Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation Questions in English

(A) Given the pressure-volume relation: $P = \frac{a}{1 + (V/b)^2}$.
At $V = b$,the pressure $P$ becomes: $P = \frac{a}{1 + (b/b)^2} = \frac{a}{1 + 1} = \frac{a}{2}$.
Using the ideal gas equation for $n = 1$ mole: $PV = nRT$.
Substituting $P = \frac{a}{2}$,$V = b$,and $n = 1$:
$(\frac{a}{2}) \times b = 1 \times R \times T$.
Solving for $T$: $T = \frac{ab}{2R}$.

(A) Initial state: $PV = n_1RT$. Given $P_1 = 2\, atm = 2.026 \times 10^5\, Pa$,$V = 8 \times 10^{-3}\, m^3$,$T = 300\, K$,$R = 8.314\, J/(mol\cdot K)$.
$n_1 = \frac{P_1V}{RT} = \frac{2.026 \times 10^5 \times 8 \times 10^{-3}}{8.314 \times 300} \approx 0.65\, mol$.
Using the provided approximation $1\, atm = 10^5\, Pa$ as per the solution logic: $n_1 = \frac{2 \times 10^5 \times 8 \times 10^{-3}}{8.31 \times 300} = 0.64\, mol$.
Since temperature $T$ and volume $V$ are constant,$P \propto n$,so $\frac{P_2}{P_1} = \frac{n_2}{n_1}$.
$P_2 = 125\, kPa = 1.25 \times 10^5\, Pa$.
$n_2 = n_1 \times \frac{P_2}{P_1} = 0.64 \times \frac{1.25 \times 10^5}{2 \times 10^5} = 0.64 \times 0.625 = 0.40\, mol$.
Leaked moles $= n_1 - n_2 = 0.64 - 0.40 = 0.24\, mol$.

(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{N}{N_A}$. Thus,$PV = \frac{N}{N_A}RT$,which implies $N = \frac{PVN_A}{RT}$.
For the right part $(N_R)$: $P_R = 2P$,$V_R = 2V$,$T_R = T$.
$N_R = \frac{(2P)(2V)N_A}{RT} = \frac{4PVN_A}{RT}$.
For the left part $(N_L)$: $P_L = P$,$V_L = V$,$T_L = T$.
$N_L = \frac{(P)(V)N_A}{RT} = \frac{PVN_A}{RT}$.
The ratio of the number of molecules in the right part to the left part is:
$\frac{N_R}{N_L} = \frac{4PVN_A / RT}{PVN_A / RT} = \frac{4}{1}$.
Therefore,the ratio is $4:1$.

(D) Using the ideal gas equation $PV = \mu RT$, where $\mu = \frac{m}{M_w}$ and $\rho = \frac{m}{V}$, we have $m = \rho V$. Substituting $\mu = \frac{\rho V}{M_w}$ into the ideal gas equation:
$PV = \left( \frac{\rho V}{M_w} \right) RT \Rightarrow T = \frac{P M_w}{\rho R}$
Given:
$P = 1.4 \times 10^9 \, \text{atm} = 1.4 \times 10^9 \times 1.013 \times 10^5 \, N \, m^{-2} \approx 1.418 \times 10^{14} \, Pa$
$\rho = 1.4 \, g \, cm^{-3} = 1400 \, kg \, m^{-3}$
$M_w = 2 \, g \, mol^{-1} = 2 \times 10^{-3} \, kg \, mol^{-1}$
$R = 8.4 \, J \, mol^{-1} \, K^{-1}$
Substituting the values:
$T = \frac{(1.4 \times 10^9 \times 1.013 \times 10^5) \times (2 \times 10^{-3})}{1400 \times 8.4}$
$T = \frac{1.418 \times 10^{14} \times 2 \times 10^{-3}}{11760} = \frac{2.836 \times 10^{11}}{11760} \approx 2.41 \times 10^7 \, K$
Thus, the temperature is approximately $2.4 \times 10^7 \, K$.

(C) Since the containers are connected,the pressure $P$ in both containers must be equal,i.e.,$P_X = P_Y$.
Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,we get $P = \frac{mRT}{MV}$.
For both containers,$P = \frac{mRT}{MV}$ is constant,so $\frac{m_X T_X}{V_X} = \frac{m_Y T_Y}{V_Y}$.
Given $V_X = 2V_Y$,$T_X = 200 \, K$,$T_Y = 400 \, K$,and $m_X = m$.
Substituting these values: $\frac{m \times 200}{2V_Y} = \frac{m_Y \times 400}{V_Y}$.
Simplifying: $\frac{100m}{V_Y} = \frac{400m_Y}{V_Y}$.
$100m = 400m_Y \Rightarrow m_Y = \frac{100m}{400} = \frac{m}{4}$.

(C) From the ideal gas equation $PV = \frac{m}{M_0}RT$,where $m$ is mass,$M_0$ is molar mass,$R$ is the gas constant,and $T$ is temperature.
Since $m$ and $T$ are constant,$P \propto \frac{1}{M_0 V}$.
Therefore,the ratio of pressures is $\frac{P_A}{P_B} = \frac{M_{0,B}}{M_{0,A}} \times \frac{V_B}{V_A}$.
Given: $M_{0,A} (N_2) = 28 \, g/mol$,$M_{0,B} (O_2) = 32 \, g/mol$,and $V_B = 2V_A$.
Substituting these values: $\frac{P_A}{P_B} = \frac{32}{28} \times \frac{2V_A}{V_A} = \frac{32}{28} \times 2 = \frac{64}{28} = \frac{16}{7}$.

