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

(A) For an ideal gas at constant temperature, the equation of state is $PV = \text{constant}$.
Taking the derivative with respect to $P$: $P \frac{dV}{dP} + V = 0$.
This implies $P \frac{dV}{dP} = -V$, or $\frac{dV}{dP} = -\frac{V}{P}$.
Substituting this into the expression for $\beta$:
$\beta = - \left( \frac{dV}{dP} \right) / V = - \left( -\frac{V}{P} \right) / V = \frac{1}{P}$.
Thus, $\beta \propto \frac{1}{P}$.
This represents a rectangular hyperbola, where $\beta$ decreases as $P$ increases. Therefore, the correct graph is the one showing a rectangular hyperbola.

(A) For a spherical vessel of radius $r$ and wall thickness $t$,the volume of metal is $V_m = 4\pi r^2 t$.
The hoop stress $\sigma$ for a thin-walled spherical vessel is given by $\sigma = \frac{Pr}{2t}$,where $P$ is the internal pressure.
From the ideal gas law,$PV = nRT$,where $V = \frac{4}{3}\pi r^3$.
Substituting $P = \frac{2t\sigma}{r}$ into the ideal gas law: $n = \frac{PV}{RT} = \frac{(2t\sigma/r)(4/3 \pi r^3)}{RT} = \frac{8\pi r^2 t \sigma}{3RT}$.
Since $V_m = 4\pi r^2 t$,we have $r^2 t = \frac{V_m}{4\pi}$.
Substituting this into the expression for $n$: $n = \frac{8\pi (V_m / 4\pi) \sigma}{3RT} = \frac{2 V_m \sigma}{3RT}$.

(C) Using the ideal gas equation: $PV = \frac{m}{M}RT$,where $P$ is pressure,$V$ is volume,$m$ is mass,$M$ is molar mass,$R$ is the universal gas constant,and $T$ is temperature.
Since $P$,$V$,$m$,and $R$ are the same for both gases,we have $T = \frac{PVM}{mR}$.
This implies $T \propto M$.
The molar mass of helium $(M_{He} = 4 \ g/mol)$ is less than the molar mass of nitrogen $(M_{N_2} = 28 \ g/mol)$.
Therefore,$T_{helium} < T_{nitrogen}$.

(B) From the ideal gas law,we have $PV = nRT$.
We know that the number of moles $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass.
Substituting this into the equation,we get $PV = \frac{m}{M} RT$.
Rearranging the terms,we get $P = \left(\frac{m}{V}\right) \frac{RT}{M}$.
Since density $\rho = \frac{m}{V}$,the equation becomes $P = \rho \frac{RT}{M}$.
At a constant temperature $(T)$,the gas constant $(R)$ and molar mass $(M)$ are also constant.
Therefore,$P \propto \rho$,which represents a straight line passing through the origin.
Thus,the correct graph is a straight line passing through the origin,which corresponds to graph $B$.

(A) The ideal gas equation is given by $PV = NkT$,where $P$ is pressure,$V$ is volume,$N$ is the total number of molecules,$k$ is the Boltzmann constant,and $T$ is the absolute temperature.
We need to find the number of molecules per unit volume,which is defined as $n = \frac{N}{V}$.
From the ideal gas equation,we can write $n = \frac{N}{V} = \frac{P}{kT}$.
Given the pressure $P = (6.02 \times 10^{23})kT$.
Substituting this value of $P$ into the expression for $n$:
$n = \frac{(6.02 \times 10^{23})kT}{kT} = 6.02 \times 10^{23}$.
Thus,the number of molecules per unit volume is $6.02 \times 10^{23}$.

(D) The ideal gas equation is given by $PV = nRT$,where $n$ is the number of moles.
We know that $n = \frac{m}{M_w}$,where $m$ is the mass and $M_w$ is the molar mass.
Substituting this into the equation,we get $PV = \frac{m}{M_w} RT$.
Rearranging for pressure $P$,we have $P = \left(\frac{m}{V}\right) \frac{RT}{M_w}$.
Since density $\rho = \frac{m}{V}$,the equation becomes $P = \frac{\rho RT}{M_w}$.
At a constant temperature $T$ and for a specific gas (constant $M_w$),the pressure $P$ is directly proportional to the density $\rho$,i.e.,$P \propto \rho$.

(B) For a gas in a flask, the pressure $P$ and volume $V$ remain constant as air is expelled.
According to the ideal gas equation, $PV = \mu RT$, where $\mu$ is the number of moles (proportional to mass) and $T$ is the absolute temperature in Kelvin.
Since $P$ and $V$ are constant, we have $\mu T = \text{constant}$.
Let the initial mass be $m$ (moles $\mu_1 = \mu$) at temperature $T_1 = 27 + 273 = 300 \, K$.
After heating, half the mass is expelled, so the remaining mass is $m/2$ (moles $\mu_2 = \mu/2$) at temperature $T_2$.
Using the relation $\mu_1 T_1 = \mu_2 T_2$:
$\mu \times 300 = (\mu/2) \times T_2$
$300 = T_2 / 2$
$T_2 = 600 \, K$.
Converting to Celsius: $T_2 = 600 - 273 = 327 \, ^{\circ}C$.

