (V_0 / (V_0/4))^{0.6} = T_0 (4)^{0.6} = T_0 (2^{1.2}) = 2.3 T_0$.
Total internal energy $U = n_1 (3/2 RT) + n_2 (5/2 RT) = 5(1.5 RT) + 1(2.5 RT) = 10 RT$.
$U_f = 10 R (2.3 T_0) = 23 RT_0$. Thus,$(2)$ is incorrect.
Work done $|W| = |\Delta U| = |n_f C_{vm} T_f - n_i C_{vm} T_i| = |6 \times (5R/6) \times (2.3 T_0 - T_0)| = |5R \times 1.3 T_0| = 6.5 RT_0$. Thus,$(3)$ is incorrect.
Correct statements are $(1)$ and $(4)$.

(D) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.
Substituting $n$,we get $PV = \frac{m}{M}RT$,which can be rewritten as $P = \frac{m}{V} \cdot \frac{RT}{M} = \frac{\rho RT}{M}$,where $\rho$ is the density.
Thus,the density is $\rho = \frac{PM}{RT}$.
For two gases at the same temperature $T$,the ratio of their densities is $\frac{\rho_1}{\rho_2} = \frac{P_1 M_1}{P_2 M_2}$.
Given the ratio of atomic masses $\frac{M_1}{M_2} = \frac{2}{3}$ and the ratio of partial pressures $\frac{P_1}{P_2} = \frac{4}{3}$.
Substituting these values,we get $\frac{\rho_1}{\rho_2} = \left(\frac{4}{3}\right) \times \left(\frac{2}{3}\right) = \frac{8}{9}$.

(C) For a mixture of $1$ mole of $H_2$ (diatomic,$C_v = 5R/2$) and $1$ mole of $He$ (monatomic,$C_v = 3R/2$):
Total internal energy $U = n_1 C_{v1} T + n_2 C_{v2} T = 1(5R/2)T + 1(3R/2)T = 4RT$.
Average energy per mole $= U / (n_1 + n_2) = 4RT / 2 = 2RT$. Thus,$(A)$ is correct.
For the mixture,$C_{v,mix} = (n_1 C_{v1} + n_2 C_{v2}) / (n_1 + n_2) = (5R/2 + 3R/2) / 2 = 2R$.
$C_{p,mix} = C_{v,mix} + R = 3R$. $\gamma_{mix} = C_{p,mix} / C_{v,mix} = 3R / 2R = 1.5 = 3/2$.
Speed of sound $v = \sqrt{\gamma RT / M}$. For $He$,$\gamma = 5/3$ and $M = 4$. For mixture,$M_{mix} = (2+4)/2 = 3$.
Ratio $v_{mix} / v_{He} = \sqrt{(\gamma_{mix} / M_{mix}) / (\gamma_{He} / M_{He})} = \sqrt{(1.5 / 3) / ((5/3) / 4)} = \sqrt{0.5 / (5/12)} = \sqrt{6/5}$. Thus,$(B)$ is correct.
$RMS$ speed $v_{rms} = \sqrt{3RT/M}$. Ratio $v_{rms,He} / v_{rms,H2} = \sqrt{M_{H2} / M_{He}} = \sqrt{2/4} = 1/\sqrt{2}$. Thus,$(D)$ is correct.

(B) For an ideal gas,the internal energy is given by $U = \frac{f}{2} nRT = \frac{f}{2} PV$. Since the partition is a thermal insulator,the total internal energy of the system is conserved when the partition is removed.
$U_{\text{initial}} = U_{\text{final}}$
$\frac{f_1}{2} P_1 V_1 + \frac{f_2}{2} P_2 V_2 = \frac{f_{\text{mix}}}{2} P_{\text{mix}} (V_1 + V_2)$
Assuming the gas is the same in both chambers,$f_1 = f_2 = f_{\text{mix}} = f$.
$P_1 V_1 + P_2 V_2 = P_{\text{mix}} (V_1 + V_2)$
Substituting the given values:
$(1 \ atm \times 2 \ L) + (2 \ atm \times 3 \ L) = P_{\text{mix}} (2 \ L + 3 \ L)$
$2 + 6 = P_{\text{mix}} (5)$
$8 = 5 P_{\text{mix}}$
$P_{\text{mix}} = \frac{8}{5} \ atm = 1.6 \ atm$.

