Ideal gas equation and Related gas laws Questions in English

(D) The ideal gas equation is $PV = nRT$.
For $n = 1 \ mol$,the equation becomes $P = (RT) \times (V^{-1})$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = P$ and $x = V^{-1}$,we get the slope $m = RT$.
Given that the slope $m = 32.8 \ L \ atm \ mol^{-1}$ and $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
Therefore,$RT = 32.8 \ L \ atm \ mol^{-1}$.
$T = \frac{32.8 \ L \ atm \ mol^{-1}}{0.082 \ L \ atm \ K^{-1} \ mol^{-1}} = 400 \ K$.

(C) The ideal gas equation is given by $PV = nRT$.
Since concentration $C = \frac{n}{V}$,we can rewrite the equation as $P = CRT$.
Therefore,$C = \frac{P}{RT}$.
Substituting the given values: $C = \frac{16.4 \ atm}{0.082 \ L \ atm \ mol^{-1} \ K^{-1} \times 200 \ K}$.
$C = \frac{16.4}{16.4} = 1.00 \ mol \ L^{-1}$.

(B) From the ideal gas equation,$PV = nRT$.
Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M} RT$.
Rearranging for density $d = \frac{m}{V}$,we get $d = \frac{PM}{RT}$.
At a constant temperature $T$ and for a specific gas with molar mass $M$,the term $\frac{M}{RT}$ is constant.
Therefore,$d \propto P$.

(A) The ideal gas equation is $PV = nRT$.
Rearranging this for the graph of $P$ versus $\frac{1}{V}$,we get $P = (nRT) \times \frac{1}{V}$.
Comparing this with the equation of a straight line $y = mx$,where $y = P$,$x = \frac{1}{V}$,and $m = nRT$.
Given that the slope $m = 1.1$.
Thus,$nRT = 1.1$.
We are given $P = 10 \ atm$.
Using $PV = nRT$,we have $V = \frac{nRT}{P}$.
Substituting the values,$V = \frac{1.1}{10} = 0.11 \ L$.

(D) The ideal gas equation is $pV = nRT$.
Dividing both sides by $V$,we get $p = (n/V)RT = CRT$,where $C$ is the concentration in $mol$ $L^{-1}$.
Given $p = 16.4$ $atm$,$C = 1$ $mol$ $L^{-1}$,and $R = 0.082$ $L$ $atm$ $mol^{-1}$ $K^{-1}$.
Substituting these values into the equation: $16.4 = 1 \times 0.082 \times T$.
$T = 16.4 / 0.082 = 200$ $K$.

(C) According to Charles's law,the volume of a gas is directly proportional to its absolute temperature $(V \propto T)$.
As the temperature decreases,the volume decreases.
Theoretically,the volume of a gas becomes zero at absolute zero,which is $0 \ K$.
Converting this to the Celsius scale: $T(^{\circ}C) = T(K) - 273.15$.
Therefore,$0 \ K = -273.15^{\circ} C$.

(A) For an ideal gas,the compressibility factor $Z = \frac{PV}{RT} = 1$.
From the given data table,we observe that for $P = 1, 2, \text{ and } 3 \ bar$,the value of $\frac{PV}{RT}$ is $1$.
For $P = 4 \ bar$ and $P = 5 \ bar$,the value of $\frac{PV}{RT}$ deviates from $1$ ($1.5$ and $2.0$ respectively).
Therefore,the gas behaves as an ideal gas in the pressure range of $1$ to $3 \ bar$.

(A) Initial concentration of gas $(X)$ is $1 \ mol \ L^{-1}$ in a $1 \ L$ flask,so initial moles $(n_1) = 1 \ mol$.
Using the ideal gas equation $PV = nRT$,the initial pressure $(P_1)$ is:
$P_1 = \frac{n_1 RT}{V} = \frac{1 \times 0.082 \times 200}{1} = 16.4 \ atm$.
After adding $0.1 \ mol$ of $X$,the new number of moles $(n_2) = 1 + 0.1 = 1.1 \ mol$.
Since temperature $(T)$ and volume $(V)$ remain constant,the pressure is directly proportional to the number of moles $(P \propto n)$.
Therefore,$\frac{P_1}{n_1} = \frac{P_2}{n_2}$.
$P_2 = P_1 \times \frac{n_2}{n_1} = 16.4 \times \frac{1.1}{1} = 18.04 \ atm$.