(A) The ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles.
Since $n = N/N_A$ (where $N$ is the number of molecules and $N_A$ is Avogadro's number),the equation becomes $PV = (N/N_A)RT$.
For container $A$: $P_A V_A = (N_A / N_{Avogadro}) R T_A \Rightarrow PV = (N_A / N_{Avogadro}) R T$.
For container $B$: $P_B V_B = (N_B / N_{Avogadro}) R T_B \Rightarrow (2P)(V/4) = (N_B / N_{Avogadro}) R (2T) \Rightarrow PV/2 = (N_B / N_{Avogadro}) R (2T)$.
Dividing the two equations:
$\frac{PV}{PV/2} = \frac{N_A}{N_B} \times \frac{T}{2T} \Rightarrow 2 = \frac{N_A}{N_B} \times \frac{1}{2}$.
Therefore,$\frac{N_A}{N_B} = 4$,which is $4:1$.

(B) The universal gas constant $R$ is given by the formula $R = \frac{PV}{nT}$.
At $STP$ (Standard Temperature and Pressure),for $n = 1 \ mol$ of an ideal gas:
$P = 1.013 \times 10^5 \ Pa$ (or $1 \ atm$),
$V = 22.4 \times 10^{-3} \ m^3$,
$T = 273.15 \ K$.
Substituting these values:
$R = \frac{(1.013 \times 10^5 \ Pa) \times (22.4 \times 10^{-3} \ m^3)}{1 \ mol \times 273.15 \ K} \approx 8.314 \ J \ mol^{-1} K^{-1}$.
Thus,the value is approximately $8.31 \ J \ mol^{-1} K^{-1}$.

(C) Using the ideal gas law,$PV = \mu RT$,we have the relation $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given: $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$,$P_2 = 2P_1$,and $V_2 = 3V_1$.
Substituting these values into the equation:
$\frac{P_1 V_1}{T_1} = \frac{(2P_1)(3V_1)}{T_2}$
$\frac{1}{T_1} = \frac{6}{T_2}$
$T_2 = 6T_1 = 6 \times 300 = 1800 \ K$.
To convert the temperature to Celsius: $T(^{\circ}C) = 1800 - 273 = 1527^{\circ}C$.

(D) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$.
Since pressure $P$,volume $V$,and molar mass $M$ are constant,we have $m_1 T_1 = m_2 T_2$.
Given: $m_1 = 13 \, g$,$T_1 = 27 + 273 = 300 \, K$,$T_2 = 52 + 273 = 325 \, K$.
Substituting the values: $13 \times 300 = m_2 \times 325$.
$m_2 = \frac{13 \times 300}{325} = \frac{3900}{325} = 12 \, g$.
The amount of gas released is $\Delta m = m_1 - m_2 = 13 \, g - 12 \, g = 1 \, g$.

(D) From the ideal gas equation,$PV = \frac{m}{M}RT$,where $M$ is the molar mass.
For a constant pressure $P$,the volume $V$ is given by $V = (\frac{mR}{MP})T$.
The slope of the $V-T$ graph is $S = \frac{mR}{MP}$.
For the initial case,the slope is $S_1 = \frac{mR}{MP}$,which corresponds to graph $D$.
For the second case,the mass is $2m$ and the pressure is $P/2$.
The new slope is $S_2 = \frac{(2m)R}{M(P/2)} = 4 \times \frac{mR}{MP} = 4S_1$.
Since the slope of graph $D$ is proportional to $1$,the slope of the new graph will be proportional to $4$.
Looking at the provided graph,the slopes increase from $E$ to $A$. Graph $D$ has a slope proportional to $1$,and graph $A$ has a slope proportional to $4$ (as $4 \times 1 = 4$).
Therefore,the correct graph is $A$.

(A) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,we have $P = \frac{\rho RT}{M}$,where $\rho$ is the density of the gas.
Thus,$\frac{P}{\rho T} = \frac{R}{M} = \text{constant}$.
This implies $\frac{\rho_1}{\rho_2} = \frac{P_1}{P_2} \times \frac{T_2}{T_1}$.
Given at the top: $P_{\text{top}} = 70 \text{ cm of Hg}$,$T_{\text{top}} = 7^{\circ}\text{C} = 280 \text{ K}$.
Given at the bottom: $P_{\text{bottom}} = 76 \text{ cm of Hg}$,$T_{\text{bottom}} = 27^{\circ}\text{C} = 300 \text{ K}$.
Therefore,the ratio of density at the top to the density at the bottom is:
$\frac{\rho_{\text{top}}}{\rho_{\text{bottom}}} = \frac{P_{\text{top}}}{P_{\text{bottom}}} \times \frac{T_{\text{bottom}}}{T_{\text{top}}} = \frac{70}{76} \times \frac{300}{280} = \frac{70}{76} \times \frac{30}{28} = \frac{70}{76} \times \frac{15}{14} = \frac{5 \times 15}{76} = \frac{75}{76}$.