(B) $C_p = C_v + R$
$\therefore C_p$ for hydrogen,$C_{p_1} = \frac{5}{2} R + R = \frac{7}{2} R$
$C_p$ for helium,$C_{p_2} = \frac{3}{2} R + R = \frac{5}{2} R$
$C_p$ for water vapour,$C_{p_3} = 3 R + R = 4 R$
Given: $n_1 = 4, n_2 = 2, n_3 = 1$
$C_p \text{ of mixture} = \frac{n_1 C_{p_1} + n_2 C_{p_2} + n_3 C_{p_3}}{n_1 + n_2 + n_3}$
$= \frac{4 \times \frac{7}{2} R + 2 \times \frac{5}{2} R + 1 \times 4 R}{4 + 2 + 1}$
$= \frac{14 R + 5 R + 4 R}{7} = \frac{23 R}{7}$

(A) For a monoatomic gas,the molar heat capacity at constant volume is $C_{v1} = \frac{3}{2}R$.
For a diatomic gas,the molar heat capacity at constant volume is $C_{v2} = \frac{5}{2}R$.
For a mixture of $n_1$ moles of gas $1$ and $n_2$ moles of gas $2$,the equivalent $C_v$ is given by $C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}$.
Substituting the values: $C_{v,mix} = \frac{1 \cdot \frac{3}{2}R + 1 \cdot \frac{5}{2}R}{1 + 1} = \frac{4R}{2} = 2R$.
Similarly,the equivalent $C_p$ is $C_{p,mix} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2}$. Since $C_p = C_v + R$,we have $C_{p1} = \frac{5}{2}R$ and $C_{p2} = \frac{7}{2}R$.
$C_{p,mix} = \frac{1 \cdot \frac{5}{2}R + 1 \cdot \frac{7}{2}R}{2} = \frac{6R}{2} = 3R$.
The adiabatic index for the mixture is $\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}} = \frac{3R}{2R} = 1.5$.

(C) Initially,each gas has pressure $P$ and volume $V$. When they are mixed,the total volume of the mixture is $V_{1} = V + V = 2V$.
Since the temperature is constant,we use Boyle's Law: $P_{1}V_{1} = P_{2}V_{2}$.
The initial pressure of the mixture $P_{1}$ is $P$ (as the pressure of each gas is $P$ and they are at the same temperature and volume).
Given the final volume $V_{2} = \frac{V}{2}$.
Substituting the values: $P \times (2V) = P_{2} \times (\frac{V}{2})$.
Solving for $P_{2}$: $P_{2} = \frac{P \times 2V}{V/2} = 4P$.

(C) Let $T^{\circ} C$ be the final equilibrium temperature of the mixture.
Since the container is closed and isolated,the heat lost by the hotter gas $(O_2)$ equals the heat gained by the cooler gas $(CO_2)$.
For $CO_2$ (a non-linear triatomic gas),the molar heat capacity at constant volume is $C_{V, CO_2} = 3R$.
For $O_2$ (a diatomic gas),the molar heat capacity at constant volume is $C_{V, O_2} = \frac{5}{2}R$.
The number of moles are: $\mu_{CO_2} = \frac{22}{44} = 0.5 \ mol$ and $\mu_{O_2} = \frac{16}{32} = 0.5 \ mol$.
Applying the principle of calorimetry: $\mu_{CO_2} C_{V, CO_2} (T - 27) = \mu_{O_2} C_{V, O_2} (37 - T)$.
$0.5 \times 3R \times (T - 27) = 0.5 \times \frac{5}{2}R \times (37 - T)$.
$3(T - 27) = 2.5(37 - T)$.
$3T - 81 = 92.5 - 2.5T$.
$5.5T = 173.5$.
$T = \frac{173.5}{5.5} \approx 31.5^{\circ} C$.