(C) For the $P-V$ graph,draw a line parallel to the $V$-axis (constant $P$). At a constant pressure,the volume $V$ is directly proportional to the temperature $T$ $(V \propto T)$. Since the curve for $T_2$ is at a higher volume than $T_1$ for the same pressure,we have $T_2 > T_1$.
For the $V-T$ graph,the equation is $V = (\frac{nR}{P})T$. The slope of the line is $\frac{nR}{P}$,which is inversely proportional to the pressure $P$. Since the slope of the line for $P_1$ is smaller than the slope of the line for $P_2$,it follows that $P_1 > P_2$.

(C) From the ideal gas equation,$PV = nRT$.
Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M} RT$.
Rearranging for density $d = \frac{m}{V}$,we get $d = \frac{PM}{RT}$.
Given: $P = 0.82 \ atm$,$M = 4 \ g \ mol^{-1}$,$R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,and $T = 300 \ K$.
Substituting the values: $d = \frac{0.82 \times 4}{0.082 \times 300} = \frac{4}{30} = 0.1333 \ g \ L^{-1} = 1.33 \times 10^{-1} \ g \ L^{-1}$.

(A) According to the ideal gas equation,we have:
$PV = nRT$
Since the number of moles $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we can substitute this into the equation:
$PV = \frac{m}{M} RT$
Rearranging the terms to solve for density $\rho = \frac{m}{V}$:
$P = \frac{m}{V} \times \frac{RT}{M}$
$P = \rho \times \frac{RT}{M}$
Therefore,the density $\rho$ is given by:
$\rho = \frac{PM}{RT}$

(B) According to Boyle's Law,for a fixed amount of gas at a constant temperature,the pressure $(P)$ is inversely proportional to the volume $(V)$ of the gas: $P \propto \frac{1}{V}$ or $PV = \text{constant}$.
$1$. Graph $(b)$: Represents $PV$ vs $P$. Since $PV$ is constant,the graph is a horizontal line parallel to the $P$-axis.
$2$. Graph $(c)$: Represents $PV$ vs $V$. Since $PV$ is constant,the graph is a horizontal line parallel to the $V$-axis.
Both graphs $(b)$ and $(c)$ correctly represent the constant product of $PV$ as dictated by Boyle's Law.

(B) The density $(d)$ of a gas is given by the formula: $d = \frac{pM}{RT}$
Given:
Pressure $(p)$ = $5 \text{ atm}$
Molar mass of $O_2$ $(M)$ = $32 \text{ g mol}^{-1}$
Gas constant $(R)$ = $0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}$
Temperature $(T)$ = $127 + 273 = 400 \text{ K}$
Substituting the values:
$d = \frac{5 \times 32}{0.082 \times 400}$
$d = \frac{160}{32.8} \approx 4.878 \text{ g L}^{-1}$
Thus,the density is approximately $4.88 \text{ g L}^{-1}$.

(A) Using the ideal gas equation,$PV = nRT$,the temperature $T$ is given by $T = \frac{PV}{nR}$.
Given values are $P = 3.32 \ bar$,$V = 5 \ dm^3 = 5 \ L$,$n = 4 \ mol$,and the gas constant $R = 0.08314 \ bar \ L \ K^{-1} \ mol^{-1}$.
Substituting these values into the equation:
$T = \frac{3.32 \times 5}{4 \times 0.08314} = \frac{16.6}{0.33256} \approx 49.91 \ K$.
Rounding to the nearest whole number,we get $T \approx 50 \ K$.

(A) According to Boyle's law,at constant temperature,$PV = k$.
$1.$ For $y = P$ and $x = V$,the relation $P = k/V$ represents a hyperbola. Thus,$(A)$ matches $2$.
$2.$ For $y = P$ and $x = 1/V$,the relation $P = k(1/V)$ represents a straight line passing through the origin. Thus,$(B)$ matches $3$.
$3.$ For $y = PV$ and $x = P$,the relation $PV = k$ represents a horizontal line parallel to the $x$-axis. Thus,$(C)$ matches $1$.

(B) Given: $n = 4 \ mol$,$V = 5 \ L$,$P = 3.32 \ bar$,$R = 0.083 \ bar \ L \ K^{-1} \ mol^{-1}$.
Using the ideal gas equation: $PV = nRT$.
Rearranging for temperature: $T = \frac{PV}{nR}$.
Substituting the values: $T = \frac{3.32 \times 5}{4 \times 0.083}$.
$T = \frac{16.6}{0.332} = 50 \ K$.
Therefore,the correct option is $B$.

