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

(N/A) Avogadro's law states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.
Mathematically,it is expressed as $V \propto n$ (at constant $T$ and $P$),where $V$ is the volume of the gas,$n$ is the amount of substance (number of moles),$T$ is the absolute temperature,and $P$ is the pressure.
This implies that the molar volume of all ideal gases at standard temperature and pressure $(STP)$ is the same,which is approximately $22.4 \ L$ per mole.

(N/A) The equation relating pressure $(P)$,volume $(V)$,and temperature $(T)$ of a gas is given by:
$PV = KT$
Where $T$ is the absolute temperature in Kelvin.
$K$ is a constant for a given gas,but it varies with the amount of gas.
$K$ is proportional to the number of molecules $(N)$ in the given sample of gas.
$\therefore K \propto N$
$\therefore K = k_{B} N$
Here,$k_{B}$ is the Boltzmann constant.
Its value is the same for all gases.
The unit of $k_{B}$ is $\text{J/K}$ and its dimension is $[M^{1} L^{2} T^{-2} K^{-1}]$.
Substituting $K$ into the first equation:
$PV = k_{B} NT$
$\therefore \frac{PV}{NT} = k_{B} = \text{constant}$
This implies that for two different gas samples:
$\frac{P_{1} V_{1}}{N_{1} T_{1}} = \frac{P_{2} V_{2}}{N_{2} T_{2}}$
If $P, V,$ and $T$ are the same for different gases,then $N$ must also be equal. This represents Avogadro's hypothesis,which states: Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

(N/A) $STP$ stands for standard temperature and pressure.
The standard temperature is defined as $0^{\circ}C$,which is equal to $0 + 273.15 = 273.15 \ K \approx 273 \ K$.
The standard pressure is defined as $1 \ atm$,which is equal to $1.01325 \times 10^{5} \ Nm^{-2} \approx 1.01 \times 10^{5} \ Nm^{-2}$.

(N/A) mole is defined as the amount of substance that contains as many elementary entities as there are atoms in $0.012 \ kg$ of carbon-$12$. At standard temperature and pressure $(STP)$,the mass contained in $22.4 \ L$ volume of an ideal gas is equal to its molecular mass. Its symbol is $mol$.
Equation $1$: In terms of mass,if $M$ is the mass of the gas and $M_{0}$ is the molar mass,the number of moles $\mu$ is given by:
$\mu = \frac{M}{M_{0}}$
Equation $2$: In terms of the number of molecules,if $N$ is the total number of molecules and $N_{A}$ is the Avogadro constant $(6.022 \times 10^{23} \ mol^{-1})$,the number of moles $\mu$ is given by:
$\mu = \frac{N}{N_{A}}$

(N/A) The number of molecules contained in $1 \text{ mole}$ of a substance is called the Avogadro number. It is represented by the symbol $N_{A}$.
$N_{A} = 6.022 \times 10^{23} \text{ mol}^{-1}$.
Modern definition: The number of atoms contained in exactly $12 \text{ g}$ of carbon-$12$ is called the Avogadro number.
Avogadro number is a universal constant applicable to all types of fundamental particles:
$1 \text{ mole of atoms} = 6.022 \times 10^{23} \text{ atoms/mole}$
$1 \text{ mole of ions} = 6.022 \times 10^{23} \text{ ions/mole}$
$1 \text{ mole of electrons} = 6.022 \times 10^{23} \text{ electrons/mole}$

(N/A) The ideal gas equation is given by $PV = \mu RT$.
Rearranging for $R$,we get $R = \frac{PV}{\mu T}$.
At standard temperature and pressure $(STP)$,for $1 \text{ mole}$ of an ideal gas $(\mu = 1)$:
$P = 1.013 \times 10^{5} \text{ N/m}^2$
$V = 22.4 \times 10^{-3} \text{ m}^3$
$T = 273.15 \text{ K}$
Substituting these values:
$R = \frac{1.013 \times 10^{5} \times 22.4 \times 10^{-3}}{1 \times 273.15} \approx 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$.
Unit of $R$: $\text{J mol}^{-1} \text{ K}^{-1}$ or $\text{N m mol}^{-1} \text{ K}^{-1}$.
Dimension of $R$:
Since $R = \frac{PV}{\mu T}$,the dimension is $\frac{[M^1 L^{-1} T^{-2}] [L^3]}{[mol] [K]} = [M^1 L^2 T^{-2} K^{-1} mol^{-1}]$.