(C) The total internal energy of a gas is given by the formula:
$U = \frac{n f R T}{2}$
where $n$ is the number of moles and $f$ is the degree of freedom.
Given,$n_{\text{diatomic}} = n_{\text{monoatomic}} = 2$.
For a monoatomic gas,the degree of freedom $f_{\text{monoatomic}} = 3$.
For a diatomic gas,the degree of freedom $f_{\text{diatomic}} = 5$.
Calculating internal energy for each:
$U_{\text{monoatomic}} = \frac{2 \times 3 \times R T}{2} = 3 R T$
$U_{\text{diatomic}} = \frac{2 \times 5 \times R T}{2} = 5 R T$
Therefore,the total internal energy of the mixture is:
$U_{\text{total}} = U_{\text{monoatomic}} + U_{\text{diatomic}} = 3 R T + 5 R T = 8 R T$.

(C) Given: $n_1 = 1 \text{ mole}$,$n_2 = 1 \text{ mole}$,$\gamma_1 = \frac{7}{5}$,$\gamma_2 = \frac{4}{3}$.
For a mixture of gases,the equivalent adiabatic index $\gamma_{\text{mix}}$ is given by the formula:
$\frac{n_1 + n_2}{\gamma_{\text{mix}} - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}$
Substituting the values:
$\frac{1 + 1}{\gamma_{\text{mix}} - 1} = \frac{1}{\frac{7}{5} - 1} + \frac{1}{\frac{4}{3} - 1}$
$\frac{2}{\gamma_{\text{mix}} - 1} = \frac{1}{\frac{2}{5}} + \frac{1}{\frac{1}{3}}$
$\frac{2}{\gamma_{\text{mix}} - 1} = \frac{5}{2} + 3 = \frac{5 + 6}{2} = \frac{11}{2}$
$\gamma_{\text{mix}} - 1 = \frac{4}{11}$
$\gamma_{\text{mix}} = 1 + \frac{4}{11} = \frac{15}{11}$.

(D) Mass of Helium,$m_H = 16 \ g$. Molar mass of Helium,$M_H = 4 \ g/mol$. Number of moles of Helium,$n_H = 16/4 = 4 \ mol$. Helium is a monoatomic gas,so its molar heat capacity at constant volume is $C_{v,H} = \frac{3}{2}R$.
Mass of Oxygen,$m_O = 16 \ g$. Molar mass of Oxygen,$M_O = 32 \ g/mol$. Number of moles of Oxygen,$n_O = 16/32 = 0.5 \ mol$. Oxygen is a diatomic gas,so its molar heat capacity at constant volume is $C_{v,O} = \frac{5}{2}R$.
The molar heat capacity of the mixture at constant volume is $C_{v,mix} = \frac{n_H C_{v,H} + n_O C_{v,O}}{n_H + n_O} = \frac{4(\frac{3}{2}R) + 0.5(\frac{5}{2}R)}{4 + 0.5} = \frac{6R + 1.25R}{4.5} = \frac{7.25R}{4.5} = \frac{29R}{18}$.
The molar heat capacity of the mixture at constant pressure is $C_{p,mix} = C_{v,mix} + R = \frac{29R}{18} + R = \frac{47R}{18}$.
The ratio of specific heats is $\gamma = \frac{C_{p,mix}}{C_{v,mix}} = \frac{47R/18}{29R/18} = \frac{47}{29} \approx 1.62$.

