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

(C) Let the initial pressure of the air inside be $p_0$ (atmospheric pressure) and initial volume be $V_i$. When a mass $m$ is attached to the piston,the piston moves down and the pressure of the gas becomes $p_f = p_0 - \frac{mg}{A}$,where $A$ is the cross-sectional area of the pipe.
Since the temperature remains constant during this expansion,we use Boyle's Law: $p_i V_i = p_f V_f$.
$p_0 V_i = (p_0 - \frac{mg}{A}) V_f$
$\frac{V_f}{V_i} = \frac{p_0}{p_0 - \frac{mg}{A}} = \frac{1}{1 - \frac{mg}{p_0 A}}$
Using the binomial approximation $(1-x)^{-1} \approx 1+x$ for small $x = \frac{mg}{p_0 A}$,we get $\frac{V_f}{V_i} \approx 1 + \frac{mg}{p_0 A}$.
Thus,$\frac{\Delta V}{V_i} = \frac{V_f - V_i}{V_i} = \frac{mg}{p_0 A}$.
When the gas is cooled from $T$ to $T-\Delta T$,it returns to its original volume $V_i$. For a constant pressure process (isobaric),$\frac{V}{T} = \text{constant}$,so $\frac{\Delta V}{V_f} = \frac{\Delta T}{T}$.
Therefore,$\frac{\Delta T}{T} = \frac{\Delta V}{V_f} \approx \frac{\Delta V}{V_i} = \frac{mg}{p_0 A}$.
Given $m = 50 \,kg$,$g = 10 \,m/s^2$,$p_0 = 10^5 \,Pa$,and $r = 0.2 \,m$ $(A = \pi r^2 = 3.14 \times 0.04 = 0.1256 \,m^2)$:
$\frac{\Delta T}{T} = \frac{50 \times 10}{10^5 \times 0.1256} = \frac{500}{12560} \approx 0.0398 \approx 0.04$.

(C) The volume of the room is $V = 5 \,m \times 5 \,m \times 4 \,m = 100 \,m^3$.
At standard temperature and pressure $(STP)$,the pressure $p = 1.013 \times 10^5 \,Pa \approx 10^5 \,Pa$ and temperature $T = 273 \,K$.
The Boltzmann constant $k_B = 1.38 \times 10^{-23} \,J/K$.
Using the ideal gas equation $pV = Nk_BT$,the number of molecules $N$ is given by $N = \frac{pV}{k_BT}$.
Substituting the values: $N = \frac{10^5 \times 100}{1.38 \times 10^{-23} \times 273} \approx \frac{10^7}{3.7674 \times 10^{-21}} \approx 2.65 \times 10^{27}$.
Rounding to the nearest order of magnitude,$N \approx 3 \times 10^{27}$ molecules.

(D) Let the initial pressure inside the tube be $P_0 = 10^5 \,N/m^2$ and the initial volume be $V_1 = A \times 20 \,cm$,where $A$ is the cross-sectional area of the tube.
When the tube is dipped such that $10 \,cm$ is outside the water,the air column inside the tube is compressed. Let the new length of the air column be $L = (20 - h) \,cm$. The new volume is $V_2 = A \times (20 - h) \,cm$.
Since the temperature is constant,we apply Boyle's Law: $P_1 V_1 = P_2 V_2$.
$10^5 \times 20 = P \times (20 - h) \quad \dots(1)$
The pressure $P$ inside the tube at the water level is given by the atmospheric pressure plus the pressure due to the water column outside the tube relative to the air-water interface inside the tube. The depth of the interface below the outer water surface is $(10 - h) \,cm = \frac{10 - h}{100} \,m$.
$P = P_0 + \rho g \Delta y = 10^5 + 10^3 \times 10 \times \frac{10 - h}{100} = 10^5 + 100(10 - h) = 10^5 + 1000 - 100h \quad \dots(2)$
Substituting $(2)$ into $(1)$:
$20 \times 10^5 = (10^5 + 1000 - 100h)(20 - h)$
$20 \times 10^5 = 20 \times 10^5 - 10^5 h + 20000 - 1000h - 2000h + 100h^2$
$0 = 100h^2 - 103000h + 20000$
Dividing by $100$:
$h^2 - 1030h + 200 = 0$
Since $h$ is very small,$h^2 \approx 0$,so $1030h \approx 200 \implies h \approx \frac{200}{1030} \approx 0.194 \,cm$.
This is closest to $0.2 \,cm$.

(B) From the ideal gas equation,$PV = nRT$,we can write:
$\frac{1}{V} = \frac{P}{nRT}$
For a fixed mass of gas,$n$ is constant. Thus,$\frac{1}{V} = \left(\frac{1}{nRT}\right)P$.
This represents a straight line passing through the origin with a slope $m = \frac{1}{nRT}$.
Since the slope $m$ is inversely proportional to temperature $T$ $(m \propto \frac{1}{T})$,the line with the largest slope corresponds to the lowest temperature.
Looking at the graph,the slope of line $A$ is the greatest,followed by $B$,and then $C$ (i.e.,$m_A > m_B > m_C$).
Therefore,the temperatures must follow the order $T_A < T_B < T_C$.