(N/A) An ideal gas is a theoretical model of a gas that satisfies the equation $PV = \mu RT$ at all pressures and temperatures.
No real gas is a perfectly ideal gas.
The provided graph shows the deviation of real gases from ideal gas behavior at three different temperatures. The horizontal line represents the ideal gas,which is parallel to the $P$ (pressure) axis.
All curves approach ideal gas behavior at low pressure and high temperature.
At low pressure and high temperature,the molecules are far apart,and molecular interactions are negligible. Hence,in this limit,a real gas behaves like an ideal gas.
The state equation for an ideal gas is $PV = \mu RT$,where $R$ is the universal gas constant.
Gas Laws:
$(1)$ Boyle's Law: At a constant temperature,the pressure of a given mass of gas varies inversely with its volume.
$\therefore P \propto \frac{1}{V} \implies PV = \text{constant}$.
$(2)$ Charles's Law: At a constant pressure,the volume of a given mass of gas is directly proportional to its absolute temperature.
$\therefore V \propto T \implies \frac{V}{T} = \text{constant}$.
$(3)$ Gay-Lussac's Law: At a constant volume,the pressure of a given mass of gas is directly proportional to its absolute temperature.
$\therefore P \propto T \implies \frac{P}{T} = \text{constant}$.

(N/A) Law: The total pressure exerted by a mixture of non-reacting ideal gases is equal to the sum of the partial pressures of the individual gases.
Let the volume of the container be $V$ and the temperature of the gases be $T$. Let the total pressure of the mixture be $P$.
Let the number of moles of the individual gases be $\mu_{1}, \mu_{2}, \ldots, \mu_{n}$.
The total number of moles is $\mu = \mu_{1} + \mu_{2} + \ldots + \mu_{n}$.
According to the ideal gas equation,$PV = \mu RT$.
Substituting the total moles: $PV = (\mu_{1} + \mu_{2} + \ldots + \mu_{n}) RT$.
Therefore,$P = \frac{\mu_{1} RT}{V} + \frac{\mu_{2} RT}{V} + \ldots + \frac{\mu_{n} RT}{V}$.
Since the partial pressure of an individual gas $i$ is defined as $P_{i} = \frac{\mu_{i} RT}{V}$,we get:
$P = P_{1} + P_{2} + \ldots + P_{n}$.
Thus,in a mixture of non-interacting gases,the total pressure of the mixture is equal to the sum of the partial pressures of the individual gases.

(C) The ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles and $R$ is the universal gas constant.
Comparing this with the given equation $PV = KT$,we find that $K = nR$.
Since $n = \frac{m}{M}$ (where $m$ is the mass of the gas and $M$ is the molar mass),we have $K = \frac{m}{M} R$.
Thus,$K$ depends on the mass of the gas $(m)$ and the molar mass $(M)$,which is characteristic of the nature of the gas.
Therefore,$K$ depends on both the nature of the gas and the mass of the gas.

(N/A) Avogadro's hypothesis states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.
Mathematically,for two gases at the same temperature $T$ and pressure $P$,if their volumes $V_1 = V_2$,then the number of molecules $N_1 = N_2$.

(N/A) The Avogadro number,denoted by $N_A$,is defined as the number of constituent particles (usually atoms or molecules) that are contained in one mole of a given substance.
It represents the number of carbon atoms in exactly $12 \ g$ of the carbon-$12$ isotope.
The value of the Avogadro number is approximately $6.022 \times 10^{23} \ mol^{-1}$.