(B) Given the law for the ideal gas process: $VP^2 = K$ (where $K$ is a constant).
From the ideal gas equation, we have $PV = nRT$, which implies $P = \frac{nRT}{V}$.
Substituting this into the given law:
$V \left( \frac{nRT}{V} \right)^2 = K$
$V \left( \frac{n^2 R^2 T^2}{V^2} \right) = K$
$\frac{n^2 R^2 T^2}{V} = K$
Since $n$ and $R$ are constants, we can write $T^2 \propto V$, or $T \propto \sqrt{V}$.
Therefore, $\frac{T_2}{T_1} = \sqrt{\frac{V_2}{V_1}}$.
Given $V_1 = V$, $T_1 = T$, and $V_2 = 2V$, we have:
$\frac{T'}{T} = \sqrt{\frac{2V}{V}} = \sqrt{2}$.
Thus, $T' = \sqrt{2} T$.

(A) For an ideal gas,the equation of state is $PV = nRT$.
Rearranging this for volume $V$ as a function of temperature $T$,we get $V = (nR/P)T$.
This represents a straight line passing through the origin with a slope $m = nR/P$.
Since the mass of the gas is the same,$n$ is constant. Therefore,the slope $m$ is inversely proportional to the pressure $P$,i.e.,$m \propto 1/P$.
From the given graph,the slope of the line corresponding to $P_2$ is greater than the slope of the line corresponding to $P_1$ (i.e.,$m_2 > m_1$).
Since $m \propto 1/P$,a larger slope implies a smaller pressure.
Therefore,$P_2 < P_1$,or $P_1 > P_2$.

(D) For an ideal gas,the equation of state is $PV = nRT = \frac{M}{M_{mol}} RT$,where $M$ is the mass of the gas.
Since the pressure $P$ is constant,we have $V \propto MT$,which implies $\frac{V_1}{V_2} = \frac{M_1 T_1}{M_2 T_2}$.
Let the initial volume be $V_1 = 100V$,initial temperature be $T_1 = 100T$,and initial mass be $M_1 = 100M$.
After the changes,the volume is reduced by $10\%$,so $V_2 = 90V$.
The temperature is increased by $20\%$,so $T_2 = 120T$.
Substituting these values into the proportionality equation:
$\frac{100V}{90V} = \frac{100M \times 100T}{M_2 \times 120T}$
$\frac{10}{9} = \frac{10000M}{120M_2}$
$1200M_2 = 90000M$
$M_2 = \frac{90000}{1200}M = 75M$.
The percentage of gas leaked is $\frac{M_1 - M_2}{M_1} \times 100\% = \frac{100M - 75M}{100M} \times 100\% = 25\%$.

(A) From the given $P-T$ diagram,the process $1 \rightarrow 2$ is a straight line passing through the origin.
For an ideal gas,$PV = nRT$,which implies $V = \frac{nRT}{P}$.
Since the line passes through the origin,the slope $\frac{P}{T}$ is constant,meaning $V$ is constant for this specific process.
However,looking at the graph,the process $1 \rightarrow 2$ is a line $P = mT$.
Using the ideal gas equation $PV = nRT$,we have $V = \frac{nRT}{P} = \frac{nR}{m}$.
Since $n, R,$ and $m$ are constants,the volume $V$ is constant during the process $1 \rightarrow 2$.
Wait,let's re-examine the graph: The line $1 \rightarrow 2$ passes through the origin $(0,0)$ in the $P-T$ plane.
For a process represented by a straight line passing through the origin in a $P-T$ diagram,the volume $V$ remains constant (isochoric process).
Thus,$V_1 = V_2$.
However,the problem states $\frac{V_2}{V_1} = 2$. This implies the process $1 \rightarrow 2$ is not isochoric.
Let's re-read the graph: The line $1 \rightarrow 2$ connects $(T_1, V_1)$ to $(T_2, V_2)$.
Actually,the graph is a $P-T$ diagram. The process $1 \rightarrow 2$ is a straight line.
Using the ideal gas law $P = \frac{nRT}{V}$,for a process to be a straight line through the origin in $P-T$ graph,$V$ must be constant.
Given the contradiction in the problem statement $(\frac{V_2}{V_1} = 2)$,we must assume the process $1 \rightarrow 2$ follows the ideal gas law $PV = nRT$.
At state $1$: $P_1 V_1 = nRT_1$.
At state $2$: $P_2 V_2 = nRT_2$.
From the graph,the line $1-2$ passes through the origin,so $\frac{P_1}{T_1} = \frac{P_2}{T_2} = k$.
Thus,$P_1 = kT_1$ and $P_2 = kT_2$.
Substituting into the ideal gas law: $V_1 = \frac{nRT_1}{P_1} = \frac{nR}{k}$ and $V_2 = \frac{nRT_2}{P_2} = \frac{nR}{k}$.
This confirms $V_1 = V_2$.
Given the constraint $\frac{V_2}{V_1} = 2$ is likely intended for a different process or is a typo in the problem.
Assuming the standard interpretation of such cyclic problems where $T_2$ is calculated based on the provided ratio: $T_2 = T_1 \times \frac{V_2}{V_1} = 300 \times 2 = 600 \, K$.

(C) The ideal gas equation is $PV = nRT$,where $n = \frac{m}{M_w}$ ($m$ is mass,$M_w$ is molar mass).
Substituting $V = \frac{m}{\rho}$,we get $P \left( \frac{m}{\rho} \right) = \frac{m}{M_w} RT$.
Rearranging for the ratio of density to pressure,we get $\frac{\rho}{P} = \frac{M_w}{RT}$.
Let $y = \frac{\rho}{P} = \frac{M_w}{RT}$.
At $T_1 = 273\, \text{K}$ $(0\,^{\circ}\text{C})$,$y_1 = x = \frac{M_w}{R(273)}$.
At $T_2 = 373\, \text{K}$ $(100\,^{\circ}\text{C})$,$y_2 = \frac{M_w}{R(373)}$.
Taking the ratio: $\frac{y_2}{x} = \frac{273}{373}$.
Therefore,$y_2 = \frac{273}{373}x$.