(B) For an ideal gas at constant temperature,the pressure $P$ is directly proportional to the mass $m$ of the gas in the container $(P \propto m)$.
Given:
Initial pressure $P_1 = 10^7 \ N/m^2$
Initial mass $m_1 = 10 \ kg$
Final pressure $P_2 = 2.5 \times 10^6 \ N/m^2$
Using the relation $\frac{P_1}{P_2} = \frac{m_1}{m_2}$:
$\frac{10^7}{2.5 \times 10^6} = \frac{10}{m_2}$
Solving for $m_2$:
$m_2 = \frac{2.5 \times 10^6 \times 10}{10^7} = 2.5 \ kg$
The mass of gas removed is $\Delta m = m_1 - m_2 = 10 \ kg - 2.5 \ kg = 7.5 \ kg$.

(C) According to Charles's Law,for a fixed mass of gas at constant pressure,the volume is directly proportional to its absolute temperature: $V \propto T$.
This can be written as $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$ and $T_2 = 327^{\circ}C = 327 + 273 = 600 \ K$.
Substituting the values: $\frac{V}{300} = \frac{V_2}{600}$.
Solving for $V_2$: $V_2 = V \times \frac{600}{300} = 2V$.
Therefore,the new volume of the gas is $2V$.

(A) Let the initial pressure be $P_1 = P$ and the initial temperature be $T_1 = T$ (in Kelvin).
Given that the pressure increases by $0.4\%$,the new pressure is $P_2 = P + 0.004P = 1.004P$.
Given that the temperature increases by $1\,^{\circ}\text{C}$,the new temperature is $T_2 = T + 1$.
Assuming the volume remains constant,according to Gay-Lussac's Law,$\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Substituting the values: $\frac{P}{T} = \frac{1.004P}{T + 1}$.
$T + 1 = 1.004T$.
$1 = 0.004T$.
$T = \frac{1}{0.004} = \frac{1000}{4} = 250\, \text{K}$.

(A) From the ideal gas equation,$PV = \mu RT$,we can write:
$P = \left( \frac{\mu R}{V} \right) T$
Comparing this with the equation of a straight line $y = mx$,where $y = P$ and $x = T$,the slope $m$ is given by:
$m = \tan \theta = \frac{\mu R}{V}$
Since the number of moles $\mu$ and the gas constant $R$ are constant,we have:
$V \propto \frac{1}{\tan \theta}$
As the angle $\theta$ increases,$\tan \theta$ increases,and therefore the volume $V$ decreases.
From the graph,the slope of line $1$ is equal to the slope of line $2$ (as they are parallel),so $V_1 = V_2$.
Similarly,the slope of line $3$ is equal to the slope of line $4$,so $V_3 = V_4$.
Since the slope of line $3$ (and $4$) is greater than the slope of line $1$ (and $2$),we have $\tan \theta_3 > \tan \theta_1$.
Therefore,$V_3 < V_1$ or $V_2 > V_3$.
Thus,the correct relation is $V_1 = V_2, V_3 = V_4$ and $V_2 > V_3$.

(A) At $S.T.P.$ (Standard Temperature and Pressure),the molar volume of an ideal gas is $22.4 \, litres$.
According to Avogadro's hypothesis,$1 \, mole$ of any gas contains $6.023 \times 10^{23}$ molecules.
Therefore,$22.4 \, litres$ of gas contains $6.023 \times 10^{23}$ molecules.
To find the number of molecules in $1 \, litre$,we divide the total number of molecules by the molar volume:
Number of molecules $= \frac{6.023 \times 10^{23}}{22.4} \approx 2.68 \times 10^{22}$ molecules.

(A) At $N.T.P.$,the molar volume of an ideal gas is $22.4 \, L$.
Since $1 \, L = 1000 \, cm^{3}$,the volume of $1 \, mole$ of gas is $22.4 \times 10^{3} \, cm^{3} = 22400 \, cm^{3}$.
According to Avogadro's hypothesis,$1 \, mole$ of any gas contains $6.023 \times 10^{23}$ molecules.
Therefore,the number of molecules in $22400 \, cm^{3}$ is $6.023 \times 10^{23}$.
Thus,the number of molecules in $1 \, cm^{3}$ is $\frac{6.023 \times 10^{23}}{22400}$.

(B) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.
Density $\rho = \frac{m}{V} = \frac{PM}{RT}$.
The ratio of density to pressure is $\frac{\rho}{P} = \frac{M}{RT}$.
Since $M$ and $R$ are constants,$\frac{\rho}{P} \propto \frac{1}{T}$.
Let the ratio at $T_1 = 10^{\circ} C = 283 \ K$ be $x_1 = x$.
Let the ratio at $T_2 = 110^{\circ} C = 383 \ K$ be $x_2$.
Since $x \cdot T = \text{constant}$,we have $x_1 T_1 = x_2 T_2$.
$x \cdot 283 = x_2 \cdot 383$.
Therefore,$x_2 = \left( \frac{283}{383} \right)x$.