(N/A) Standard Temperature and Pressure $(STP)$ is defined by the International Union of Pure and Applied Chemistry $(IUPAC)$ as follows:
$1$. Standard Temperature: The standard temperature is defined as $273.15 \ K$ (which is equivalent to $0 \ ^\circ C$).
$2$. Standard Pressure: The standard pressure is defined as $10^5 \ Pa$ (or $1 \ bar$).
Note: In older conventions,$STP$ was often defined as $273.15 \ K$ and $1 \ atm$ $(101325 \ Pa)$,but the current $IUPAC$ standard is $1 \ bar$.

(N/A) $1$ mole is defined as the amount of substance that contains exactly $6.022 \times 10^{23}$ elementary entities (atoms,molecules,ions,or other particles). This number is known as Avogadro's number $(N_A)$.
Two different methods to represent a mole are:
$1$. In terms of mass: $A$ mole is the amount of substance whose mass in grams is numerically equal to its molar mass (atomic or molecular mass).
$2$. In terms of number of particles: $A$ mole is the amount of substance that contains $N_A = 6.022 \times 10^{23}$ particles.

(A) The universal gas constant, denoted by $R$, is a physical constant that appears in the ideal gas law equation, $PV = nRT$.
It represents the work done by one mole of an ideal gas when its temperature is increased by $1 \ K$ at constant pressure.
The value of the universal gas constant in $SI$ units is $R = 8.314 \ J \ mol^{-1} \ K^{-1}$.

(N/A) Boyle's Law: For a fixed mass of an ideal gas at a constant temperature, the pressure $P$ is inversely proportional to its volume $V$. Mathematically, $P \propto 1/V$ or $PV = \text{constant}$.
Charles's Law: For a fixed mass of an ideal gas at a constant pressure, the volume $V$ is directly proportional to its absolute temperature $T$. Mathematically, $V \propto T$ or $V/T = \text{constant}$.

(N/A) Gay-Lussac's Law states that for a given mass of an ideal gas, the pressure $(P)$ of the gas is directly proportional to its absolute temperature $(T)$, provided the volume $(V)$ remains constant.
Mathematically, this is expressed as:
$P \propto T$ (at constant $V$ and mass)
$P = kT$ or $\frac{P}{T} = \text{constant}$
where $P$ is the pressure, $T$ is the absolute temperature in Kelvin, and $k$ is a proportionality constant.

(B) Molar mass is defined as the mass of one mole of a substance.
It is typically expressed in units of $g/mol$ or $kg/mol$.
Mathematically,it is the ratio of the mass of a sample to the amount of substance in moles $(n = m/M)$.

(A) According to Charles' law, the volume of a gas is directly proportional to its absolute temperature $(V \propto T)$ at constant pressure.
As the temperature decreases, the volume of the gas decreases.
The theoretical minimum temperature is reached when the volume of the gas becomes zero.
This temperature is known as absolute zero, which is equal to $-273.15^{\circ} C$ or $0 \ K$.

(A) For an ideal gas,the equation of state is $PV = nRT$.
At a constant temperature $(T)$,$n$ and $R$ are constants,so the product $PV$ remains constant regardless of the pressure $P$.
Therefore,the graph of $PV$ versus $P$ for an ideal gas is a horizontal straight line parallel to the $P$-axis.
Looking at the provided graph,line $A$ is a horizontal line,which indicates that $PV$ is constant for all values of $P$.
Thus,graph $A$ represents an ideal gas.

(N/A) When a vehicle moves for a long period of time,the tyre becomes warm due to friction between the tyre and the road surface.
As a result,the temperature of the air trapped inside the tyre increases.
Since the volume of the tyre remains approximately constant,according to Gay-Lussac's Law ($P \propto T$ at constant volume),an increase in temperature leads to a proportional increase in the pressure of the air inside the tyre.

(C) According to Avogadro's Law,equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.
Since both oxygen and hydrogen are at $N.T.P.$ and have the same volume $(1 \text{ cm}^3)$,they will contain the same number of molecules.