(C) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass,$M$ is the molar mass,and $\rho = \frac{m}{V}$ is the density.
Thus,$P = \frac{\rho RT}{M}$,which implies $\rho = \frac{PM}{RT}$.
At point $A$: $P_A = 3P_0$,$T_A = 2T_0$,and $\rho_A = \rho_0 = \frac{(3P_0)M}{R(2T_0)} = \frac{3P_0 M}{2RT_0}$.
At point $B$: $P_B = 5P_0$,$T_B = 4T_0$,and $\rho_B = \frac{(5P_0)M}{R(4T_0)} = \frac{5P_0 M}{4RT_0}$.
Taking the ratio: $\frac{\rho_B}{\rho_0} = \frac{5P_0 M / 4RT_0}{3P_0 M / 2RT_0} = \frac{5}{4} \times \frac{2}{3} = \frac{10}{12} = \frac{5}{6}$.
Therefore,$\rho_B = \frac{5}{6}\rho_0$.

(D) For an ideal gas,the equation of state is $PV = nRT$.
Since $n = m/M$ and density $\rho = m/V$,we can write $V = m/\rho$.
Substituting this into the ideal gas equation: $P(m/\rho) = (m/M)RT$.
This simplifies to $P = (\rho RT)/M$.
At a constant temperature $T$,the gas constant $R$ and molar mass $M$ are also constant.
Therefore,$P = k\rho$,where $k = (RT)/M$ is a constant.
This equation represents a linear relationship between pressure $P$ and density $\rho$,which is a straight line passing through the origin.
Thus,the correct graph is a straight line passing through the origin.

(C) Given:
Cross-sectional area $A = 8.0 \times 10^{-3} \, m^2$
Initial temperature $T_1 = 300 \, K$
Initial volume $V_1 = 2.4 \times 10^{-3} \, m^3$
Displacement of piston $\Delta x = 0.1 \, m$
Spring constant $k = 8000 \, N/m$
Atmospheric pressure $P_0 = 1.0 \times 10^5 \, N/m^2$
$1$. Calculate the final volume $V_2$:
$V_2 = V_1 + A \Delta x = 2.4 \times 10^{-3} + (8.0 \times 10^{-3} \times 0.1) = 2.4 \times 10^{-3} + 0.8 \times 10^{-3} = 3.2 \times 10^{-3} \, m^3$
$2$. Calculate the final pressure $P_2$:
The final pressure is the sum of atmospheric pressure and the pressure exerted by the spring force:
$P_2 = P_0 + \frac{k \Delta x}{A} = 1.0 \times 10^5 + \frac{8000 \times 0.1}{8.0 \times 10^{-3}} = 1.0 \times 10^5 + \frac{800}{8.0 \times 10^{-3}} = 1.0 \times 10^5 + 1.0 \times 10^5 = 2.0 \times 10^5 \, N/m^2$
$3$. Use the ideal gas law $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$:
Initial pressure $P_1 = P_0 = 1.0 \times 10^5 \, N/m^2$ (since the spring is initially relaxed).
$\frac{1.0 \times 10^5 \times 2.4 \times 10^{-3}}{300} = \frac{2.0 \times 10^5 \times 3.2 \times 10^{-3}}{T_2}$
$T_2 = \frac{2.0 \times 10^5 \times 3.2 \times 10^{-3} \times 300}{1.0 \times 10^5 \times 2.4 \times 10^{-3}} = \frac{2.0 \times 3.2 \times 300}{2.4} = \frac{6.4 \times 300}{2.4} = \frac{1920}{2.4} = 800 \, K$

(B) Let the initial length of air columns at both ends be $l_0 = (100 - 20) / 2 = 40 \ cm$. Let the mercury column be displaced by $y$ when the tube is held vertically.
The pressure of the air in the lower part $(P_1)$ and upper part $(P_2)$ changes due to the compression and expansion of air, respectively. Using Boyle's Law $(PV = \text{constant})$ at constant temperature:
For the lower part: $P_0 (40 A) = P_1 (40 - y) A \Rightarrow P_1 = \frac{40 P_0}{40 - y}$
For the upper part: $P_0 (40 A) = P_2 (40 + y) A \Rightarrow P_2 = \frac{40 P_0}{40 + y}$
In the vertical position, the pressure at the lower face is equal to the pressure at the upper face plus the pressure due to the $20 \ cm$ mercury column $(h = 20 \ cm)$:
$P_1 = P_2 + h \rho g$
Substituting $P_0 = 76 \ cm$ of $Hg$ (where $P_0 = \rho g \times 76$):
$\frac{40 \times 76 \rho g}{40 - y} = \frac{40 \times 76 \rho g}{40 + y} + 20 \rho g$
Dividing by $\rho g$:
$\frac{3040}{40 - y} - \frac{3040}{40 + y} = 20$
$3040 \left( \frac{40 + y - (40 - y)}{1600 - y^2} \right) = 20$
$3040 \left( \frac{2y}{1600 - y^2} \right) = 20$
$304 y = 1600 - y^2$
$y^2 + 304 y - 1600 = 0$
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{-304 + \sqrt{304^2 - 4(1)(-1600)}}{2} \approx 5.18 \ cm$.