(B) The process occurs at a constant volume because the cylinder is rigid.
According to Gay-Lussac's Law,for a fixed amount of gas at constant volume,the pressure is directly proportional to the absolute temperature: $P \propto T$ or $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given:
Initial pressure $P_1 = 20 \,atm$
Initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \,K$
Maximum pressure the cylinder can withstand $P_2 = 100 \,atm$
We need to find the temperature $T_2$ at which the pressure reaches $100 \,atm$.
Substituting the values into the formula:
$\frac{20}{300} = \frac{100}{T_2}$
$T_2 = \frac{100 \times 300}{20}$
$T_2 = 5 \times 300 = 1500 \,K$
Thus,the danger of explosion sets in at $1500 \,K$.

(A) Since the temperature remains constant, we use Boyle's Law, $PV = \text{constant}$.
Initially, the gas occupies volume $V$ at pressure $P$. When the pump performs one stroke, the gas expands to fill the total volume $(V + v)$.
Let $P_1$ be the pressure after the first stroke:
$P \cdot V = P_1 \cdot (V + v)$
$P_1 = P \left( \frac{V}{V + v} \right)$
After the second stroke, the gas at pressure $P_1$ in volume $V$ expands to volume $(V + v)$:
$P_1 \cdot V = P_2 \cdot (V + v)$
$P_2 = P_1 \left( \frac{V}{V + v} \right) = P \left( \frac{V}{V + v} \right)^2$
Following this pattern, after $n$ strokes, the pressure $P_n$ will be:
$P_n = P \left( \frac{V}{V + v} \right)^n$

(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$ is the number of moles.
For the first container: $m(O_2) = 32 \,g$,$M(O_2) = 32 \,g/mol$. Thus,$n_1 = \frac{32}{32} = 1 \,mol$.
The pressure is $P = \frac{n_1 RT}{V} = \frac{1 \cdot RT}{V} = \frac{RT}{V}$.
For the second container: $m(H_2) = 4 \,g$,$M(H_2) = 2 \,g/mol$. Thus,$n_2 = \frac{4}{2} = 2 \,mol$.
The temperature is $2T$. The pressure $P'$ is given by $P' = \frac{n_2 R(2T)}{V} = \frac{2 \cdot R(2T)}{V} = \frac{4RT}{V}$.
Substituting $\frac{RT}{V} = P$,we get $P' = 4P$.

(B) Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$ (mass/molar mass).
Since the volume $V$ and molar mass $M$ are constant,we have $P = \frac{m}{M} \frac{RT}{V}$,which implies $P \propto m \cdot T$ or $m \propto \frac{P}{T}$.
Initial state: $m_i = 28 \,g$,$P_i = 10 \,atm$,$T_i = 57 + 273 = 330 \,K$.
Final state: $m_f$,$P_f = 5 \,atm$,$T_f = 27 + 273 = 300 \,K$.
Taking the ratio: $\frac{m_f}{m_i} = \frac{P_f}{P_i} \times \frac{T_i}{T_f}$.
$\frac{m_f}{28} = \frac{5}{10} \times \frac{330}{300} = \frac{1}{2} \times \frac{11}{10} = \frac{11}{20}$.
$m_f = 28 \times \frac{11}{20} = 15.4 \,g$.
The amount of gas leaked out is $\Delta m = m_i - m_f = 28 - 15.4 = 12.6 \,g$.
$12.6 \,g = \frac{126}{10} = \frac{63}{5} \,g$.

(B) Let $P_0$ be the atmospheric pressure.
Initially,the tube is half-immersed,so $40 \,cm$ of the tube is inside the mercury. The air trapped inside the tube at the top has a length of $40 \,cm$ at pressure $P_0$.
When the tube is taken out,$20 \,cm$ of mercury remains. The total length of the tube is $80 \,cm$. Since $20 \,cm$ is occupied by mercury,the remaining length for the air column is $80 - 20 = 60 \,cm$.
Using Boyle's Law $(P_1 V_1 = P_2 V_2)$:
$P_0 \times 40 = P_1 \times 60 \Rightarrow P_1 = \frac{2}{3} P_0$.
The pressure inside the tube $P_1$ is related to atmospheric pressure by the mercury column height: $P_0 = P_1 + h$,where $h = 20 \,cm$.
Substituting $P_1$: $P_0 = \frac{2}{3} P_0 + 20$.
$\frac{1}{3} P_0 = 20 \Rightarrow P_0 = 60 \,cm$ of $Hg$.

(C) Assuming the volume of the tyre remains constant,we use Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given:
$P_1 = 270\,kPa$
$T_1 = 27^{\circ}C = 27 + 273 = 300\,K$
$T_2 = 36^{\circ}C = 36 + 273 = 309\,K$
Substituting the values:
$P_2 = \frac{P_1 \times T_2}{T_1} = \frac{270 \times 309}{300}$.
$P_2 = 0.9 \times 309 = 278.1\,kPa$.
Rounding to the nearest integer,the pressure is $278\,kPa$.