(C) According to Gay-Lussac's Law,for a fixed mass of gas at constant volume,the pressure is directly proportional to its absolute temperature: $P \propto T$.
Therefore,$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$.
Given:
$P_{1} = 1 \text{ atm}$
$T_{1} = -173^{\circ} C = (-173 + 273) K = 100 K$
$P_{2} = 2 \text{ atm}$
Substituting the values:
$T_{2} = \frac{P_{2} \times T_{1}}{P_{1}} = \frac{2 \times 100}{1} = 200 K$.
Converting back to Celsius:
$t_{2} = T_{2} - 273 = 200 - 273 = -73^{\circ} C$.

(B) According to Boyle's Law,for a fixed mass of an ideal gas at a constant temperature,the pressure $P$ is inversely proportional to the volume $V$,which is expressed as $P \propto 1/V$ or $P = k(1/V)$,where $k$ is a constant.
This equation is in the form of $y = mx$,where $y = P$,$x = 1/V$,and the slope $m = k$.
Therefore,the graph of $P$ versus $1/V$ is a straight line passing through the origin.

(A) According to Charles's Law,the volume $V$ of a given mass of an ideal gas is directly proportional to its absolute temperature $T$ at constant pressure,expressed as $V \propto T$ or $V = kT$. At absolute zero temperature $(T = 0 \ K)$,the volume of the gas becomes $V = k \times 0 = 0$. Therefore,the theoretical volume of an ideal gas at absolute zero is $0$.

(N/A) According to the ideal gas equation,$PV = nRT$.
If both pressure $P$ and volume $V$ are kept constant,then the product $PV$ is constant.
Since $n$ (number of moles) and $R$ (universal gas constant) are also constants,the temperature $T$ must remain constant.
Therefore,it is mathematically impossible to increase the temperature $T$ while keeping both $P$ and $V$ constant for a fixed amount of gas.

(N/A) The universal gas constant $R$ represents the work done per unit mole of gas per unit change in temperature (in Kelvin). It is defined by the ideal gas equation $PV = nRT$. Its value is $8.314 \text{ J mol}^{-1} \text{ K}^{-1}$.

(N/A) Molar volume is defined as the volume occupied by $1$ mole of a substance (usually a gas) at a given temperature and pressure.
Mathematically,$V_m = V/n$,where $V$ is the volume and $n$ is the number of moles.
The $SI$ unit of molar volume is $m^3/mol$.

(B) The number of moles ($\mu$ or $n$) is defined as the ratio of the total number of molecules $(N)$ to the Avogadro constant $(N_{A})$.
Mathematically,it is expressed as:
$\mu = \frac{N}{N_{A}}$
where $N_{A} \approx 6.022 \times 10^{23} \text{ mol}^{-1}$.

(A) The number of moles ($\mu$ or $n$) of a substance is defined as the ratio of the given mass $(M)$ of the substance to its molar mass $(M_{0})$.
Therefore,the formula for the number of moles is given by:
$\mu = \frac{M}{M_{0}}$

(A) The ratio of the universal gas constant $(R)$ to the Avogadro number $(N_A)$ is defined as the Boltzmann constant $(k_B)$.
$k_B = \frac{R}{N_A}$
Given:
$R = 8.314 \text{ J/(mol K)}$
$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$
Calculating the value:
$k_B = \frac{8.314}{6.022 \times 10^{23}} \approx 1.38 \times 10^{-23} \text{ J/K}$

(N/A) Boltzmann's constant $(k_B)$ is defined as the ratio of the universal gas constant $(R)$ to the Avogadro number $(N_A)$.
Mathematically,it is expressed as: $k_B = \frac{R}{N_A}$.
Its value is approximately $1.38 \times 10^{-23} \ J/K$.