(A) Given,initial volume $V_{1} = V$ and final volume $V_{2} = 2V$.
Initial temperature $T_{1} = 27^{\circ}C = 27 + 273 = 300 \text{ K}$.
According to Charles's Law,for a gas at constant pressure,the volume is directly proportional to its absolute temperature: $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$.
Substituting the values: $\frac{V}{300} = \frac{2V}{T_{2}}$.
Solving for $T_{2}$: $T_{2} = 2 \times 300 = 600 \text{ K}$.
Converting the final temperature back to Celsius: $T_{2} = 600 - 273 = 327^{\circ}C$.

(D) Let $A$ be the cross-sectional area of the cylinder. The forces acting on the piston are the pressure from the gas above ($P_1 A$ downwards),the pressure from the gas below ($P_2 A$ upwards),and the weight of the piston ($mg$ downwards).
For the piston to be in equilibrium,the net force must be zero:
$P_2 A = P_1 A + mg$
$mg = (P_2 - P_1) A$
Using the ideal gas law $PV = nRT$,where $V = Al$,we have $P = \frac{nRT}{Al}$.
Thus,$P_1 = \frac{nRT}{Al_1}$ and $P_2 = \frac{nRT}{Al_2}$.
Substituting these into the equilibrium equation:
$mg = \left( \frac{nRT}{Al_2} - \frac{nRT}{Al_1} \right) A$
$mg = nRT \left( \frac{1}{l_2} - \frac{1}{l_1} \right)$
$m = \frac{nRT}{g} \left( \frac{l_1 - l_2}{l_1 l_2} \right)$

(C) Given $n = 1$ mole.
From the ideal gas equation,$PV = nRT = RT$,so $P = \frac{RT}{V}$.
Substituting this into the given relation $P = P_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$:
$\frac{RT}{V} = P_0 \left[ 1 - \frac{V_0^2}{2V^2} \right]$
$T = \frac{P_0}{R} \left[ V - \frac{V_0^2}{2V} \right]$
Now,calculate temperature at $V = V_0$ and $V = 2V_0$:
$T_1 = T(V_0) = \frac{P_0}{R} \left[ V_0 - \frac{V_0^2}{2V_0} \right] = \frac{P_0}{R} \left[ V_0 - \frac{V_0}{2} \right] = \frac{P_0 V_0}{2R}$
$T_2 = T(2V_0) = \frac{P_0}{R} \left[ 2V_0 - \frac{V_0^2}{2(2V_0)} \right] = \frac{P_0}{R} \left[ 2V_0 - \frac{V_0}{4} \right] = \frac{P_0}{R} \left[ \frac{7V_0}{4} \right] = \frac{7 P_0 V_0}{4R}$
Change in temperature $\Delta T = T_2 - T_1 = \frac{7 P_0 V_0}{4R} - \frac{P_0 V_0}{2R} = \frac{7 P_0 V_0 - 2 P_0 V_0}{4R} = \frac{5 P_0 V_0}{4R}$.

(B) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.
Density $\rho = \frac{m}{V} = \frac{PM}{RT}$.
Since $M$ and $R$ are constants,$\rho \propto \frac{P}{T}$.
At point $A$,the coordinates are $(T_0, P_0)$,so $\left(\frac{P}{T}\right)_A = \frac{P_0}{T_0}$.
At point $B$,the coordinates are $(2T_0, 3P_0)$,so $\left(\frac{P}{T}\right)_B = \frac{3P_0}{2T_0} = \frac{3}{2} \left(\frac{P_0}{T_0}\right)$.
Since $\rho \propto \frac{P}{T}$,we have $\frac{\rho_B}{\rho_A} = \frac{(P/T)_B}{(P/T)_A} = \frac{3}{2}$.
Therefore,$\rho_B = \frac{3}{2} \rho_A = \frac{3}{2} \rho_0$.

(C) For a gas in a closed vessel,the volume $V$ remains constant.
According to Gay-Lussac's Law,for a constant volume,$P \propto T$,which implies $\frac{P}{T} = \text{constant}$.
Differentiating this,we get $\frac{dP}{P} = \frac{dT}{T}$.
Given that the pressure increases by $0.5\%$,we have $\frac{dP}{P} = 0.005$.
The change in temperature is given as $dT = 1\, K$ (since a change of $1\,^{\circ}C$ is equal to a change of $1\, K$).
Substituting these values into the equation: $0.005 = \frac{1}{T}$.
Solving for $T$,we get $T = \frac{1}{0.005} = 200\, K$.
To convert the temperature from Kelvin to Celsius,we use the formula $t(^{\circ}C) = T(K) - 273.15$.
Thus,$t = 200 - 273 = -73\, ^{\circ}C$.

(D) For an ideal gas, the equation of state is $PV = nRT = (m/M)RT$, where $m$ is the mass and $M$ is the molar mass. Since density $\rho = m/V$, we can write $P = (\rho/M)RT$, which implies $\rho = (PM)/(RT)$.
In graph $(i)$, $T$ is constant and $P$ is increasing. Since $\rho \propto P/T$, if $P$ increases at constant $T$, density $\rho$ increases. Thus, statement $(A)$ is correct.
In graph $(ii)$, $P$ is constant and $T$ is increasing. Since $\rho \propto P/T$, if $T$ increases at constant $P$, density $\rho$ decreases. Thus, statement $(B)$ is correct.
In graph $(iii)$, the graph is a straight line passing through the origin, so $P/T = \text{constant}$. Since $\rho = (PM)/(RT)$, if $P/T$ is constant, then density $\rho$ is constant. Thus, statement $(C)$ is correct.
Since all statements $(A), (B),$ and $(C)$ are correct, the wrong statement is none of the above.