(D) Given the relation $PT^2 = C$ (where $C$ is a constant).
Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.
Substituting $P$ into the given equation: $\left(\frac{nRT}{V}\right)T^2 = C$.
This simplifies to $\frac{nRT^3}{V} = C$, which implies $V = \left(\frac{nR}{C}\right)T^3$.
Let $K = \frac{nR}{C}$, so $V = KT^3$.
The volume expansion coefficient $\gamma$ is defined as $\gamma = \frac{1}{V} \frac{dV}{dT}$.
Differentiating $V$ with respect to $T$: $\frac{dV}{dT} = 3KT^2$.
Now, substitute into the formula for $\gamma$: $\gamma = \frac{1}{KT^3} \times (3KT^2) = \frac{3}{T}$.

(A) For an isochoric process (constant volume),the ideal gas equation is given by $PV = nRT$,which can be rearranged as $P = (\frac{nR}{V})T$.
Here,$T$ is the absolute temperature in Kelvin,which is related to the temperature $t$ in Celsius by $T = t + 273.15$.
Substituting this into the equation,we get $P = (\frac{nR}{V})(t + 273.15)$.
This represents a straight line equation of the form $y = mx + c$,where the intercept on the temperature axis occurs when pressure $P = 0$.
Setting $P = 0$,we get $0 = (\frac{nR}{V})(t + 273.15)$,which implies $t + 273.15 = 0$,or $t = -273.15^{\circ}\,C$.
Rounding to the nearest integer,the temperature at point '$K$' is $-273^{\circ}\,C$,which represents absolute zero.

(A) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$ is the number of moles.
Since the vessels have the same size $(V_A = V_B)$ and are at the same temperature $(T_A = T_B)$,the pressure $P$ is directly proportional to the number of moles $n$ $(P \propto n)$.
Therefore,$\frac{P_A}{P_B} = \frac{n_A}{n_B}$.
The molar mass of hydrogen $(H_2)$ is $M_A = 2 \ g/mol$,and the molar mass of oxygen $(O_2)$ is $M_B = 32 \ g/mol$.
The number of moles in vessel $A$ is $n_A = \frac{1 \ g}{2 \ g/mol} = 0.5 \ mol$.
The number of moles in vessel $B$ is $n_B = \frac{1 \ g}{32 \ g/mol} = \frac{1}{32} \ mol$.
Thus,$\frac{P_A}{P_B} = \frac{0.5}{1/32} = 0.5 \times 32 = 16$.

(A) The ideal gas equation in terms of number density $n = N/V$ is given by $P = nkT$.
Given:
Number density $n = 2.0 \times 10^{25} \text{ m}^{-3}$
Pressure $P = 1.38 \text{ atm} = 1.38 \times 1.01325 \times 10^5 \text{ Pa} \approx 1.4 \times 10^5 \text{ Pa}$
Boltzmann constant $k = 1.38 \times 10^{-23} \text{ J K}^{-1}$
Using the formula $P = nkT$:
$T = \frac{P}{nk}$
$T = \frac{1.38 \times 1.01325 \times 10^5}{2.0 \times 10^{25} \times 1.38 \times 10^{-23}}$
$T = \frac{1.01325 \times 10^5}{2.0 \times 10^2}$
$T = \frac{101325}{200} \approx 506.6 \text{ K}$
Rounding to the nearest given option,the temperature is approximately $500 \text{ K}$.

(D) For an ideal gas,the equation of state is $PV = nRT$.
Rearranging for $V$ as a function of $T$,we get $V = \left(\frac{nR}{P}\right)T$.
Comparing this with the equation of a straight line $y = mx$,where $y = V$ and $x = T$,the slope of the $V-T$ graph is given by $\text{Slope} = \frac{nR}{P}$.
Since $n$ and $R$ are constants,the slope is inversely proportional to the pressure,i.e.,$\text{Slope} \propto \frac{1}{P}$.
From the given figure,the slope of the line corresponding to $P_2$ is greater than the slope of the line corresponding to $P_1$,i.e.,$(\text{Slope})_2 > (\text{Slope})_1$.
Since $\text{Slope} \propto \frac{1}{P}$,a higher slope implies a lower pressure. Therefore,$P_2 < P_1$ or $P_1 > P_2$.

(B) For an ideal gas,the equation of state is given by:
$PV = nRT$
Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have:
$PV = \frac{m}{M} RT$
Rearranging for pressure $P$:
$P = \left( \frac{m}{V} \right) \frac{RT}{M}$
Since density $\rho = \frac{m}{V}$,we get:
$P = \left( \frac{\rho R}{M} \right) T$
This represents a straight line passing through the origin in a $P-T$ graph,where the slope is $\frac{\rho R}{M}$.
For a constant temperature $T$,from the graph,we can see that $P_1 > P_2 > P_3$.
Since $P \propto \rho$ for a constant $T$ and $M$,it follows that $\rho_1 > \rho_2 > \rho_3$.
Therefore,the correct option is $\rho_1 > \rho_2$.

(C) From the ideal gas equation,$PV = nRT$,we have $T = (P/nR)V$.
For a fixed amount of gas ($n$ is constant),the slope of the $T-V$ graph is given by $m = P/nR$.
Since $n$ and $R$ are constants,the slope $m$ is directly proportional to the pressure $P$.
Therefore,a steeper slope in the $T-V$ graph corresponds to a higher pressure.
Looking at the graph,the slope of the curve for $P_1$ is the steepest,followed by $P_2$,and then $P_3$.
Thus,the correct relation is $P_1 > P_2 > P_3$.