(A) Boyle's law states that for a fixed mass of an ideal gas at constant temperature, the pressure is inversely proportional to the volume $(PV = \text{constant})$. Thus, $(a)$ matches with $(iii)$.
Charles's law states that for a fixed mass of an ideal gas at constant pressure, the volume is directly proportional to the absolute temperature $(V/T = \text{constant})$. Thus, $(b)$ matches with $(i)$.
Therefore, the correct matching is $(a-iii, b-i)$.

(A-III, B-I) For an ideal gas, the equation of state is $PV = \mu RT$, which implies $\frac{PV}{\mu T} = R$. Since $R$ (the universal gas constant) is always constant for an ideal gas, the top graph showing $\frac{PV}{\mu T}$ vs $P$ as a horizontal line represents the property of an ideal gas. Thus, $(a-iii)$.
For the bottom graph, we have $\frac{V}{T}$ as a constant. According to Charles's Law, $\frac{V}{T} = \text{constant}$ when the pressure $P$ is constant. Thus, the graph represents a process at constant pressure. Thus, $(b-i)$.

(B) According to Charles's Law,if the pressure of a given mass of gas remains constant,then the volume is directly proportional to its absolute temperature: $V \propto T$.
Therefore,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given:
$V_1 = 100$ $cc$
$T_1 = 27^{\circ} C = 27 + 273 = 300$ $K$
$T_2 = 327^{\circ} C = 327 + 273 = 600$ $K$
Substituting the values into the formula:
$V_2 = V_1 \times \left(\frac{T_2}{T_1}\right)$
$V_2 = 100 \times \left(\frac{600}{300}\right)$
$V_2 = 100 \times 2 = 200$ $cc$.
Thus,the volume at $327^{\circ} C$ will be $200$ $cc$.

(N/A) Boyle's law states that for a fixed mass of an ideal gas at a constant temperature, the pressure is inversely proportional to the volume $(PV = \text{constant})$.
In this scenario, when air is pumped into the cycle tyre, the mass (number of moles) of the air inside the tyre increases.
Since the mass of the gas is not constant, the conditions required for Boyle's law are not satisfied.
Therefore, Boyle's law is not applicable in this case.

(N/A) The volume of a car tyre is approximately fixed. During driving,the friction between the tyre and the road generates heat,which increases the temperature of the air inside the tyre. Since the volume $V$ remains constant,according to Gay-Lussac's Law,the pressure $P$ is directly proportional to the absolute temperature $T$ $(P \propto T)$. Therefore,as the temperature increases,the air pressure inside the tyre also increases.

(C) For an ideal gas,the equation of state is given by $PV = nRT$.
At constant pressure,$V = (nR/P)T$.
The coefficient of volume expansion $\alpha_V$ is defined as $\alpha_V = \frac{1}{V} (\frac{\partial V}{\partial T})_P$.
Substituting the expression for $V$,we get $\alpha_V = \frac{1}{V} (\frac{nR}{P}) = \frac{1}{V} (\frac{V}{T}) = \frac{1}{T}$.
Thus,the value of $\alpha_V$ depends only on the temperature $T$ and is inversely proportional to it.

(A) For an ideal gas,the equation of state is $PV = nRT$.
Case $1$: Constant temperature ($T$ is constant).
Differentiating $PV = nRT$ with respect to pressure,we get $P \Delta V + V \Delta P = 0$,which implies $\Delta V = -\frac{V \Delta P}{P}$.
The magnitude of the change in volume is $|\Delta V| = \frac{V |\Delta P|}{P}$.
Case $2$: Constant pressure ($P$ is constant).
Differentiating $PV = nRT$ with respect to temperature,we get $P \Delta V = nR \Delta T$,which implies $\Delta V = \frac{nR \Delta T}{P}$.
The magnitude of the change in volume is $|\Delta V| = \frac{nR |\Delta T|}{P}$.
Since the magnitude of the change in volume is the same in both cases:
$\frac{V |\Delta P|}{P} = \frac{nR |\Delta T|}{P}$
$V |\Delta P| = nR |\Delta T|$
$|\Delta T| = \frac{V}{nR} |\Delta P|$
From the ideal gas law,$\frac{V}{nR} = \frac{T}{P}$.
Substituting this into the equation,we get $|\Delta T| = \frac{T}{P} |\Delta P|$.
Given $|\Delta T| = C |\Delta P|$,we have $C = \frac{T}{P}$.
Given $T = 300 \ K$ and $P = 2 \ atm$,$C = \frac{300}{2} = 150 \ K/atm$.