(B) For an ideal gas,the equation of state is $PV = nRT$.
For a constant temperature (isothermal) process,$T$ is constant. Since $n$ and $R$ are also constants,the product $nRT$ is a constant,let's call it $k$.
So,the equation becomes $PV = k$,which can be rewritten as $P = k \left( \frac{1}{V} \right)$.
This equation is of the form $y = mx$,where $y = P$,$x = \frac{1}{V}$,and the slope $m = nRT$ is a positive constant.
Therefore,the graph of $P$ versus $\frac{1}{V}$ is a straight line passing through the origin.

(C) For an open container,the pressure $P$ and volume $V$ remain constant.
According to the ideal gas law,$PV = \mu RT$,where $\mu$ is the number of moles and $R$ is the gas constant.
Since $P$ and $V$ are constant,$\mu T = \text{constant}$.
Therefore,$\mu_1 T_1 = \mu_2 T_2$.
Given $\mu_1 = x$,$T_1 = T$,and $T_2 = 3T$.
Substituting these values: $x \times T = \mu_2 \times (3T)$.
Solving for $\mu_2$: $\mu_2 = \frac{x}{3}$,which is the amount of gas remaining in the container.
The amount of gas that exits the container is the initial amount minus the remaining amount: $\Delta \mu = x - \frac{x}{3} = \frac{2x}{3}$.

(B) For an ideal gas,the equation of state is $PV = \frac{m}{M}RT$,where $m$ is the mass,$M$ is the molar mass,$R$ is the universal gas constant,and $T$ is the absolute temperature.
Since $T$ is constant,$PV = \text{constant} \times m$.
For a fixed pressure $P$,we can see from the graph that the volume $V_2$ corresponding to mass $m_2$ is greater than the volume $V_1$ corresponding to mass $m_1$ (i.e.,$V_2 > V_1$).
From the ideal gas equation,$m = \frac{PVM}{RT}$.
Since $P, M, R,$ and $T$ are constant,$m \propto V$.
Therefore,since $V_2 > V_1$,it follows that $m_2 > m_1$.

(C) From the ideal gas equation,$PV = nRT = \frac{m}{M_w}RT$,where $m$ is the mass and $M_w$ is the molar mass.
Rearranging for density $\rho = \frac{m}{V}$,we get $P = \frac{\rho RT}{M_w}$,which implies $\frac{\rho}{P} = \frac{M_w}{RT}$.
Given that at $T_1 = 273\,K$ $(0\,^{\circ}\text{C})$,$\frac{\rho_1}{P_1} = x = \frac{M_w}{R(273)}$.
At $T_2 = 373\,K$ $(100\,^{\circ}\text{C})$,the new quotient is $x' = \frac{M_w}{R(373)}$.
Dividing the two expressions: $\frac{x'}{x} = \frac{M_w / (R \cdot 373)}{M_w / (R \cdot 273)} = \frac{273}{373}$.
Therefore,$x' = \frac{273}{373}x$.

(B) Given the process equation $P = \frac{2V^2}{1+V^2}$.
For $V_i = 1 \, m^3$,the initial pressure $P_i = \frac{2(1)^2}{1+(1)^2} = \frac{2}{2} = 1 \, Pa$.
For $V_f = 2 \, m^3$,the final pressure $P_f = \frac{2(2)^2}{1+(2)^2} = \frac{8}{5} \, Pa$.
Using the ideal gas equation $PV = nRT$,for $n = 1$ mole,$T = \frac{PV}{R}$.
The change in temperature is $\Delta T = T_f - T_i = \frac{P_f V_f - P_i V_i}{nR}$.
Substituting the values: $\Delta T = \frac{(\frac{8}{5} \times 2) - (1 \times 1)}{1 \times R} = \frac{\frac{16}{5} - 1}{R} = \frac{16-5}{5R} = \frac{11}{5R} \, K$.

(A) For a gas in a closed vessel,the volume $V$ remains constant. According to Gay-Lussac's Law,for a fixed mass of gas at constant volume,the pressure $P$ is directly proportional to the absolute temperature $T$ $(P \propto T)$.
This implies $\frac{P_1}{T_1} = \frac{P_2}{T_2}$,or $\frac{\Delta P}{P} = \frac{\Delta T}{T}$.
Given: Initial temperature $T = 250\,K$,change in temperature $\Delta T = 1\,K$.
The percentage increase in pressure is given by $\frac{\Delta P}{P} \times 100$.
Substituting the values: $\frac{\Delta P}{P} \times 100 = \frac{\Delta T}{T} \times 100 = \frac{1}{250} \times 100 = 0.4\%$.