(A) Given: $n_{He} = 2$ moles, $n_{H_2} = 2$ moles, $T = \frac{972}{5} \,K$, $R = \frac{25}{3} \,J \,mol^{-1} \,K^{-1}$.
$1$. Calculate the molar mass of the mixture $(M_{mix})$:
$M_{mix} = \frac{n_1 M_1 + n_2 M_2}{n_1 + n_2} = \frac{2 \times 4 + 2 \times 2}{2 + 2} = \frac{8 + 4}{4} = 3 \,g/mol = 3 \times 10^{-3} \,kg/mol$.
$2$. Calculate the adiabatic index $(\gamma_{mix})$ of the mixture:
For Helium (monatomic), $f_1 = 3$. For Hydrogen (diatomic), $f_2 = 5$.
$f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2} = \frac{2 \times 3 + 2 \times 5}{2 + 2} = \frac{16}{4} = 4$.
$\gamma_{mix} = 1 + \frac{2}{f_{mix}} = 1 + \frac{2}{4} = 1.5$.
$3$. Calculate the speed of sound $(v)$:
$v = \sqrt{\frac{\gamma_{mix} R T}{M_{mix}}} = \sqrt{\frac{1.5 \times \frac{25}{3} \times \frac{972}{5}}{3 \times 10^{-3}}} = \sqrt{\frac{1.5 \times 5 \times 972}{3 \times 10^{-3}}} = \sqrt{\frac{7.5 \times 972}{3 \times 10^{-3}}} = \sqrt{2.5 \times 324000} = \sqrt{810000} = 900 \,m/s$.
Given $v = n \times 100 \,m/s$, therefore $n = 9$.

(D) For a monatomic gas,the degrees of freedom $f_1 = 3$. For a diatomic gas,the degrees of freedom $f_2 = 5$.
The adiabatic constant $\gamma$ for a mixture is given by $\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}}$.
Given $n_1 = 1$ (monatomic) and $n_2 = n$ (diatomic).
$C_{v1} = \frac{3}{2}R$,$C_{p1} = \frac{5}{2}R$.
$C_{v2} = \frac{5}{2}R$,$C_{p2} = \frac{7}{2}R$.
Substituting these into the formula:
$\gamma_{mix} = \frac{1(\frac{5}{2}R) + n(\frac{7}{2}R)}{1(\frac{3}{2}R) + n(\frac{5}{2}R)} = \frac{5 + 7n}{3 + 5n}$.
Given $\gamma_{mix} = \frac{13}{9}$,we have $\frac{5 + 7n}{3 + 5n} = \frac{13}{9}$.
Cross-multiplying: $9(5 + 7n) = 13(3 + 5n)$.
$45 + 63n = 39 + 65n$.
$45 - 39 = 65n - 63n$.
$6 = 2n$,which gives $n = 3$.

(D) The total internal energy $U$ of a gaseous mixture is the sum of the internal energies of its individual components.
For a gas with $n$ moles and degrees of freedom $f$,the internal energy is given by $U = n \cdot \frac{f}{2} RT$.
Oxygen $(O_2)$ is a diatomic gas. Neglecting vibrational modes,its degrees of freedom $f_1 = 5$.
Internal energy of oxygen: $U_1 = n_1 \cdot \frac{f_1}{2} RT = 2 \cdot \frac{5}{2} RT = 5 RT$.
Argon $(Ar)$ is a monatomic gas,so its degrees of freedom $f_2 = 3$.
Internal energy of argon: $U_2 = n_2 \cdot \frac{f_2}{2} RT = 4 \cdot \frac{3}{2} RT = 6 RT$.
Total internal energy $U = U_1 + U_2 = 5 RT + 6 RT = 11 RT$.

(C) The total number of moles $n$ in the container is the sum of the moles of each gas: $n = 2 + 5 + 3 = 10 \ moles$.
The temperature $T$ in Kelvin is $T = 0 \ ^{\circ}C + 273.15 = 273.15 \ K$. For simplicity,we use $T = 273 \ K$.
The volume $V$ is $16.62 \ m^3$.
Using the ideal gas equation $PV = nRT$,we can solve for pressure $P$:
$P = \frac{nRT}{V}$
$P = \frac{10 \times 8.31 \times 273}{16.62}$
$P = \frac{83.1 \times 273}{16.62}$
$P = 5 \times 273 = 1365 \ Pa$.
Thus,the pressure in the container is $1365 \ Pa$.