(C) The coefficient of volume expansion is defined as $\gamma = \frac{1}{V} \left( \frac{dV}{dT} \right)$.
Given the process equation: $PT^2 = \text{constant}$.
Using the ideal gas law $PV = nRT$, we can write $P = \frac{nRT}{V}$.
Substituting $P$ into the process equation: $\left( \frac{nRT}{V} \right) T^2 = \text{constant}$.
This simplifies to $\frac{T^3}{V} = \text{constant}$, or $V = k T^3$ (where $k$ is a constant).
Differentiating $V$ with respect to $T$: $\frac{dV}{dT} = 3k T^2$.
Now, substitute these into the expression for $\gamma$:
$\gamma = \frac{1}{V} \left( \frac{dV}{dT} \right) = \frac{1}{kT^3} (3kT^2) = \frac{3}{T}$.

(B) For a gas in a container of fixed volume,the volume $V$ is constant. According to Gay-Lussac's Law,for a fixed amount of gas at constant volume,the pressure $P$ is directly proportional to the absolute temperature $T$ ($P \propto T$ or $\frac{P_1}{T_1} = \frac{P_2}{T_2}$).
Given initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$.
Let the initial pressure be $P_1 = P$.
We want the final pressure to be $P_2 = 2P$.
Using the relation $\frac{P_1}{T_1} = \frac{P_2}{T_2}$:
$\frac{P}{300} = \frac{2P}{T_2}$
$T_2 = 2 \times 300 = 600 \ K$.
To convert the temperature back to Celsius: $T_2(^{\circ} C) = 600 - 273 = 327^{\circ} C$.

(C) For an ideal gas in a closed vessel,the volume $V$ remains constant,which represents an isochoric process.
According to Gay-Lussac's Law,for a constant volume,the pressure $P$ is directly proportional to the absolute temperature $T$ $(P \propto T)$.
This implies $\frac{P_2}{P_1} = \frac{T_2}{T_1}$,or in terms of small changes,$\frac{\Delta P}{P} = \frac{\Delta T}{T}$.
Given that the pressure increases by $0.4 \%$,we have $\frac{\Delta P}{P} = \frac{0.4}{100}$.
The change in temperature is given as $\Delta T = 1 \ K$ (since a change of $1^{\circ} C$ is equivalent to a change of $1 \ K$).
Substituting these values into the equation: $\frac{0.4}{100} = \frac{1}{T}$.
Solving for $T$,we get $T = \frac{100}{0.4} = 250 \ K$.

(B) The total number of moles of the gas remains constant because the system is closed.
Let the volume of the smaller vessel be $V$ and the volume of the larger vessel be $2V$.
Initial moles in the larger vessel: $n_1 = \frac{P_1 V_1}{R T_1} = \frac{8 \times 2V}{R \times 1000} = \frac{16V}{1000R}$.
Initial moles in the smaller vessel: $n_2 = \frac{P_2 V_2}{R T_2} = \frac{7 \times V}{R \times 500} = \frac{14V}{1000R}$.
Total initial moles: $n_{total} = n_1 + n_2 = \frac{16V + 14V}{1000R} = \frac{30V}{1000R}$.
At steady state,the total volume is $V_f = V + 2V = 3V$ and the temperature is $T_f = 600 \ K$.
Using the ideal gas law for the final state: $n_{total} = \frac{P_f V_f}{R T_f}$.
$\frac{30V}{1000R} = \frac{P_f \times 3V}{R \times 600}$.
Canceling $V$ and $R$ from both sides:
$\frac{30}{1000} = \frac{3 P_f}{600}$.
$\frac{30}{1000} = \frac{P_f}{200}$.
$P_f = \frac{30 \times 200}{1000} = 6 \ kPa$.

(B) From the ideal gas equation,$PV = nRT$,we can write $V = (\frac{nR}{P})T$.
Comparing this with the equation of a straight line $y = mx$,where $y = V$ and $x = T$,the slope $m = \frac{nR}{P}$.
Since $n$ and $R$ are constants,the slope is inversely proportional to pressure,i.e.,$\text{slope} \propto \frac{1}{P}$.
In the given $V-T$ graph,the line representing the process from state-$1$ to state-$2$ is steeper than the line passing through the origin to state-$1$.
This means the slope increases as we move from state-$1$ to state-$2$.
Since $\text{slope} \propto \frac{1}{P}$,an increase in slope implies a decrease in pressure.
Therefore,the pressure decreased.

(C) The volume $V$ of the vessel remains constant.
Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is the mass of gas,$M$ is the molar mass).
Since $V$ and $M$ are constant,we have $\frac{P}{mT} = \text{constant}$.
Therefore,$\frac{P_1}{m_1 T_1} = \frac{P_2}{m_2 T_2}$.
Given: $P_1 = P$,$m_1 = 8 \ g$,$T_1 = 400 \ K$,$P_2 = P/4$,$T_2 = 200 \ K$.
Substituting the values: $\frac{P}{8 \times 400} = \frac{P/4}{m_2 \times 200}$.
$\frac{1}{3200} = \frac{1}{800 \times m_2}$.
$m_2 = \frac{3200}{800} = 4 \ g$.
The mass of oxygen that leaked out is $\Delta m = m_1 - m_2 = 8 \ g - 4 \ g = 4 \ g$.