(C) The ideal gas equation is given by $PV = nRT$.
Since $n = \frac{m}{M_w}$,we have $PV = \frac{m}{M_w} RT$.
Rearranging for density $\rho = \frac{m}{V}$,we get $\rho = \frac{PM_w}{RT}$.
Given: Pressure $P = 249\; kPa = 249 \times 10^{3}\; Pa$,Temperature $T = 27^{\circ} C = 300\; K$,Gas constant $R = 8.3\; J\; mol^{-1} K^{-1}$,and Molar mass of hydrogen gas $(H_2)$ $M_w = 2 \times 10^{-3}\; kg/mol$.
Substituting the values:
$\rho = \frac{249 \times 10^{3} \times 2 \times 10^{-3}}{8.3 \times 300}$
$\rho = \frac{498}{2490} = 0.2\; kg/m^{3}$.

(D) The ideal gas equation is given by $PV = \mu RT$.
Here,$\mu$ is the number of moles,which is defined as $\mu = \frac{M}{M_{0}}$,where $M$ is the total mass of the gas and $M_{0}$ is the molar mass.
Substituting this into the ideal gas equation,we get $PV = \left(\frac{M}{M_{0}}\right) RT$.
Rearranging the terms,we get $P = \left(\frac{M}{V}\right) \frac{RT}{M_{0}}$.
Since mass density $\rho$ is defined as $\rho = \frac{M}{V}$,we can substitute it into the equation to get $P = \frac{\rho RT}{M_{0}}$.
Thus,$\rho$ represents the mass density and $M_{0}$ represents the molar mass.

(A) For an ideal gas,the combined gas law is given by the equation: $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$.
Given values are:
Initial pressure $P_{1} = 1 \, bar$
Initial volume $V_{1} = 30 \, m^{3}$
Initial temperature $T_{1} = 320 \, K$
Final volume $V_{2} = 10 \, m^{3}$
Final temperature $T_{2} = 280 \, K$
Substituting these values into the equation:
$\frac{1 \times 30}{320} = \frac{P_{2} \times 10}{280}$
Solving for $P_{2}$:
$P_{2} = \frac{1 \times 30 \times 280}{320 \times 10}$
$P_{2} = \frac{3 \times 280}{320} = \frac{840}{320} = 2.625 \, bar$.
Thus,the final pressure of the gas is $2.625 \, bar$.

(C) For an ideal gas,the equation of state is given by the ideal gas law:
$PV = nRT$
where $P$ is pressure,$V$ is volume,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.
Since $n$ and $R$ are constants for a given amount of gas,we have:
$PV \propto T$
This implies that the graph of $PV$ versus $T$ should be a straight line passing through the origin,as $PV$ is directly proportional to $T$.
Therefore,the correct graph is the one showing a linear increase of $PV$ with respect to $T$ passing through the origin.

(C) The number of moles $n$ is given by the mass divided by the molar mass of water $(H_{2}O)$.
Given mass $= 4.5 \, kg = 4.5 \times 10^{3} \, g$.
Molar mass of water $= 18 \, g/mol$.
$n = \frac{4.5 \times 10^{3}}{18} = 0.25 \times 10^{3} = 250 \, mol$.
At $STP$,the volume of $1 \, mole$ of an ideal gas is $22.4 \, L = 22.4 \times 10^{-3} \, m^{3}$.
Therefore,the total volume $V = n \times 22.4 \times 10^{-3} \, m^{3}$.
$V = 250 \times 22.4 \times 10^{-3} \, m^{3} = 5.6 \, m^{3}$.