(C) The internal energy $U$ of a gas is given by $U = \frac{f}{2}nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.
For oxygen $(O_2)$,which is a diatomic gas,the degrees of freedom $f_1 = 5$ (neglecting vibrational modes).
For argon $(Ar)$,which is a monatomic gas,the degrees of freedom $f_2 = 3$.
The total internal energy $U_{total} = U_{O_2} + U_{Ar}$.
$U_{total} = \frac{f_1}{2}n_1RT + \frac{f_2}{2}n_2RT$.
Given $n_1 = 3$ moles and $n_2 = 4$ moles.
$U_{total} = \frac{5}{2}(3)RT + \frac{3}{2}(4)RT$.
$U_{total} = 7.5RT + 6RT = 13.5RT$.

(C) For a gas mixture,the adiabatic index $\gamma_{\text{mix}}$ is given by $\gamma_{\text{mix}} = \frac{C_{p(\text{mix})}}{C_{V(\text{mix})}}$.
Using the formula for mixture properties: $\gamma_{\text{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{V1} + n_2 C_{V2}}$.
For oxygen $(O_2)$,which is diatomic: $n_1 = \frac{4 \ g}{32 \ g/mol} = \frac{1}{8} \ mol$. $C_{p1} = \frac{7}{2}R$,$C_{V1} = \frac{5}{2}R$.
For helium (He),which is monatomic: $n_2 = \frac{4 \ g}{4 \ g/mol} = 1 \ mol$. $C_{p2} = \frac{5}{2}R$,$C_{V2} = \frac{3}{2}R$.
Substituting these values into the formula:
$\gamma_{\text{mix}} = \frac{(\frac{1}{8} \times \frac{7}{2}R) + (1 \times \frac{5}{2}R)}{(\frac{1}{8} \times \frac{5}{2}R) + (1 \times \frac{3}{2}R)}$
$\gamma_{\text{mix}} = \frac{\frac{7}{16} + \frac{5}{2}}{\frac{5}{16} + \frac{3}{2}} = \frac{\frac{7+40}{16}}{\frac{5+24}{16}} = \frac{47}{29}$.

(A) According to Dalton's Law of Partial Pressures,the total pressure exerted by a mixture of non-reacting gases is the sum of the partial pressures of the individual gases.
For gas $A$,the pressure is $P$ when it occupies volume $V$ at temperature $T$.
For gas $B$,the pressure is $P$ when it occupies volume $V$ at temperature $T$.
When both gases are mixed into the same volume $V$ at the same temperature $T$,the total pressure $P_{mix}$ is the sum of the individual pressures.
$P_{mix} = P_A + P_B = P + P = 2 P$.

(B) For an ideal monoatomic gas,the internal energy is given by $U = n C_v T$,where $C_v = \frac{3}{2} R$.
Since the gases are mixed in an isolated system,the total internal energy is conserved: $U_{mix} = U_1 + U_2$.
$(n_1 + n_2) C_v T_{mix} = n_1 C_v T_1 + n_2 C_v T_2$.
Since $C_v$ is the same for both monoatomic gases,it cancels out:
$(n_1 + n_2) T_{mix} = n_1 T_1 + n_2 T_2$.
Given: $n_1 = 2 \text{ mol}$,$T_1 = 27 + 273 = 300 \text{ K}$,$n_2 = 4 \text{ mol}$,$T_2 = 327 + 273 = 600 \text{ K}$.
$(2 + 4) T_{mix} = (2 \times 300) + (4 \times 600)$.
$6 T_{mix} = 600 + 2400 = 3000$.
$T_{mix} = \frac{3000}{6} = 500 \text{ K}$.
Converting back to Celsius: $T_{mix} = 500 - 273 = 227^{\circ} C$.