(C) The variation in the volume of a gas with temperature was first studied by Jacques Charles $(1787)$ and is known as Charles's law.
Charles's law states that,"The volume of a given mass of a gas is directly proportional to its absolute temperature at constant pressure."
Mathematically,$V \propto T$ or $\frac{V}{T} = k$ at constant pressure.
Hence,the correct option is $(C)$.

(D) According to Boyle's law,at constant temperature,$P_1V_1 = P_2V_2$.
Let the initial volume be $V_1 = V$.
The new volume $V_2 = V + 0.10V = 1.10V$.
Substituting into the equation: $P_1V = P_2(1.10V)$.
$P_2 = P_1 / 1.10 \approx 0.9091 P_1$.
The change in pressure is $P_2 - P_1 = 0.9091 P_1 - P_1 = -0.0909 P_1$.
This represents a decrease of approximately $9.09 \%$.
Given the options provided,the closest value is a decrease of $10 \%$,which corresponds to option $(D)$.

(B) Charles' law states that the volume of a fixed mass of an ideal gas is directly proportional to its absolute temperature at constant pressure.
At constant pressure,$V \propto T$.
Or,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Hence,the correct option is $(B)$.

(B) Given: Amount of gas $n = 4.0 \ mol$,Volume $V = 5 \ dm^3$,Pressure $p = 3.32 \ bar$,and Gas constant $R = 0.083 \ bar \ dm^3 \ K^{-1} \ mol^{-1}$.
Using the ideal gas equation $p \ V = n \ R \ T$,we can solve for temperature $T$:
$T = \frac{p \ V}{n \ R}$
Substituting the values:
$T = \frac{3.32 \times 5}{4.0 \times 0.083}$
$T = \frac{16.6}{0.332} = 50 \ K$.
Therefore,the temperature of the gas is $50 \ K$.

(A) Using the ideal gas equation $PV = nRT$,where $n = \frac{w}{M}$.
For gas $A$: $P_A V = \frac{w_A}{M_A} RT \implies 1 \times V = \frac{2}{M_A} RT \implies V = \frac{2RT}{M_A} \quad (1)$.
When gas $B$ is added,the total pressure $P_{total} = P_A + P_B = 1.5 \ atm$. Since $P_A = 1 \ atm$,then $P_B = 0.5 \ atm$.
For gas $B$: $P_B V = \frac{w_B}{M_B} RT \implies 0.5 \times V = \frac{3}{M_B} RT \implies V = \frac{6RT}{M_B} \quad (2)$.
Equating $(1)$ and $(2)$: $\frac{2RT}{M_A} = \frac{6RT}{M_B}$.
$\frac{2}{M_A} = \frac{6}{M_B} \implies \frac{M_A}{M_B} = \frac{2}{6} = \frac{1}{3}$.
Thus,the ratio of molar masses $M_A : M_B$ is $1: 3$.

(A) Using the ideal gas law $PV = nRT$,we find the moles of each gas at constant temperature $T = 300 \ K$.
For $H_2$: $n_1 = \frac{P_1 V_1}{RT} = \frac{1 \ bar \times 1 \ L}{RT} = \frac{1}{RT}$.
For $O_2$: $n_2 = \frac{P_2 V_2}{RT} = \frac{2 \ bar \times 2 \ L}{RT} = \frac{4}{RT}$.
Total moles $n_{total} = n_1 + n_2 = \frac{1+4}{RT} = \frac{5}{RT}$.
In the $10 \ L$ flask,the total pressure $P_{total}$ is given by $P_{total} = \frac{n_{total} RT}{V_{final}}$.
$P_{total} = \frac{(5/RT) \times RT}{10 \ L} = \frac{5}{10} \ bar = 0.5 \ bar$.

(A) For an ideal gas,the product of pressure and volume $(PV)$ is proportional to the number of moles $(n)$ at a constant temperature $(T)$.
Using the principle of conservation of moles,the total pressure of the mixture $(P_{mix})$ in the final volume $(V_{final} = 100 \ L)$ is given by the sum of the partial pressures of $A$ and $B$.
Initial state of $A$: $P_1 = 1 \ bar$,$V_1 = 2 \ L$.
Initial state of $B$: $P_2 = p_B \ bar$,$V_2 = 4 \ L$.
Final state of mixture: $P_{mix} = 0.1 \ bar$,$V_{final} = 100 \ L$.
According to Boyle's Law $(P_1V_1 + P_2V_2 = P_{mix}V_{final})$:
$(1 \ bar \times 2 \ L) + (p_B \ bar \times 4 \ L) = 0.1 \ bar \times 100 \ L$
$2 + 4p_B = 10$
$4p_B = 10 - 2$
$4p_B = 8$
$p_B = 2 \ bar$.
Therefore,the value of $p_B$ is $2$.