(C) For a gas in a closed vessel,the volume $V$ remains constant.
According to the ideal gas law,$pV = nRT$.
Since $V$,$n$,and $R$ are constant,we have $p \propto T$,which implies $\frac{\Delta p}{p} = \frac{\Delta T}{T}$.
Given that the pressure increases by $0.4 \%$,we have $\frac{\Delta p}{p} = \frac{0.4}{100}$.
The change in temperature is $\Delta T = 1^{\circ} C = 1 \ K$.
Substituting these values into the equation: $\frac{0.4}{100} = \frac{1}{T}$.
Solving for $T$,we get $T = \frac{100}{0.4} = 250 \ K$.

(A) The ideal gas equation is given by $PV = nRT$.
Rearranging this for the volume $V$ as a function of temperature $T$,we get $V = (\frac{nR}{P})T$.
This represents a straight line passing through the origin with a slope $m = \frac{nR}{P}$.
From the graph,the slope of the line for $P_{2}$ is greater than the slope of the line for $P_{1}$ (since the line for $P_{2}$ is steeper).
Therefore,$\frac{nR}{P_{2}} > \frac{nR}{P_{1}}$.
This implies that $\frac{1}{P_{2}} > \frac{1}{P_{1}}$,which further simplifies to $P_{1} > P_{2}$.

(C) The molar mass of $H_{2}$ is $2 \,g/mol$. Number of moles of $H_{2} = \frac{16 \,g}{2 \,g/mol} = 8 \,moles$.
The molar mass of $O_{2}$ is $32 \,g/mol$. Number of moles of $O_{2} = \frac{128 \,g}{32 \,g/mol} = 4 \,moles$.
Total number of moles $n = 8 + 4 = 12 \,moles$.
At $STP$,the molar volume of an ideal gas is $22.4 \,L = 22.4 \times 10^{3} \,cm^{3}$.
Total volume $V = n \times 22.4 \times 10^{3} \,cm^{3} = 12 \times 22.4 \times 10^{3} \,cm^{3} = 268.8 \times 10^{3} \,cm^{3} = 26.88 \times 10^{4} \,cm^{3}$.
Rounding to the nearest option,the volume is approximately $27 \times 10^{4} \,cm^{3}$.

(D) Let $n_{1}$ and $n_{2}$ be the number of moles of gas present in containers $C_{1}$ and $C_{2}$ respectively,before the valve is opened.
Using the ideal gas equation $pV = nRT$:
$n_{1} = \frac{pV}{R(300)}$
$n_{2} = \frac{5p(4V)}{R(400)} = \frac{20pV}{400R} = \frac{pV}{20R}$
When the valve is opened,gas flows until the pressure in both $C_{1}$ and $C_{2}$ becomes equal to $p_{0}$.
Let $n_{1}'$ and $n_{2}'$ be the new number of moles:
$n_{1}' = \frac{p_{0}V}{R(300)}$
$n_{2}' = \frac{p_{0}(4V)}{R(400)} = \frac{p_{0}V}{100R}$
Since the total number of moles is conserved $(n_{1} + n_{2} = n_{1}' + n_{2}')$:
$\frac{pV}{300R} + \frac{20pV}{400R} = \frac{p_{0}V}{300R} + \frac{p_{0}V}{100R}$
$\frac{p}{300} + \frac{p}{20} = p_{0} \left( \frac{1}{300} + \frac{3}{300} \right)$
$\frac{p + 15p}{300} = p_{0} \left( \frac{4}{300} \right)$
$16p = 4p_{0} \Rightarrow p_{0} = 4p$
Now,the ratio of moles in the hot container $(C_{2})$ to the cold container $(C_{1})$ is:
$\frac{n_{2}'}{n_{1}'} = \frac{p_{0}(4V) / R(400)}{p_{0}V / R(300)} = \frac{4/400}{1/300} = \frac{1/100}{1/300} = 3$
Thus,the number of moles in the hot container is thrice that of the cold one.