(C) According to the ideal gas equation,
$PV = nRT$
Rearranging for $V$ in terms of $T$:
$V = \left( \frac{nR}{P} \right) T$
This represents a straight line passing through the origin,where the slope $m$ is given by:
$m = \frac{V}{T} = \frac{nR}{P}$
Since $n$ and $R$ are constants,the slope is inversely proportional to the pressure $P$ $(m \propto \frac{1}{P})$.
From the given figure,the angle $\theta_2 > \theta_1$,which implies that the slope of the line for $P_2$ is greater than the slope of the line for $P_1$:
$(\text{Slope})_2 > (\text{Slope})_1$
Since slope is inversely proportional to pressure,a higher slope corresponds to a lower pressure:
$P_2 < P_1$

(C) According to the ideal gas equation,the molecular weight $M$ of an ideal gas is given by:
$M = \frac{\rho R T}{P}$
where $P$ is the pressure,$T$ is the temperature,$\rho$ is the density,and $R$ is the universal gas constant.
For gases $A$ and $B$:
$M_A = \frac{\rho_A R T_A}{P_A}$ and $M_B = \frac{\rho_B R T_B}{P_B}$
The ratio of their molecular weights is:
$\frac{M_A}{M_B} = \left( \frac{\rho_A}{\rho_B} \right) \left( \frac{T_A}{T_B} \right) \left( \frac{P_B}{P_A} \right)$
Given:
$\frac{\rho_A}{\rho_B} = 1.5 = \frac{3}{2}$
$T_A = T_B \implies \frac{T_A}{T_B} = 1$
$P_A = 2 P_B \implies \frac{P_B}{P_A} = \frac{1}{2}$
Substituting these values:
$\frac{M_A}{M_B} = \left( \frac{3}{2} \right) \times (1) \times \left( \frac{1}{2} \right) = \frac{3}{4} = 0.75$

(D) The ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles.
We know that $n = \frac{\text{mass of gas}}{M}$,where $M$ is the molar mass.
Also,the molar mass $M = m N_A$,where $m$ is the mass of one molecule and $N_A$ is Avogadro's number.
Substituting these into the ideal gas equation: $PV = \left(\frac{\text{mass}}{m N_A}\right) RT$.
Rearranging for density $\rho = \frac{\text{mass}}{V}$,we get $\rho = \frac{P m N_A}{RT}$.
Since the Boltzmann constant $k = \frac{R}{N_A}$,we have $R = N_A k$.
Substituting $R$ in the density equation: $\rho = \frac{P m N_A}{(N_A k) T} = \frac{Pm}{kT}$.

(C) From the ideal gas equation,$PV = nRT$.
Rearranging for the universal gas constant,we get $R = \frac{PV}{nT}$.
The unit of pressure $P$ is $Pascal$ $(Pa)$ or $Joule/m^3$,and volume $V$ is $m^3$. Thus,the product $PV$ has the unit of energy,which is $Joule$ $(J)$.
The unit of amount of substance $n$ is $mole$ $(mol)$ and temperature $T$ is $Kelvin$ $(K)$.
Substituting these units,we get the unit of $R = \frac{J}{mol \cdot K} = J\,K^{-1}mol^{-1}$.

(C) For an ideal gas, the equation of state is $PV = nRT$.
An isotherm is a process where the temperature $T$ remains constant.
Since $n$ and $R$ are constants, for an isotherm, $PV = \text{constant}$.
Given initial conditions: $P_i = 3$ and $V_i = 4$.
Therefore, the product $P_i V_i = 3 \times 4 = 12$.
We check the product $PV$ for each process:
Process $A$: $P \times V = 5 \times 7 = 35 \neq 12$.
Process $B$: $P \times V = 4 \times 6 = 24 \neq 12$.
Process $C$: $P \times V = 12 \times 1 = 12$.
Process $D$: $P \times V = 6 \times 3 = 18 \neq 12$.
Thus, only process $C$ satisfies the condition $P_f V_f = P_i V_i$.

(C) Let the initial pressure be $P = 1 \text{ atm}$ and initial temperature be $T = 300 \text{ K}$.
The total number of moles of gas in the system is constant.
Initial moles: $\mu = \frac{PV}{RT} + \frac{PV}{RT} = \frac{2PV}{RT}$.
Let the final common pressure be $P'$.
Final moles: $\mu' = \frac{P'V}{RT_1} + \frac{P'V}{RT_2}$,where $T_1 = 600 \text{ K}$ and $T_2 = 300 \text{ K}$.
Since $\mu = \mu'$,we have: $\frac{2PV}{RT} = \frac{P'V}{RT_1} + \frac{P'V}{RT_2}$.
Canceling $V/R$ from both sides: $\frac{2P}{T} = P' \left( \frac{1}{T_1} + \frac{1}{T_2} \right) = P' \left( \frac{T_1 + T_2}{T_1 T_2} \right)$.
Solving for $P'$: $P' = \frac{2P T_1 T_2}{T(T_1 + T_2)}$.
Substituting the values: $P' = \frac{2 \times 1 \times 600 \times 300}{300 \times (600 + 300)} = \frac{2 \times 600}{900} = \frac{1200}{900} = \frac{4}{3} \approx 1.33 \text{ atm}$.

(C) The ideal gas equation is given by $PV = \mu RT$,where $\mu$ is the number of moles.
Given:
Pressure $P = 21 \times 10^4 \ N/m^2$
Volume $V = 83 \ L = 83 \times 10^{-3} \ m^3$
Temperature $T = 27^\circ C = 27 + 273 = 300 \ K$
Gas constant $R = 8.3 \ J/mol/K$
Substituting the values into the equation:
$\mu = \frac{PV}{RT} = \frac{21 \times 10^4 \times 83 \times 10^{-3}}{8.3 \times 300}$
$\mu = \frac{21 \times 10^4 \times 83 \times 10^{-3}}{8.3 \times 300} = \frac{21 \times 830}{2490} = \frac{17430}{2490} = 7 \ gm-mole$.