(B) The internal energy $U$ of $n$ moles of a gas is given by $U = n \frac{f}{2} RT$,where $f$ is the degree of freedom.
For monoatomic gases $(He, Ar)$,$f = 3$.
For diatomic gases $(N_2, H_2)$,$f = 5$ (ignoring vibrational modes).
Total internal energy $U_{total} = U_{He} + U_{Ar} + U_{N_2} + U_{H_2}$.
$U_{total} = (3 \times \frac{3}{2} RT) + (1 \times \frac{3}{2} RT) + (5 \times \frac{5}{2} RT) + (3 \times \frac{5}{2} RT)$.
$U_{total} = (4.5 + 1.5 + 12.5 + 7.5) RT = 26 RT$.

(D) The internal energy $U$ of a gas is given by the formula $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature.
For oxygen $(O_2)$,which is a diatomic gas,the degrees of freedom $f_1 = 5$ (neglecting vibrational modes). The number of moles $n_1 = 2$.
Internal energy of oxygen $U_1 = \frac{5}{2} \times 2 \times RT = 5 RT$.
For Argon $(Ar)$,which is a monatomic gas,the degrees of freedom $f_2 = 3$. The number of moles $n_2 = 4$.
Internal energy of Argon $U_2 = \frac{3}{2} \times 4 \times RT = 6 RT$.
The total internal energy of the system is $U_{total} = U_1 + U_2 = 5 RT + 6 RT = 11 RT$.

(B) Mass of helium, $m_{He} = 4 \,g$. Molar mass of $He = 4 \,g/mol$. Number of moles $n_1 = 4/4 = 1 \,mol$.
Mass of oxygen, $m_{O_2} = 32 \,g$. Molar mass of $O_2 = 32 \,g/mol$. Number of moles $n_2 = 32/32 = 1 \,mol$.
Helium is a monoatomic gas, so degrees of freedom $f_1 = 3$.
Oxygen is a diatomic gas, so degrees of freedom $f_2 = 5$.
The specific heat ratio $\gamma_{mix}$ is given by $\frac{C_{p,mix}}{C_{V,mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{V1} + n_2 C_{V2}}$.
Using $C_V = \frac{f}{2}R$ and $C_p = (1 + \frac{f}{2})R$:
$C_{V,mix} = \frac{n_1(f_1/2)R + n_2(f_2/2)R}{n_1 + n_2} = \frac{1(3/2)R + 1(5/2)R}{1 + 1} = \frac{4R}{2} = 2R$.
$C_{p,mix} = C_{V,mix} + R = 2R + R = 3R$.
Therefore, $\gamma_{mix} = \frac{3R}{2R} = \frac{3}{2}$.

(A) For a gas mixture,the molar heat capacities at constant volume and constant pressure are given by:
$C_V = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2}$ and $C_p = \frac{n_1 C_{p_1} + n_2 C_{p_2}}{n_1 + n_2}$
For a monoatomic gas,the degrees of freedom $f_1 = 3$,so $C_{V_1} = \frac{3}{2}R$ and $C_{p_1} = \frac{5}{2}R$.
For a rigid diatomic gas,the degrees of freedom $f_2 = 5$,so $C_{V_2} = \frac{5}{2}R$ and $C_{p_2} = \frac{7}{2}R$.
The adiabatic exponent $\gamma = \frac{C_p}{C_V} = 1.5 = \frac{3}{2}$.
Substituting the values:
$\frac{C_p}{C_V} = \frac{n_1(\frac{5}{2}R) + n_2(\frac{7}{2}R)}{n_1(\frac{3}{2}R) + n_2(\frac{5}{2}R)} = \frac{5n_1 + 7n_2}{3n_1 + 5n_2} = \frac{3}{2}$
Cross-multiplying gives:
$2(5n_1 + 7n_2) = 3(3n_1 + 5n_2)$
$10n_1 + 14n_2 = 9n_1 + 15n_2$
$n_1 = n_2$
Therefore,the ratio $\frac{n_1}{n_2} = 1$.