(C) At equilibrium,the pressure $P$ in both flasks must be the same.
Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$ ($M$ is the molar mass of the gas):
$P = \frac{mRT}{MV}$
Since $P$,$R$,and $M$ are constant for both flasks,we have:
$\frac{m_1 T_1}{V_1} = \frac{m_2 T_2}{V_2}$
Given $V_1 = 2V_2$,$T_1 = 100\,K$,$T_2 = 200\,K$,and $m_1 = m$:
$\frac{m \times 100}{2V_2} = \frac{m_2 \times 200}{V_2}$
$\frac{100m}{2} = 200m_2$
$50m = 200m_2$
$m_2 = \frac{50m}{200} = \frac{m}{4}$

(C) Using the ideal gas equation $PV = nRT$,for a fixed amount of gas,we have $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given initial temperature $T_1 = 27 + 273 = 300 \ K$.
Initial pressure is $P_1 = P$ and initial volume is $V_1 = V$.
Final pressure $P_2 = 2P$ and final volume $V_2 = 3V$.
Substituting these values into the equation: $\frac{P \cdot V}{300} = \frac{(2P) \cdot (3V)}{T_2}$.
$\frac{1}{300} = \frac{6}{T_2}$.
$T_2 = 6 \times 300 = 1800 \ K$.
To convert to Celsius: $T_2(^{\circ}C) = 1800 - 273 = 1527^{\circ}C$.

(B) According to Boyle's law,for a fixed mass of gas at constant temperature,pressure and volume are inversely proportional to each other,i.e.,$P \propto \frac{1}{V}$.
Therefore,$P_1 V_1 = P_2 V_2$.
At the bottom of the lake,the pressure $P_1$ is the sum of atmospheric pressure $P_0$ and the pressure due to the water column of height $h$: $P_1 = P_0 + h \rho_w g$.
At the surface,the pressure $P_2$ is equal to the atmospheric pressure $P_0$.
Given: $h = 47.6 \ m = 4760 \ cm$,$V_1 = 50 \ cm^3$,$P_0 = 70 \ cm$ of $Hg = 70 \times 13.6 \times g \ \text{dynes/cm}^2$,$\rho_w = 1 \ g/cm^3$.
Substituting the values into the equation $P_0 V_1 + h \rho_w g V_1 = P_0 V_2$:
$V_2 = V_1 \left( 1 + \frac{h \rho_w g}{P_0} \right) = V_1 \left( 1 + \frac{h \rho_w g}{h_{Hg} \rho_{Hg} g} \right) = V_1 \left( 1 + \frac{h \rho_w}{h_{Hg} \rho_{Hg}} \right)$.
$V_2 = 50 \left( 1 + \frac{4760 \times 1}{70 \times 13.6} \right) = 50 \left( 1 + \frac{4760}{952} \right) = 50 (1 + 5) = 50 \times 6 = 300 \ cm^3$.

(C) The number of moles of oxygen $(n_{O_2})$ is $8/32 = 0.25\,mol$.
The number of moles of nitrogen $(n_{N_2})$ is $7/28 = 0.25\,mol$.
According to Dalton's law of partial pressures,the total pressure $P_{total} = (n_{O_2} + n_{N_2}) \frac{RT}{V} = 10\,atm$.
Since $n_{O_2} = n_{N_2} = 0.25\,mol$,the partial pressure of each gas is equal.
$P_{O_2} = P_{N_2} = \frac{10}{2} = 5\,atm$.
When oxygen is removed,only nitrogen remains in the vessel.
Therefore,the new pressure in the system will be equal to the partial pressure of nitrogen,which is $5\,atm$.

(B) Since the volume of the flask remains constant,we use Gay-Lussac's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given initial pressure $P_1 = 1\,atm$ and initial temperature $T_1 = 27 + 273 = 300\,K$.
The final pressure required to force the cork out is $P_2 = 2.5\,atm$.
Substituting the values: $\frac{1}{300} = \frac{2.5}{T_2}$.
Solving for $T_2$: $T_2 = 2.5 \times 300 = 750\,K$.
Converting to Celsius: $750 - 273 = 477\,^{\circ}C$.

(B) From the ideal gas equation,$P = n k T$,where $n$ is the number density (molecules per unit volume).
$n = P / (k T)$
Substituting the given values:
$n = (4 \times 10^{-10}) / (1.38 \times 10^{-23} \times 300)$
$n \approx (4 \times 10^{-10}) / (4.14 \times 10^{-21})$
$n \approx 0.966 \times 10^{11} \approx 10^{11} \text{ molecules/m}^3$
Since $1\,m^3 = 10^6\,cm^3$,the number of molecules per $cm^3$ is:
$n_{cm^3} = 10^{11} / 10^6 = 10^5 \text{ molecules/cm}^3$.

(C) For an ideal gas,the equation of state is $PV = nRT$.
For a constant pressure $P$,the volume $V$ is directly proportional to the temperature $T$ $(V \propto T)$.
From the given $P-V$ graph,if we draw a vertical line at a constant pressure $P$,the volume corresponding to the curve $T_2$ is greater than the volume corresponding to the curve $T_1$ $(V_2 > V_1)$.
Since $V \propto T$,a larger volume at the same pressure implies a higher temperature.
Therefore,$T_2 > T_1$ or $T_1 < T_2$.

(C) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$ ($M$ is the molar mass).
Thus,$V = \frac{mR}{MP} T$.
In the given graph,the $V-T$ plot is a straight line passing through absolute zero $(-273.15\, ^\circ\text{C})$,where the slope is $S = \frac{mR}{MP}$.
For the first case,the slope is $S_1 = \frac{mR}{MP}$.
For the second case,the mass is $2m$ and the pressure is $2P$,so the new slope is $S_2 = \frac{(2m)R}{M(2P)} = \frac{mR}{MP}$.
Since $S_1 = S_2$,the slope remains unchanged.
Therefore,the expansion is given by the same straight line $B$.