(D) The density $(d)$ of an ideal gas is given by the formula $d = \frac{PM}{RT}$,where $P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature in Kelvin.
From this relation,$d \propto \frac{P}{T}$.
To maximize density,we need the highest pressure $(P)$ and the lowest temperature $(T)$.
Comparing the options:
$A$: $T = 273 \ K, P = 2 \ bar \implies \frac{P}{T} = \frac{2}{273} \approx 0.0073$
$B$: $T = 546 \ K, P = 1 \ bar \implies \frac{P}{T} = \frac{1}{546} \approx 0.0018$
$C$: $T = 546 \ K, P = 2 \ bar \implies \frac{P}{T} = \frac{2}{546} \approx 0.0036$
$D$: $T = 273 \ K, P = 3 \ bar \implies \frac{P}{T} = \frac{3}{273} \approx 0.0109$
Thus,the density is maximum at $0^{\circ} C$ and $3 \ bar$.

(C) According to the ideal gas equation,$PV = nRT$,where $n$ is the number of moles.
Since the number of molecules is directly proportional to the number of moles $(n = \frac{N}{N_A})$,the number of molecules will be equal if the number of moles is equal.
For equal volumes $(V)$ of different gases,the number of moles $(n)$ will be equal if the temperature $(T)$ and pressure $(P)$ of all the flasks are the same.
Therefore,the correct condition is that the temperature and pressure of all the flasks are the same.

(B) For gas $X$: $P_X V = \frac{m_X}{M_X} RT \Rightarrow 2V = \frac{1}{M_X} RT$ ...$(i)$
For gas $Y$: $P_Y V = \frac{m_Y}{M_Y} RT$
Given total pressure $P_{total} = 3 \ atm$,so $P_Y = P_{total} - P_X = 3 - 2 = 1 \ atm$.
Substituting values for gas $Y$: $1 \cdot V = \frac{2}{M_Y} RT$ ...$(ii)$
Dividing equation $(i)$ by $(ii)$: $\frac{2}{1} = \frac{1/M_X}{2/M_Y} = \frac{M_Y}{2M_X}$
$4M_X = M_Y$ or $M_Y = 4M_X$.

(C) The ideal gas equation is $PV = nRT$,which can be written as $PV = (m/M)RT$.
Rearranging for density $(d = m/V)$,we get $d = (PM) / (RT)$.
Given $P = 76 \ cm \ Hg = 1 \ atm$,$T = 273 \ K$,$d = 1.94 \ g/dm^3$ (which is $1.94 \ g/L$),and $R = 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
Substituting the values: $1.94 = (1 \times M) / (0.0821 \times 273)$.
$M = 1.94 \times 0.0821 \times 273 \approx 43.5 \ g/mol$.
The molar mass of $CO_2$ is $12 + 2 \times 16 = 44 \ g/mol$.
Thus,the gas is $CO_2$.

(B) Given:
$T = 27 + 273 = 300 \ K$
$V = 10 \ L$
$Molar mass (He) = 4 \ g \ mol^{-1}$
$Molar mass (Ne) = 20 \ g \ mol^{-1}$
$P = 1.23 \ atm$
$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$
Using the ideal gas equation $PV = nRT$,where $n = n_{He} + n_{Ne} = \frac{W_{He}}{M_{He}} + \frac{W_{Ne}}{M_{Ne}}$:
$PV = (\frac{W_{He}}{4} + \frac{W_{Ne}}{20})RT$
$1.23 \times 10 = (\frac{W_{He}}{4} + \frac{W_{Ne}}{20}) \times 0.082 \times 300$
$12.3 = (\frac{W_{He}}{4} + \frac{W_{Ne}}{20}) \times 24.6$
$\frac{W_{He}}{4} + \frac{W_{Ne}}{20} = \frac{12.3}{24.6} = 0.5$
Multiplying by $20$:
$5W_{He} + W_{Ne} = 10$ (Equation $i$)
We are given the total mass:
$W_{He} + W_{Ne} = 4$ (Equation $ii$)
Subtracting Equation $ii$ from Equation $i$:
$(5W_{He} + W_{Ne}) - (W_{He} + W_{Ne}) = 10 - 4$
$4W_{He} = 6 \Rightarrow W_{He} = 1.5 \ g$
$W_{Ne} = 4 - 1.5 = 2.5 \ g$
Mass $\%$ of neon $= \frac{W_{Ne}}{\text{Total mass}} \times 100 = \frac{2.5}{4} \times 100 = 62.5 \%$