(B) Given: Initial pressure $P_1 = 720 \, kPa$,Initial temperature $T_1 = 40^\circ C = 273 + 40 = 313 \, K$.
Final temperature $T_2 = 353^\circ C = 273 + 353 = 626 \, K$.
Since $\frac{1}{4}^{th}$ of the gas is released,the remaining mass $m_2 = \frac{3}{4} m_1$.
Using the ideal gas law $PV = nRT = \frac{m}{M} RT$,we have $P \propto mT$ (since $V$ and $M$ are constant).
Therefore,$\frac{P_2}{P_1} = \frac{m_2}{m_1} \times \frac{T_2}{T_1}$.
Substituting the values: $\frac{P_2}{720} = \frac{3}{4} \times \frac{626}{313} = \frac{3}{4} \times 2 = 1.5$.
$P_2 = 1.5 \times 720 = 1080 \, kPa$.

(B) Since the volume of the pressure cooker remains constant,we use Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given: $P_1 = 1$ atm,$T_1 = 30^\circ C = 30 + 273 = 303$ $K$,and $P_2 = 3$ atm.
Substituting the values: $\frac{1}{303} = \frac{3}{T_2}$.
Solving for $T_2$: $T_2 = 3 \times 303 = 909$ $K$.
To convert the temperature to Celsius: $T_2(^\circ C) = 909 - 273 = 636^\circ C$.

(D) From the ideal gas equation,$PV = nRT = (m/M)RT$.
Since the temperature $T$,volume $V$,and molar mass $M$ are constant,we have $P \propto m$.
Therefore,$\frac{P_1}{P_2} = \frac{m_1}{m_2}$.
Given $P_1 = 10^7\, N/m^2$,$m_1 = 10\, kg$,and $P_2 = 2.5 \times 10^6\, N/m^2$.
Substituting the values: $\frac{10^7}{2.5 \times 10^6} = \frac{10}{m_2}$.
$4 = \frac{10}{m_2} \implies m_2 = \frac{10}{4} = 2.5\, kg$.
The mass of the gas taken out is $\Delta m = m_1 - m_2 = 10 - 2.5 = 7.5\, kg$.

(D) For an open-mouthed vessel,the pressure $P$ remains constant as it is equal to the atmospheric pressure.
The volume $V$ of the vessel is also constant.
From the ideal gas equation $PV = \mu RT = \left( \frac{m}{M} \right) RT$,where $m$ is the mass of the gas and $M$ is the molar mass,we have $m \propto \frac{1}{T}$ at constant $P$ and $V$.
Therefore,$m_1 T_1 = m_2 T_2$,which implies $\frac{T_2}{T_1} = \frac{m_1}{m_2}$.
Initial temperature $T_1 = 60 + 273 = 333 \ K$.
Since $1/4^{th}$ of the air escapes,the remaining mass $m_2 = m_1 - \frac{1}{4}m_1 = \frac{3}{4}m_1$.
Substituting the values: $\frac{T_2}{333} = \frac{m_1}{(3/4)m_1} = \frac{4}{3}$.
$T_2 = 333 \times \frac{4}{3} = 444 \ K$.
Converting to Celsius: $T = 444 - 273 = 171^oC$.

(D) From the ideal gas equation,the number of molecules $N$ is given by $N = \frac{PV}{kT}$.
For compartment $I$,the parameters are $P, V, T$. Thus,the number of molecules $N_I = \frac{PV}{kT}$.
For compartment $II$,the parameters are $2P, 2V, T$. Thus,the number of molecules $N_{II} = \frac{(2P)(2V)}{kT} = 4 \frac{PV}{kT} = 4N_I$.
The ratio of the number of molecules in compartments $I$ and $II$ is $\frac{N_I}{N_{II}} = \frac{N_I}{4N_I} = \frac{1}{4}$.

(A) Given:
Volume $V = 7 \text{ L} = 7 \times 10^{-3} \text{ m}^{3}$
Temperature $T = 0^{\circ}C = 273 \text{ K}$
Pressure $P = 1.3 \times 10^{5} \text{ Pa}$
Universal gas constant $R = 8.314 \text{ J/(mol K)}$
Using the ideal gas equation $PV = nRT$,we find the number of moles $n$:
$n = \frac{PV}{RT} = \frac{1.3 \times 10^{5} \times 7 \times 10^{-3}}{8.314 \times 273} \approx 0.4 \text{ moles}$
The number of molecules $N$ is given by $N = n \times N_{A}$,where $N_{A} = 6.022 \times 10^{23} \text{ molecules/mol}$.
$N = 0.4 \times 6.022 \times 10^{23} \approx 2.4 \times 10^{23} \text{ molecules}$.

(C) Since the pistons are connected by a wire,the volume $V$ of the gas remains constant.
Using the ideal gas law for a constant volume process: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given $P_1 = P_0$,$T_1 = T_0$,and $T_2 = 2T_0$,we have:
$\frac{P_0}{T_0} = \frac{P'}{2T_0} \Rightarrow P' = 2P_0$.
The pressure inside the tube is now $2P_0$,while the external atmospheric pressure is $P_0$.
The net force acting outwards on each piston is $F_{net} = (P' - P_0)A = (2P_0 - P_0)A = P_0 A$.
Since the wire connects the two pistons,the tension $T$ in the wire must balance this net outward force.
Therefore,the tension in the wire is $T = P_0 A$.