(A) According to the ideal gas law,$PV = nRT$. Since the volume $V$ and temperature $T$ are constant for both gases in the container,the number of moles $n$ is directly proportional to the partial pressure $p$ of the gas $(n \propto p)$.
Initially,the pressure of nitrogen is $p_{N_2} = 100 \ kPa$.
After adding oxygen,the total pressure becomes $300 \ kPa$. Therefore,the partial pressure of oxygen is $p_{O_2} = P_{total} - p_{N_2} = 300 \ kPa - 100 \ kPa = 200 \ kPa$.
The ratio of the number of $N_2$ molecules to $O_2$ molecules is equal to the ratio of their moles,which is equal to the ratio of their partial pressures:
$\frac{n_{N_2}}{n_{O_2}} = \frac{p_{N_2}}{p_{O_2}} = \frac{100 \ kPa}{200 \ kPa} = 0.5$.

(D) $1$. Calculate the number of moles $(n)$ for each gas:
$n_{O_2} = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$
$n_{N_2} = \frac{28 \ g}{28 \ g/mol} = 1 \ mol$
$n_{CO_2} = \frac{132 \ g}{44 \ g/mol} = 3 \ mol$
$2$. The total number of moles initially is $n_{total} = 2 + 1 + 3 = 6 \ mol$.
$3$. According to the ideal gas law,$PV = nRT$. Since $V, R, T$ are constant,$P \propto n$.
$4$. Therefore,$P = k \times 6$,where $k$ is a constant.
$5$. After removing oxygen,the remaining moles are $n_{remaining} = 1 (N_2) + 3 (CO_2) = 4 \ mol$.
$6$. The new pressure $P'$ is $P' = k \times 4$.
$7$. Taking the ratio: $\frac{P'}{P} = \frac{4}{6} = \frac{2}{3}$.
$8$. Thus,$P' = \frac{2 P}{3}$.

(C) The average kinetic energy per molecule of an ideal gas is given by the formula $\langle KE \rangle = \frac{f}{2} K_B T$,where $f$ is the degrees of freedom,$K_B$ is the Boltzmann constant,and $T$ is the absolute temperature.
Since both hydrogen and nitrogen gases are in the same vessel at thermal equilibrium,they are at the same temperature $T = 30^{\circ} C$.
For both diatomic gases (hydrogen and nitrogen),the number of degrees of freedom $f$ is $5$ (at moderate temperatures).
Therefore,the average kinetic energy per molecule for hydrogen is $\langle KE \rangle_{H_2} = \frac{5}{2} K_B T$.
Similarly,the average kinetic energy per molecule for nitrogen is $\langle KE \rangle_{N_2} = \frac{5}{2} K_B T$.
The ratio of the average kinetic energies per molecule is $\frac{\langle KE \rangle_{H_2}}{\langle KE \rangle_{N_2}} = \frac{\frac{5}{2} K_B T}{\frac{5}{2} K_B T} = 1: 1$.
The mass ratio is irrelevant because the average kinetic energy per molecule depends only on the temperature.

(A) According to Dalton's Law of Partial Pressures,the total pressure exerted by a mixture of non-reacting gases is the sum of the partial pressures of the individual gases.
For gas $A$,the pressure is $P$ at volume $V$ and temperature $T$.
For gas $B$,the pressure is $P$ at volume $V$ and temperature $T$.
When mixed in a container of volume $V$ at temperature $T$,the total pressure $P_{mix}$ is given by the sum of the individual pressures:
$P_{mix} = P_A + P_B = P + P = 2 P$.

(A) According to Dalton's law of partial pressure,the total pressure $p$ of a mixture of gases is the sum of the partial pressures of the individual gases: $p = p_1 + p_2 + p_3$.
Using the ideal gas equation $pV = nRT$,the pressure of each gas is given by $p_i = \frac{n_i RT}{V}$.
Since the temperature $T$ and volume $V$ are the same for all gases in the mixture,the total pressure is:
$p = \frac{n_1 RT}{V} + \frac{n_2 RT}{V} + \frac{n_3 RT}{V}$
$p = \frac{(n_1 + n_2 + n_3) RT}{V}$.

(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 $p_{mix} = \frac{n_{mix} RT}{V_{mix}} = \frac{(2/RT) \times RT}{1} = 2 \,atm$.