(B) The equation of state for an ideal gas is $pV = nRT$,which can be rearranged as $V = (nR/p) \cdot T$.
This represents the equation of a straight line passing through the origin with a slope $m = nR/p$.
From the relationship,we can see that the slope is inversely proportional to the pressure,i.e.,$\text{slope} \propto 1/p$.
By observing the given $V-T$ graph,the slope of the line for $p_3$ is the greatest,followed by $p_2$,and the slope for $p_1$ is the smallest.
Since $\text{slope} \propto 1/p$,a larger slope corresponds to a smaller pressure.
Therefore,the relationship between the pressures is $p_1 > p_2 > p_3$.

(B) Let $n$ be the total number of moles of gas in both bulbs initially. Since the bulbs have identical volumes $V$ and are at the same temperature $T$,the number of moles in each bulb is $n_1 = n_2 = n / 2$.
When bulb $2$ is heated to $2 T$ and bulb $1$ is at $T$,the pressure $p$ in both bulbs must be equal for the system to be in equilibrium.
Using the ideal gas law $p V = n R T$,the number of moles in bulb $1$ $(n_1')$ and bulb $2$ $(n_2')$ are:
$n_1' = \frac{p V}{R T}$
$n_2' = \frac{p V}{R (2 T)} = \frac{p V}{2 R T}$
Since the total number of moles is conserved,$n_1' + n_2' = n = n_1 + n_2 = n / 2 + n / 2 = n$.
$\frac{p V}{R T} + \frac{p V}{2 R T} = n \Rightarrow \frac{3 p V}{2 R T} = n \Rightarrow \frac{p V}{R T} = \frac{2 n}{3}$.
Thus,the final number of moles in bulb $2$ is $n_2' = \frac{1}{2} \times \frac{p V}{R T} = \frac{1}{2} \times \frac{2 n}{3} = \frac{n}{3}$.
The ratio of the final mass (or final moles) in bulb $2$ to the initial mass (or initial moles) in bulb $2$ is:
Ratio $= \frac{n_2'}{n_2} = \frac{n / 3}{n / 2} = \frac{2}{3}$.

(C) Initially,the pressure inside the cylinder is the atmospheric pressure $p_0$. When a mass $m$ is attached to the piston and it moves down by a distance $\Delta l$,let the new pressure be $p$.
Since the process is isothermal (assuming temperature remains constant),we have $p_0 V_0 = p V$.
$p_0 (A l) = p A (l + \Delta l)$
$p = p_0 \frac{l}{l + \Delta l}$
In equilibrium,the downward force due to the mass $m$ is balanced by the upward force due to the pressure difference between the atmosphere and the gas inside.
$(p_0 - p) A = m g$
Substituting $p$:
$(p_0 - p_0 \frac{l}{l + \Delta l}) A = m g$
$p_0 A (1 - \frac{l}{l + \Delta l}) = m g$
$p_0 A (\frac{\Delta l}{l + \Delta l}) = m g$
Given $p_0 = 10^5 \,Pa$,$m = 50 \,kg$,$g = 10 \,m/s^2$,$r = 0.2 \,m$,$A = \pi r^2 = \pi (0.2)^2 = 0.04 \pi \,m^2$.
$10^5 \times 0.04 \pi \times \frac{\Delta l}{l + \Delta l} = 50 \times 10 = 500$
$4000 \pi \times \frac{\Delta l}{l + \Delta l} = 500$
$\frac{\Delta l}{l + \Delta l} = \frac{500}{4000 \pi} = \frac{1}{8 \pi} \approx \frac{1}{8 \times 3.14} \approx \frac{1}{25.12} \approx 0.0398$
Since $\frac{\Delta l}{l + \Delta l} \approx 0.04$,we have $\frac{\Delta l}{l} \approx 0.04$ (assuming $\Delta l \ll l$).
Thus,the correct option is $C$.