(D) According to the ideal gas equation:
$P V = n R T$
Since the volume $V$ is constant for each vessel,we can write:
$P = \left( \frac{n R}{V} \right) T$
This equation is of the form $y = m x$,where $y = P$,$x = T$,and the slope $m = \frac{n R}{V}$.
Since the mass is given,$n$ is constant. However,the volumes $V$ of the small and large vessels are different.
Therefore,the slope $m = \frac{n R}{V}$ will be different for the two vessels.
Thus,the $P-T$ curves are linear,passing through the origin,but with different slopes.

(B) Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass of argon.
For the first state: $PV = \frac{4}{M}RT$ ... $(i)$
For the second state,the temperature is $(T + 50) \ K$,the mass is $(4.0 - 0.8) = 3.2 \ g$,and the pressure $p$ remains constant: $PV = \frac{3.2}{M}R(T + 50)$ ... (ii)
Equating $(i)$ and (ii) since $PV$ is constant:
$\frac{4}{M}RT = \frac{3.2}{M}R(T + 50)$
$4T = 3.2(T + 50)$
$4T = 3.2T + 160$
$0.8T = 160$
$T = \frac{160}{0.8} = 200 \ K$.

(A) According to the Ideal Gas Law,$PV = nRT$. Since the volume $V$ and temperature $T$ are constant,the pressure $P$ is directly proportional to the number of moles $n$ $(P \propto n)$.
Let the molecular weights of the two gases be $M_A$ and $M_B$.
For the first gas $A$,the number of moles is $n_A = \frac{2}{M_A}$. Given $P_A = 1 \, atm$,we have $\frac{2}{M_A} \propto 1 \dots (i)$.
When the second gas $B$ is added,the total pressure becomes $1.5 \, atm$. The partial pressure of gas $B$ is $P_B = P_{total} - P_A = 1.5 - 1 = 0.5 \, atm$.
For the second gas $B$,the number of moles is $n_B = \frac{3}{M_B}$. Thus,$\frac{3}{M_B} \propto 0.5 \dots (ii)$.
Dividing equation $(ii)$ by equation $(i)$:
$\frac{3/M_B}{2/M_A} = \frac{0.5}{1}$
$\frac{3}{M_B} \times \frac{M_A}{2} = 0.5$
$\frac{M_A}{M_B} = 0.5 \times \frac{2}{3} = \frac{1}{3}$
Therefore,the ratio of the molecular weights $M_A : M_B$ is $1 : 3$.

(C) Initially,the air column length $L_i = 0.05 \, m = 5 \, cm$. The pressure of the air $P_i = P_{atm} = 76 \, cm$ of $Hg$.
When the tube is raised by $0.43 \, m = 43 \, cm$,let the new length of the air column be $x \, cm$. The mercury level inside the tube rises by $(48 - x) \, cm$ relative to the initial position,where $48 \, cm$ is the total displacement of the tube relative to the mercury surface.
The final pressure of the air inside the tube is $P_f = P_{atm} - (48 - x) = 76 - 48 + x = (28 + x) \, cm$ of $Hg$.
Since the temperature remains constant,we apply Boyle's Law: $P_i V_i = P_f V_f$.
Assuming the cross-sectional area $A$ is constant:
$76 \times 5 \times A = (28 + x) \times x \times A$
$380 = 28x + x^2$
$x^2 + 28x - 380 = 0$
Solving the quadratic equation:
$x = \frac{-28 \pm \sqrt{28^2 - 4(1)(-380)}}{2} = \frac{-28 \pm \sqrt{784 + 1520}}{2} = \frac{-28 \pm \sqrt{2304}}{2} = \frac{-28 \pm 48}{2}$
Taking the positive root: $x = \frac{20}{2} = 10 \, cm = 0.1 \, m$.

(B) Initial mass of air is $M_1 = M$ at temperature $T_1 = 60 + 273 = 333 \ K$.
Since the vessel is open,the pressure remains constant (atmospheric pressure).
According to the ideal gas law,$PV = nRT = \frac{M}{m}RT$,where $m$ is the molar mass of air.
Since $P$,$V$,and $m$ are constant,we have $M_1 T_1 = M_2 T_2$.
After heating,one-fourth of the air escapes,so the remaining mass is $M_2 = M - \frac{M}{4} = \frac{3M}{4}$.
Substituting the values: $M \times 333 = \frac{3M}{4} \times T_2$.
$T_2 = 333 \times \frac{4}{3} = 444 \ K$.
Converting back to Celsius: $T = 444 - 273 = 171^{\circ}C$.

(D) The ideal gas law $PV = nRT$ must hold both before and after the leakage of the gas.
Since the volume $V$ of the flask remains constant,we have $\frac{P_i}{n_i T_i} = \frac{P_f}{n_f T_f}$.
Given initial conditions: $P_i = 10\, atm$,$T_i = 57 + 273 = 330\, K$,$m_i = 28\, g$,$n_i = \frac{28}{28} = 1\, mol$.
Given final conditions: $P_f = \frac{10}{2} = 5\, atm$,$T_f = 27 + 273 = 300\, K$.
Using the relation $n_f = \frac{P_f}{P_i} \cdot \frac{T_i}{T_f} \cdot n_i$,we get:
$n_f = \frac{5}{10} \cdot \frac{330}{300} \cdot 1 = \frac{1}{2} \cdot \frac{11}{10} = \frac{11}{20}\, mol$.
The final mass $m_f = n_f \times M = \frac{11}{20} \times 28 = 15.4\, g$.
The mass of $N_2$ gas that leaked out is $\Delta m = m_i - m_f = 28 - 15.4 = 12.6\, g$.
Converting $12.6\, g$ to fraction: $12.6 = \frac{126}{10} = \frac{63}{5}\, g$.