(D) From the ideal gas equation,$PV = \mu RT$,where $\mu$ is the number of moles.
Rearranging this,we get $PV = (\mu R)T$.
Comparing this with the equation of a straight line $y = mx$,where $y = PV$ and $x = T$,the slope of the graph is $m = \mu R$.
Since $R$ is a universal gas constant,the slope is directly proportional to the number of moles $\mu$.
For a given mass $m$,the number of moles is $\mu = \frac{m}{M}$,where $M$ is the molar mass.
Thus,$\mu \propto \frac{1}{M}$.
The molar masses are $M(H_2) = 2 \text{ g/mol}$,$M(He) = 4 \text{ g/mol}$,and $M(O_2) = 32 \text{ g/mol}$.
Since $M(H_2) < M(He) < M(O_2)$,the number of moles follows the order $\mu(H_2) > \mu(He) > \mu(O_2)$.
Consequently,the slopes follow the order $\text{Slope}(H_2) > \text{Slope}(He) > \text{Slope}(O_2)$.
Looking at the graph,the slope of line $C$ is the greatest,followed by $B$,and then $A$.
Therefore,$C$ corresponds to $H_2$,$B$ corresponds to $He$,and $A$ corresponds to $O_2$.
This matches option $D$.

(C) For a gas in a closed vessel,the volume $V$ remains constant. According to Gay-Lussac's Law,$P \propto T$,which means $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Let the initial pressure be $P$ and the initial temperature be $T$. Then $P_1 = P$ and $T_1 = T$.
The pressure is increased by $2.5 \%$,so $P_2 = P + 0.025P = 1.025P$.
The temperature increases by $4 \ K$,so $T_2 = T + 4$.
Substituting these into the relation $\frac{P}{T} = \frac{1.025P}{T + 4}$:
$T + 4 = 1.025T$
$4 = 1.025T - T$
$4 = 0.025T$
$T = \frac{4}{0.025} = \frac{4000}{25} = 160 \ K$.
Thus,the initial temperature of the gas is $160 \ K$.

(B) From the ideal gas equation,$PV = nRT$,where $n$ is the number of moles. The number of molecules $N$ is given by $N = nN_A$,where $N_A$ is Avogadro's number.
Thus,$N = \frac{PV}{RT} N_A$.
For jar $A$: $N_A = \frac{PV}{RT} N_A$.
For jar $B$: $N_B = \frac{(2P)(V/4)}{(T/4)} N_A = \frac{2PV/4}{T/4} N_A = \frac{2PV}{T} N_A = 2 \left( \frac{PV}{RT} \right) N_A = 2N_A$.
Therefore,the ratio of the number of molecules in jar $A$ to jar $B$ is $N_A / N_B = 1 / 2$,which is $1: 2$.

(C) According to Boyle's Law,for a given mass of gas at a constant temperature,$P_1 V_1 = P_2 V_2$.
Let the initial pressure be $P_1 = P$ and the initial volume be $V_1 = V$.
The volume is reduced by $5 \%$,so the final volume $V_2 = V - 0.05V = 0.95V$.
Substituting these into the equation: $P \times V = P_2 \times 0.95V$.
Solving for $P_2$: $P_2 = P / 0.95 = (100/95)P = (20/19)P \approx 1.0526P$.
The increase in pressure is $\Delta P = P_2 - P = 1.0526P - P = 0.0526P$.
The percentage increase in pressure is $(\Delta P / P) \times 100 = 0.0526 \times 100 = 5.26 \%$.

(C) According to Boyle's Law,for a given mass of gas at constant temperature,$P_1 V_1 = P_2 V_2$.
Let the initial volume be $V_1 = V$.
The volume is increased by $7 \%$,so the final volume $V_2 = V + 0.07V = 1.07V$.
Substituting these into the equation: $P_1 V = P_2 (1.07V)$.
$P_2 = \frac{P_1}{1.07} \approx 0.9346 P_1$.
The decrease in pressure is $\Delta P = P_1 - P_2 = P_1 - 0.9346 P_1 = 0.0654 P_1$.
To express this as a percentage: $\frac{\Delta P}{P_1} \times 100 = 0.0654 \times 100 = 6.54 \%$.

(D) From the ideal gas equation,$PV = nRT = \frac{m}{M} RT$,where $m$ is the mass and $M$ is the molar mass.
Since density $\varrho = \frac{m}{V}$,we can write $P = \frac{\varrho RT}{M}$,which implies $P \propto \varrho T$.
At point $A$: $P_A = P_0$,$T_A = T_0$,and $\varrho_A = \varrho_0$. Thus,$P_0 = k \cdot \varrho_0 T_0$ (where $k = \frac{R}{M}$).
At point $B$: $P_B = Y$,$T_B = 3T_0$,and $\varrho_B = \frac{4}{3} \varrho_0$.
Taking the ratio of the two equations:
$\frac{P_B}{P_A} = \frac{\varrho_B T_B}{\varrho_A T_A}$
$\frac{Y}{P_0} = \frac{(\frac{4}{3} \varrho_0) (3T_0)}{\varrho_0 T_0}$
$\frac{Y}{P_0} = \frac{4}{3} \times 3 = 4$
Therefore,$Y = 4 P_0$.