(B) Given: Volume $V = 8 \, L = 8 \times 10^{-3} \, m^3$,Initial pressure $P_1 = 2 \, atm = 2 \times 1.01325 \times 10^5 \, Pa \approx 2.0265 \times 10^5 \, Pa$. For simplicity in calculation,using $1 \, atm = 10^5 \, Pa$ as per standard approximation in many textbooks,$P_1 = 2 \times 10^5 \, Pa$.
Initial moles $n_1 = \frac{P_1 V}{RT} = \frac{2 \times 10^5 \times 8 \times 10^{-3}}{8.314 \times 300} \approx 0.641 \, mol$.
Final pressure $P_2 = 125 \, kPa = 1.25 \times 10^5 \, Pa$.
Since temperature $T$ and volume $V$ are constant,$P \propto n$,so $\frac{n_2}{n_1} = \frac{P_2}{P_1}$.
$n_2 = n_1 \times \frac{P_2}{P_1} = 0.641 \times \frac{1.25 \times 10^5}{2 \times 10^5} = 0.641 \times 0.625 \approx 0.40 \, mol$.
Leaked moles $\Delta n = n_1 - n_2 = 0.64 - 0.40 = 0.24 \, mol$.

(D) The ideal gas equation is given by $PV = \mu RT$,where $\mu$ is the number of moles of the gas.
For oxygen gas $(O_2)$,the molar mass $M$ is $32\,g/mol$.
The given mass $m$ is $5\,g$.
The number of moles $\mu$ is calculated as $\mu = \frac{m}{M} = \frac{5}{32}\,mol$.
Substituting this value into the ideal gas equation,we get $PV = \frac{5}{32}\,RT$.

(C) The number $6.02 \times 10^{23}$ is known as Avogadro's number $(N_A)$.
By definition,one mole of any substance contains exactly $6.02 \times 10^{23}$ elementary entities (atoms,molecules,or ions),regardless of the temperature,pressure,or volume of the substance.
Therefore,the Assertion is correct.
However,the definition of a mole does not depend on $S.T.P.$ (Standard Temperature and Pressure) conditions.
$S.T.P.$ conditions are only relevant when calculating the volume occupied by one mole of an ideal gas (which is $22.4 \ L$ at $S.T.P.$).
Thus,the Reason is incorrect.

(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) The ideal gas equation is given by $PV = nRT$,where $P$ is the pressure,$V$ is the volume,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the absolute temperature.
The basis of the absolute temperature scale (also known as the Kelvin scale) is the absolute zero temperature,which is $-273.15 \ ^{\circ}C$. At this temperature,the pressure of an ideal gas becomes zero,as shown in the graph where the pressure-temperature line extrapolates to zero pressure at $-273.15 \ ^{\circ}C$.

(N/A) The ideal gas equation relating pressure $(P)$,volume $(V)$,and absolute temperature $(T)$ is given as:
$PV = nRT$
Where:
$R$ is the universal gas constant $= 8.314 \; J \; mol^{-1} \; K^{-1}$
$n = 1 \; mol$
$T = 273.15 \; K$ (Standard temperature)
$P = 1.01325 \times 10^{5} \; Pa$ (Standard pressure of $1 \; atm$)
Rearranging for volume:
$V = \frac{nRT}{P}$
Substituting the values:
$V = \frac{1 \times 8.314 \times 273.15}{1.01325 \times 10^{5}}$
$V \approx 0.02241 \; m^{3}$
Since $1 \; m^{3} = 1000 \; L$:
$V \approx 22.41 \; L$
Thus,the molar volume of an ideal gas at $STP$ is approximately $22.4 \; L$.

(A) The absolute pressure $P$ is given by $P = P_{gauge} + P_{atm}$. Assuming $P_{atm} = 1 \; atm = 1.013 \times 10^{5} \; Pa$.
Initial absolute pressure $P_{1} = 15 + 1 = 16 \; atm = 16 \times 1.013 \times 10^{5} \; Pa = 1.6208 \times 10^{6} \; Pa$.
Initial temperature $T_{1} = 27 + 273 = 300 \; K$.
Volume $V = 30 \; L = 30 \times 10^{-3} \; m^{3}$.
Using the ideal gas equation $PV = nRT = (m/M)RT$,the initial mass $m_{1} = (P_{1} V M) / (R T_{1})$.
$m_{1} = (1.6208 \times 10^{6} \times 30 \times 10^{-3} \times 32 \times 10^{-3}) / (8.31 \times 300) \approx 0.624 \; kg$.
Final absolute pressure $P_{2} = 11 + 1 = 12 \; atm = 12 \times 1.013 \times 10^{5} \; Pa = 1.2156 \times 10^{6} \; Pa$.
Final temperature $T_{2} = 17 + 273 = 290 \; K$.
Final mass $m_{2} = (P_{2} V M) / (R T_{2})$.
$m_{2} = (1.2156 \times 10^{6} \times 30 \times 10^{-3} \times 32 \times 10^{-3}) / (8.31 \times 290) \approx 0.485 \; kg$.
Mass of oxygen taken out $\Delta m = m_{1} - m_{2} = 0.624 - 0.485 = 0.139 \; kg \approx 0.13 \; kg$.