(A) Initial state: $T_1 = 27 + 273 = 300\,K$,$V_1 = V_2 = 100\,cm^3$,$P_1 = P_2 = P$.
Final state: One part is at $T_1' = 300\,K$,other at $T_2' = 100 + 273 = 373\,K$.
Let the piston move by distance $x$. The new volumes are $V_1' = 100 - 10.85x$ and $V_2' = 100 + 10.85x$.
Since the piston is in equilibrium,the final pressures are equal: $P_1' = P_2' = P'$.
Using the ideal gas law $PV = nRT$,and noting that the number of moles $n$ remains constant:
$P_1 V_1 / T_1 = P_1' V_1' / T_1' \implies P(100) / 300 = P'(100 - 10.85x) / 300 \implies P' = P \cdot 100 / (100 - 10.85x)$.
$P_2 V_2 / T_2 = P_2' V_2' / T_2' \implies P(100) / 300 = P'(100 + 10.85x) / 373$.
Equating $P'$ from both: $P \cdot 100 / (100 - 10.85x) = P \cdot 100 \cdot 373 / (300(100 + 10.85x))$.
$300(100 + 10.85x) = 373(100 - 10.85x)$.
$30000 + 3255x = 37300 - 4047.05x$.
$7302.05x = 7300 \implies x \approx 1\,cm$.

(B) Given: Mass $w = 12\,g$,Initial Volume $V_1 = 4\times 10^{-3}\,m^3$,Initial Temperature $T_1 = 7 + 273 = 280\,K$.
Density $\rho = \frac{w}{V}$. Since pressure $P$ is constant,from the ideal gas law $PV = nRT = \frac{w}{M}RT$,we have $P = \frac{\rho RT}{M}$.
Since $P$ and $M$ are constant,$\rho_1 T_1 = \rho_2 T_2$.
Initial density $\rho_1 = \frac{w}{V_1} = \frac{12}{4 \times 10^{-3}} = 3000\,g/m^3 = 3 \times 10^{-3}\,g/cc$.
Final density $\rho_2 = 6 \times 10^{-4}\,g/cc = 0.6 \times 10^{-3}\,g/cc$.
Using $\rho_1 T_1 = \rho_2 T_2$:
$T_2 = \frac{\rho_1 T_1}{\rho_2} = \frac{3 \times 10^{-3} \times 280}{0.6 \times 10^{-3}} = \frac{3 \times 280}{0.6} = 5 \times 280 = 1400\,K$.

(C) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$ ($M$ is the molar mass).
At constant pressure $P$,the equation becomes $V = \left(\frac{mR}{MP}\right)T$.
The slope of the $V-T$ graph is given by $S = \frac{mR}{MP}$.
In the first case,the slope is $S_1 = \frac{mR}{MP}$.
In the second case,the mass is $2m$ and the pressure is $2P$. The new slope $S_2$ is:
$S_2 = \frac{(2m)R}{M(2P)} = \frac{mR}{MP} = S_1$.
Since the slope remains the same,the expansion of the gas will be represented by the same straight line $B$.
Therefore,the correct option is $C$.

(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$ ($M$ is the molar mass of the gas).
Since the containers are connected by a thin tube,the pressure $P$ in both containers must be the same at equilibrium.
For container $X$: $V_X = 2V$,$T_X = 200\,K$,and mass $m_X = m$.
Thus,$P(2V) = \left(\frac{m}{M}\right) R(200) \implies P = \frac{mR(200)}{2VM} = \frac{100mR}{VM}$.
For container $Y$: $V_Y = V$,$T_Y = 400\,K$,and mass $m_Y = m_Y$.
Thus,$P(V) = \left(\frac{m_Y}{M}\right) R(400) \implies P = \frac{400m_Y R}{VM}$.
Equating the expressions for $P$: $\frac{100mR}{VM} = \frac{400m_Y R}{VM}$.
Solving for $m_Y$: $100m = 400m_Y \implies m_Y = \frac{100m}{400} = \frac{m}{4}$.

(C) For an ideal gas,the equation of state is $PV = nRT$.
At a constant temperature $T$,the product $PV = nRT$ is constant for a given number of moles $n$.
From the graph,for a fixed volume $V$,the pressure $P_B > P_A$.
Since $P_B V = n_B RT$ and $P_A V = n_A RT$,we have $P_B V > P_A V$,which implies $n_B RT > n_A RT$.
Therefore,$n_B > n_A$.
Since $n_B > n_A$,the statement $n_A > n_B$ (option $C$) is incorrect.
Regarding the masses $M_A$ and $M_B$,the relationship depends on the molar masses of the gases,which are not specified. Thus,$M_A > M_B$ or $M_A < M_B$ could both be true depending on the gas type. However,$n_A > n_B$ is definitely incorrect based on the graph.