(A) The units of $PV$ can be calculated as follows:
Pressure $(P)$ is defined as force per unit area,so its $SI$ unit is $N/m^2$ or $kg \cdot m^{-1} \cdot s^{-2}$.
Volume $(V)$ has the $SI$ unit of $m^3$.
Therefore,the unit of the product $PV$ is:
Unit $= (kg \cdot m^{-1} \cdot s^{-2}) \times (m^3) = kg \cdot m^2 \cdot s^{-2}$.
Since $1 \text{ Joule} = 1 \text{ kg} \cdot m^2 \cdot s^{-2}$,this unit is equivalent to the unit of energy (or work).

(B) According to Charles's Law,the volume $(V)$ of a fixed mass of an ideal gas is directly proportional to its absolute temperature $(T)$ when the pressure is held constant.
Mathematically,this is expressed as $V \propto T$ or $V = kT$,where $k$ is a constant.
This relationship represents a straight line passing through the origin on a $V$ versus $T$ graph.
Among the given options,graph $Q$ shows a straight line passing through the origin,which correctly represents the variation of volume with temperature at constant pressure.

(D) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.
For oxygen gas $(O_2)$: $P_1 V_1 = \frac{m_1}{M_1} RT_1$ ... $(i)$
For hydrogen gas $(H_2)$: $P_2 V_2 = \frac{m_2}{M_2} RT_2$ ... (ii)
Given that $P_1 = P_2$,$V_1 = V_2$,and $m_1 = m_2$,dividing equation $(i)$ by (ii) gives:
$\frac{P_1 V_1}{P_2 V_2} = \frac{m_1 / M_1}{m_2 / M_2} \cdot \frac{T_1}{T_2}$
$1 = \frac{M_2}{M_1} \cdot \frac{T_1}{T_2}$
$\frac{T_1}{T_2} = \frac{M_1}{M_2}$
Substituting the molar masses $M_1 (O_2) = 32 \text{ g/mol}$ and $M_2 (H_2) = 2 \text{ g/mol}$:
$\frac{T_1}{T_2} = \frac{32}{2} = \frac{16}{1}$
Therefore,the ratio of their absolute temperatures is $16: 1$.

(D) From the ideal gas equation,we have $PV = nRT$.
Here,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature.
We know that the number of moles $n = \frac{N}{N_A}$,where $N$ is the total number of molecules and $N_A$ is Avogadro's number.
Substituting this into the ideal gas equation: $PV = \left(\frac{N}{N_A}\right) RT$.
Rearranging the terms,we get $PV = N \left(\frac{R}{N_A}\right) T$.
Since the Boltzmann constant $k = \frac{R}{N_A}$,the equation becomes $PV = NkT$.
Therefore,$\frac{PV}{kT} = N$,which represents the total number of molecules of the gas.

(B) The ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles. Since the number of molecules $N$ is proportional to the number of moles $(N = nN_A)$,the ratio of the number of molecules is equal to the ratio of the number of moles.
For the first sample: $P_1 = P, V_1 = V, T_1 = T$. Thus,$n_1 = \frac{PV}{RT}$.
For the second sample: $P_2 = 2P, V_2 = V/4, T_2 = 2T$. Thus,$n_2 = \frac{(2P)(V/4)}{R(2T)} = \frac{PV/2}{2RT} = \frac{PV}{4RT}$.
The ratio of the number of molecules is $\frac{N_1}{N_2} = \frac{n_1}{n_2} = \frac{PV/RT}{PV/4RT} = \frac{1}{1/4} = 4:1$.

(A) The ideal gas equation is given by $PV = \mu RT$,where $\mu$ is the number of moles of the gas.
$\mu = \frac{\text{Given mass}}{\text{Molecular mass}} = \frac{m}{M}$.
Given,$m = 2 \ g$ and for oxygen $(O_2)$,$M = 32 \ g/mol$.
Substituting these values into the ideal gas equation:
$PV = \left( \frac{2}{32} \right) RT$.
$PV = \frac{1}{16} RT$.

(B) The pressure $P$ exerted by an ideal gas on the wall of a container is defined as the force $F$ per unit area $A$,given by $P = F/A$.
Since the area $A$ of the container wall is constant,we have $P \propto F$.
From the ideal gas equation,$PV = nRT$,where $n$ is the number of moles,$R$ is the universal gas constant,$T$ is the temperature,and $V$ is the volume.
For a closed container,the volume $V$ is constant. Thus,$P = (nR/V)T$,which implies $P \propto T$.
Comparing the two relations,$F \propto P$ and $P \propto T$,we get $F \propto T^1$.
Therefore,comparing this with $F \propto T^x$,we find that $x = 1$.