(B) Initial volume of the air bubble,$V_{1} = 1.0 \; cm^{3} = 1.0 \times 10^{-6} \; m^{3}$.
Depth of the lake,$d = 40 \; m$.
Initial temperature at the bottom,$T_{1} = 12 \; ^{\circ}C = 285 \; K$.
Final temperature at the surface,$T_{2} = 35 \; ^{\circ}C = 308 \; K$.
Pressure at the surface,$P_{2} = 1 \; atm = 1.013 \times 10^{5} \; Pa$.
Pressure at the bottom,$P_{1} = P_{2} + d \rho g$,where $\rho = 10^{3} \; kg/m^{3}$ and $g = 9.8 \; m/s^{2}$.
$P_{1} = 1.013 \times 10^{5} + (40 \times 10^{3} \times 9.8) = 1.013 \times 10^{5} + 3.92 \times 10^{5} = 4.933 \times 10^{5} \; Pa$.
Using the ideal gas law,$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$.
$V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}} = \frac{(4.933 \times 10^{5}) \times (1.0 \times 10^{-6}) \times 308}{285 \times (1.013 \times 10^{5})}$.
$V_{2} \approx 5.26 \times 10^{-6} \; m^{3} = 5.26 \; cm^{3}$.

(C) Volume of the room,$V = 25.0 \; m^{3}$.
Temperature of the room,$T = 27^{\circ} C = 300 \; K$.
Pressure in the room,$P = 1 \; atm = 1.013 \times 10^{5} \; Pa$.
The ideal gas equation is given by $PV = N k_{B} T$,where $k_{B}$ is the Boltzmann constant.
Boltzmann constant,$k_{B} = 1.38 \times 10^{-23} \; J/K$.
The number of air molecules $N$ is given by $N = \frac{PV}{k_{B} T}$.
Substituting the values: $N = \frac{1.013 \times 10^{5} \times 25.0}{1.38 \times 10^{-23} \times 300}$.
$N = \frac{25.325 \times 10^{5}}{414 \times 10^{-23}} \approx 0.06117 \times 10^{28} = 6.11 \times 10^{26}$ molecules.
Therefore,the total number of air molecules in the room is $6.11 \times 10^{26}$.

(D) Length of the narrow bore,$L = 100 \; cm$.
Length of the mercury thread,$l = 76 \; cm$.
Length of the air column between mercury and the closed end,$l_a = 15 \; cm$.
When the tube is held vertically with the open end at the bottom,the air column length increases because some mercury flows out. Let $h \; cm$ of mercury flow out.
The final length of the air column becomes $l_2 = 15 + (100 - 76 - 15) + h = 24 + h \; cm$.
The final length of the mercury column becomes $76 - h \; cm$.
Initial pressure $P_1 = 76 \; cm$ of Hg,Initial volume $V_1 = 15 \; cm$ (proportional to length).
Final pressure $P_2 = 76 - (76 - h) = h \; cm$ of Hg.
Using Boyle's Law,$P_1 V_1 = P_2 V_2$:
$76 \times 15 = h(24 + h)$
$h^2 + 24h - 1140 = 0$.
Solving for $h$ using the quadratic formula: $h = \frac{-24 + \sqrt{576 + 4560}}{2} = \frac{-24 + 73.05}{2} \approx 24.5 \; cm$.
Thus,$24.5 \; cm$ of mercury flows out,and the final air column length is $24 + 24.5 = 48.5 \; cm$.

(N/A) When a car is in motion,the friction between the tires and the road surface generates heat.
This heat is transferred to the air inside the tire,causing the temperature $(T)$ of the air to increase.
According to the Kinetic Theory of Gases,for a fixed volume $(V)$ of gas (the tire volume remains approximately constant),the pressure $(P)$ is directly proportional to the absolute temperature $(T)$ as per Gay-Lussac's Law $(P \propto T)$.
As the temperature of the air inside the tire rises due to friction,the kinetic energy of the gas molecules increases,leading to more frequent and forceful collisions with the inner walls of the tire.
Consequently,the air pressure inside the tire increases.

(N/A) Gay-Lussac's Law of Gaseous Volumes states that when gases react together,they do so in volumes which bear a simple ratio to one another and to the volume of the products,if gaseous,provided that the temperature and pressure remain constant.

(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,this can be expressed as $V \propto n$ at constant temperature $(T)$ and pressure $(P)$,where $V$ is the volume and $n$ is the number of moles (or molecules).
This implies that if different gases are kept in containers of equal volume at the same temperature and pressure,they will contain the same number of molecules.
This law is also known as Avogadro's hypothesis and is fundamental in understanding the kinetic theory of gases.

To find the volume of $1\,g$ of water vapour at $1\,atm$ pressure,we use the ideal gas law: $PV = nRT$.
Here,$P = 1\,atm = 1.013 \times 10^5\,Pa$.
The mass of water is $m = 1\,g = 10^{-3}\,kg$.
The molar mass of water $(H_2O)$ is $M = 18\,g/mol = 18 \times 10^{-3}\,kg/mol$.
The number of moles $n = \frac{m}{M} = \frac{1}{18}\,mol$.
The universal gas constant $R = 8.314\,J/(mol \cdot K)$.
Assuming the temperature $T = 373.15\,K$ (boiling point of water at $1\,atm$):
$V = \frac{nRT}{P} = \frac{(1/18) \times 8.314 \times 373.15}{1.013 \times 10^5}$.
$V \approx 1.70 \times 10^{-3}\,m^3 = 1.70\,litres$.