(A) According to Avogadro's Law,equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. Since both gases have the same volume $(1 \, cm^3)$ at $N.T.P$,the number of molecules is the same in both.
The internal energy of an ideal gas depends only on its temperature $(U = \frac{f}{2} nRT)$. Since both gases are at the same temperature and have the same number of moles (due to equal volume at $N.T.P$),their internal energies are the same.
The average velocity (root mean square velocity) of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Since the molar mass $(M)$ of hydrogen $(H_2)$ and oxygen $(O_2)$ are different,their average velocities will be different. Therefore,statement $B$ is incorrect.
Since statement $B$ is incorrect,statement $D$ is also incorrect. Thus,both $A$ and $C$ are technically correct,but in the context of standard multiple-choice questions,the most fundamental property derived from $N.T.P$ conditions is the number of molecules.

(C) Since the temperature $T$ and volume $V$ of the vessel remain constant,we use the ideal gas law $PV = nRT$.
From this,$n = \frac{PV}{RT}$. Since $V, R,$ and $T$ are constant,$n \propto P$.
Initial state: $P_1 = 5 \text{ atm}$,$n_1 = n$.
Final state: $P_2 = 4 \text{ atm}$,$n_2 = n'$.
Therefore,$\frac{n'}{n} = \frac{P_2}{P_1} = \frac{4}{5} = 0.8$.
This means $n' = 0.8n$.
The amount of gas that escaped is $\Delta n = n - n' = n - 0.8n = 0.2n$.
Percentage of gas escaped = $\frac{0.2n}{n} \times 100\% = 20\%$.

(D) Given: Initial temperature $T_i = 17 + 273 = 290 \ K$.
Final temperature $T_f = 27 + 273 = 300 \ K$.
Atmospheric pressure $P = 1 \times 10^5 \ Pa$.
Volume of the room $V = 30 \ m^3$.
The number of molecules $N$ is given by the ideal gas law $PV = N k_B T$,where $k_B$ is the Boltzmann constant $(k_B = 1.38 \times 10^{-23} \ J/K)$.
Thus,$N = \frac{PV}{k_B T}$.
Initial number of molecules $N_i = \frac{PV}{k_B T_i}$.
Final number of molecules $N_f = \frac{PV}{k_B T_f}$.
The change in the number of molecules is $N_f - N_i = \frac{PV}{k_B} \left( \frac{1}{T_f} - \frac{1}{T_i} \right)$.
Substituting the values: $N_f - N_i = \frac{1 \times 10^5 \times 30}{1.38 \times 10^{-23}} \left( \frac{1}{300} - \frac{1}{290} \right)$.
$N_f - N_i = \frac{30 \times 10^5}{1.38 \times 10^{-23}} \left( \frac{290 - 300}{300 \times 290} \right)$.
$N_f - N_i = \frac{30 \times 10^5}{1.38 \times 10^{-23}} \left( \frac{-10}{87000} \right) \approx -2.5 \times 10^{25}$.

(D) Initial state: The air column length is $L_1 = 8 \ cm$ and the pressure is atmospheric,$P_1 = 76 \ cm$ of $Hg$.
Final state: The tube is raised by $46 \ cm$. Let the new length of the air column be $L_2$. Let the height of the mercury column inside the tube above the external mercury level be $x$. The total length of the tube above the external mercury level is $8 + 46 = 54 \ cm$. Thus,$L_2 = 54 - x$.
The pressure of the trapped air in the final state is $P_2 = P_0 - x = 76 - x$.
Using Boyle's Law $(P_1 V_1 = P_2 V_2)$ and assuming constant cross-sectional area $A$:
$76 \times A \times 8 = (76 - x) \times A \times (54 - x)$
$608 = 4104 - 76x - 54x + x^2$
$x^2 - 130x + 3496 = 0$
Solving the quadratic equation using the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{130 \pm \sqrt{16900 - 13984}}{2} = \frac{130 \pm \sqrt{2916}}{2} = \frac{130 \pm 54}{2}$
$x_1 = 92$ (impossible as $x < 54$) or $x_2 = 38 \ cm$.
Therefore,the length of the air column is $L_2 = 54 - 38 = 16 \ cm$.

(D) Let the initial pressure be $P_0$ and the volume of each bulb be $V$. The initial temperature of both bulbs is $T_0 = 273 \ K$.
The total number of moles of gas is constant:
$n_{total} = n_1 + n_2 = \frac{P_0 V}{R T_0} + \frac{P_0 V}{R T_0} = \frac{2 P_0 V}{R T_0}$
When one bulb is in ice $(T_1 = 273 \ K)$ and the other is in a hot bath $(T_2 = T)$,the new pressure is $P = 1.5 P_0$.
The new number of moles is:
$n'_{total} = \frac{P V}{R T_1} + \frac{P V}{R T_2} = \frac{1.5 P_0 V}{R (273)} + \frac{1.5 P_0 V}{R T}$
Since $n_{total} = n'_{total}$:
$\frac{2 P_0 V}{R (273)} = \frac{1.5 P_0 V}{R (273)} + \frac{1.5 P_0 V}{R T}$
Dividing both sides by $\frac{P_0 V}{R}$:
$\frac{2}{273} = \frac{1.5}{273} + \frac{1.5}{T}$
$\frac{0.5}{273} = \frac{1.5}{T}$
$T = \frac{1.5 \times 273}{0.5} = 3 \times 273 = 819 \ K$
Converting to Celsius:
$T(^{\circ}C) = 819 - 273 = 546^{\circ}C$.