(D) According to the ideal gas equation,$PV = Nk_B T$,where $N$ is the number of molecules and $k_B$ is the Boltzmann constant.
For jar $A$: $PV = N_A k_B T$ --- $(i)$
For jar $B$: $(2P) \times (V/4) = N_B k_B (2T)$
Simplifying the left side: $(1/2) PV = N_B k_B (2T)$
$PV = 4 N_B k_B T$ --- $(ii)$
Equating $(i)$ and $(ii)$: $N_A k_B T = 4 N_B k_B T$
$N_A = 4 N_B$
Therefore,the ratio $N_A / N_B = 4/1$ or $4: 1$.

(A) From the ideal gas equation,$PV = nRT$,where $n = m/M$ (mass/molar mass).
Substituting $n$,we get $PV = (m/M)RT$.
Since density $\rho = m/V$,we can write $P = (\rho/M)RT$,which implies $\rho = (PM)/(RT)$.
Therefore,$\rho \propto P/T$.
Given: Initial state is $(\rho_0, P_0, T_0)$ and final state is $(\rho', 3P_0, 2T_0)$.
Using the proportionality $\rho' / \rho_0 = (P'/P_0) \times (T_0/T')$,we get:
$\rho' = \rho_0 \times (3P_0 / P_0) \times (T_0 / 2T_0)$.
$\rho' = \rho_0 \times 3 \times (1/2) = \frac{3}{2} \rho_0$.

(B) Using the ideal gas equation $PV = nRT$,we can find the initial and final number of moles of the gas.
Initial number of moles,$n_1 = \frac{PV}{RT}$.
Final number of moles,$n_2 = \frac{P^{\prime}V}{RT}$.
The number of moles that have leaked is the difference between the initial and final moles: $\Delta n = n_1 - n_2$.
Substituting the values,$\Delta n = \frac{PV}{RT} - \frac{P^{\prime}V}{RT} = \frac{V}{RT}(P - P^{\prime})$.
Thus,the correct option is $B$.

(D) Given: Initial pressure $P_1 = 270 \text{ kPa}$,Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \text{ K}$.
Final temperature $T_2 = 37^{\circ}C = 37 + 273 = 310 \text{ K}$.
Assuming the volume of the tyre remains constant,according to Gay-Lussac's Law,$\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Substituting the values: $\frac{270}{300} = \frac{P_2}{310}$.
$P_2 = \frac{270 \times 310}{300} = 0.9 \times 310 = 279 \text{ kPa}$.
Therefore,the final pressure is $279 \text{ kPa}$.

(D) The ideal gas equation is given by $PV = nRT$.
Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have $PV = \frac{m}{M} RT$.
Rearranging this,we get $P = \frac{m}{V} \frac{RT}{M} = \varrho \frac{RT}{M}$,where $\varrho$ is the density.
Thus,$\varrho = \frac{PM}{RT}$.
For the initial state: $\varrho_0 = \frac{P_0 M}{R T_0}$.
For the final state: $\varrho' = \frac{P' M}{R T'} = \frac{(3 P_0) M}{R (2 T_0)}$.
Substituting $\varrho_0$ into the expression for $\varrho'$,we get $\varrho' = \frac{3}{2} \left( \frac{P_0 M}{R T_0} \right) = \frac{3}{2} \varrho_0$.

(C) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.
For oxygen $(O_2)$: $PV = \frac{m}{M_{O_2}} RT_{O_2}$ --- $(i)$
For hydrogen $(H_2)$: $PV = \frac{m}{M_{H_2}} RT_{H_2}$ --- $(ii)$
Since $P$,$V$,and $m$ are the same for both gases,we equate the two expressions:
$\frac{m}{M_{O_2}} RT_{O_2} = \frac{m}{M_{H_2}} RT_{H_2}$
$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$
$\frac{T_{O_2}}{T_{H_2}} = \frac{M_{O_2}}{M_{H_2}}$
Given $M_{O_2} = 32$ and $M_{H_2} = 2$:
$\frac{T_{O_2}}{T_{H_2}} = \frac{32}{2} = \frac{16}{1}$
Therefore,the ratio of their absolute temperatures is $16: 1$.

(B) Given: Initial temperature $T_1 = 100^{\circ} C = 373 \ K$. The volume is constant in a closed vessel. According to Gay-Lussac's Law,for a fixed mass of gas at constant volume,$P \propto T$.
Therefore,$\frac{P_2}{P_1} = \frac{T_2}{T_1}$.
Given that the pressure increases by $4 \%$,$P_2 = P_1 + 0.04 P_1 = 1.04 P_1$.
Substituting this into the ratio: $\frac{T_2}{T_1} = 1.04$.
$T_2 = 1.04 \times 373 \ K = 387.92 \ K$.
The increase in temperature $\Delta T = T_2 - T_1 = 387.92 \ K - 373 \ K = 14.92 \ K$.
Since a change in temperature in Kelvin is equal to a change in temperature in Celsius,$\Delta T = 14.92^{\circ} C$.

(C) For an ideal gas,the equation of state is $PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$.
Rearranging this,we get $P = \frac{m}{V} \cdot \frac{RT}{M} = \rho \frac{RT}{M}$,where $\rho$ is the density and $M$ is the molecular weight.
Thus,$M = \frac{\rho RT}{P}$.
Since the temperature $T$ is the same for both gases,we have:
$\frac{M_A}{M_B} = \frac{\rho_A}{\rho_B} \times \frac{P_B}{P_A}$
Given that $P_A = 2P_B$ and $\rho_A = 1.5\rho_B = \frac{3}{2}\rho_B$,we substitute these values:
$\frac{M_A}{M_B} = \frac{1.5\rho_B}{\rho_B} \times \frac{P_B}{2P_B} = 1.5 \times \frac{1}{2} = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$
Therefore,the ratio of the molecular weights of $A$ and $B$ is $3: 4$.

(C) Since the temperature is constant,we use Boyle's Law: $P_1 V_1 = P_2 V_2$.
Let the initial pressure be $P$ and initial volume be $V$.
The pressure is decreased by $20 \%$,so the new pressure $P_2 = P - 0.20P = 0.80P$.
Substituting into the equation: $P \times V = 0.80P \times V_2$.
Solving for $V_2$: $V_2 = \frac{V}{0.80} = 1.25V$.
The change in volume is $\Delta V = V_2 - V = 1.25V - V = 0.25V$.
The percentage change in volume is $\frac{\Delta V}{V} \times 100 = 0.25 \times 100 = 25 \%$.
Since the result is positive,the volume increases by $25 \%$.

(D) According to the ideal gas equation,$PV = Nk_B T$,where $N$ is the number of molecules and $k_B$ is the Boltzmann constant.
For jar $P$,the equation is: $PV = N_P k_B T$ ... $(i)$
For jar $Q$,the equation is: $(2P) \left( \frac{V}{4} \right) = N_Q k_B (2T)$
Simplifying the left side: $\frac{PV}{2} = N_Q k_B (2T)$
$\Rightarrow PV = 4 N_Q k_B T$ ... $(ii)$
Comparing equations $(i)$ and $(ii)$:
$N_P k_B T = 4 N_Q k_B T$
$N_P = 4 N_Q$
Therefore,the ratio of the number of molecules in jar $P$ to those in jar $Q$ is $\frac{N_P}{N_Q} = 4:1$.

(D) According to Boyle's Law at constant temperature,$P \propto \frac{1}{V}$,which implies $P_1 V_1 = P_2 V_2$.
Let the initial pressure be $P_1 = P$.
The new pressure is $P_2 = P + 0.05P = 1.05P$.
Substituting these into the equation: $P V_1 = 1.05P V_2$.
Solving for the new volume: $V_2 = \frac{V_1}{1.05} \approx 0.9524 V_1$.
The change in volume is $\Delta V = V_2 - V_1 = 0.9524 V_1 - V_1 = -0.0476 V_1$.
The percentage decrease in volume is $\frac{|\Delta V|}{V_1} \times 100 = 0.0476 \times 100 = 4.76 \%$.

(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.
Thus,$PV = \frac{m}{M}RT$,which implies $m = \frac{PVM}{RT}$.
For container $A$:
$m_A = \frac{PVM}{RT} \quad (1)$
For container $B$:
$m_B = \frac{(2P)(2V)M}{R(T/2)} = \frac{4PVM}{RT/2} = \frac{8PVM}{RT} \quad (2)$
Taking the ratio of $m_A$ to $m_B$:
$\frac{m_A}{m_B} = \frac{PVM/RT}{8PVM/RT} = \frac{1}{8}$.
Therefore,the ratio is $1: 8$.

(C) According to Boyle's Law, at constant temperature, $PV = \text{constant}$.
Let the initial pressure be $P_1 = P$ and the initial volume be $V_1 = V$.
If the pressure is decreased by $20 \%$, the new pressure $P_2$ is $P - 0.20P = 0.8P$.
Using the relation $P_1 V_1 = P_2 V_2$:
$P \cdot V = (0.8P) \cdot V_2$
$V_2 = \frac{PV}{0.8P} = \frac{V}{0.8} = 1.25V$.
The change in volume is $\Delta V = V_2 - V_1 = 1.25V - V = 0.25V$.
The percentage change in volume is $\frac{\Delta V}{V_1} \times 100 = \frac{0.25V}{V} \times 100 = 25 \%$.
Since the value is positive, the volume increases by $25 \%$.

(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.
Thus,$\frac{PV}{T} = \frac{m}{M}R$.
Initially,$P_1 = 10 \text{ atm}$,$T_1 = 27 + 273 = 300 \text{ K}$,and mass is $m$.
So,$\frac{10V}{300} = \frac{m}{M}R \implies \frac{m}{M}R = \frac{10V}{300} = \frac{V}{30}$.
After removing half the mass,the new mass is $m' = \frac{m}{2}$.
The new temperature is $T_2 = 87 + 273 = 360 \text{ K}$.
Using the gas equation for the final state: $P_2 V = n' R T_2 = \frac{m}{2M} R T_2$.
Substituting $\frac{m}{M}R = \frac{V}{30}$:
$P_2 V = \frac{1}{2} \left( \frac{V}{30} \right) \times 360$.
$P_2 = \frac{360}{60} = 6 \text{ atm}$.