(B) For an ideal gas,the kinetic energy $(KE)$ is given by $KE = \frac{3}{2} PV$.
Since the graph of $P$ versus $V^{-1}$ is a straight line passing through the origin,we have $P = m \times V^{-1}$,where $m$ is the slope.
Comparing this with the ideal gas equation $PV = RT$,we get $P = RT \times V^{-1}$,so the slope $m = RT$.
However,the problem defines the slope $m$ in terms of $KE$ as $P = \frac{2 \ KE}{3} \times V^{-1}$.
Thus,the slope $m = \frac{2 \ KE}{3} = 2000 \ J \ mol^{-1}$.
Solving for $KE$: $KE = \frac{2000 \times 3}{2} = 3000 \ J \ mol^{-1}$.

(D) For an ideal gas,the equation is $PV = nRT$,which can be rearranged as $P = (nRT) \times (\frac{1}{V})$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = P$,$x = \frac{1}{V}$,and $c = 0$,the slope $m$ is equal to $nRT$.
Since $n$ and $R$ are constants,the slope $m$ is directly proportional to the temperature $T$ $(m \propto T)$.
Given the condition $T_1 > T_2 > T_3$,it follows that the slopes must satisfy the order $m_1 > m_2 > m_3$.

(A) The ideal gas equation is given by $PV = nRT$.
Concentration is defined as the number of moles per unit volume,i.e.,$C = \frac{n}{V}$.
From the ideal gas equation,we can rearrange the terms to find $\frac{n}{V}$:
$\frac{n}{V} = \frac{P}{RT}$.
Therefore,the concentration is $\frac{P}{RT} \ mol/L$.

(A) For one mole of an ideal gas,the ideal gas equation is $pV = RT$,which can be rearranged as $V = (R/p)T$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V$ and $x = T$,the slope $m$ is given by $m = R/p$.
Since $R$ is a constant,the slope $m$ is inversely proportional to the pressure $p$ $(m \propto 1/p)$.
Given the condition $p_1 < p_2 < p_3$,it follows that $1/p_1 > 1/p_2 > 1/p_3$.
Therefore,the relation between the slopes is $m_1 > m_2 > m_3$.

(C) For an open vessel,the pressure $P$ and volume $V$ remain constant. According to the ideal gas equation $PV = nRT$,we have $n \propto \frac{1}{T}$.
Therefore,$n_1 T_1 = n_2 T_2$.
Given: $T_1 = 27 + 273 = 300 \ K$ and $T_2 = 727 + 273 = 1000 \ K$.
The fraction of air remaining is $\frac{n_2}{n_1} = \frac{T_1}{T_2}$.
Substituting the values: $\frac{n_2}{n_1} = \frac{300}{1000} = \frac{3}{10}$.

(B) For an ideal gas,the equation is $PV = nRT$.
Rearranging for volume $V$ as a function of moles $n$: $V = n \times (\frac{RT}{P})$.
Comparing this with the equation of a straight line $Y = mx + C$,where $Y = V$ and $x = n$,the slope $m$ is given by $\frac{RT}{P}$.
Given $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,and $P = 1 \ atm$.
$\text{Slope} = \frac{0.0821 \times 300}{1} = 24.6 \ L \ mol^{-1}$.

(D) Using the ideal gas equation $PV = nRT = \frac{wRT}{M}$,we have $n = \frac{PV}{RT}$.
For gas $A$: $n_A = \frac{5V}{RT} = \frac{4}{M_A} \implies M_A = \frac{4RT}{5V}$.
When gas $B$ is added,the total pressure becomes $10 \ atm$. The partial pressure of gas $B$ is $P_B = P_{total} - P_A = 10 \ atm - 5 \ atm = 5 \ atm$.
For gas $B$: $n_B = \frac{5V}{RT} = \frac{16}{M_B} \implies M_B = \frac{16RT}{5V}$.
Taking the ratio: $\frac{M_A}{M_B} = \frac{4RT/5V}{16RT/5V} = \frac{4}{16} = \frac{1}{4}$.
Therefore,$4 M_A = M_B$.

(B) Using the ideal gas equation: $PV = nRT = \frac{m}{M} RT$
$\therefore m = \frac{PVM}{RT}$
Since both containers contain the same gas $(CO_2)$,the molar mass $(M)$ is constant.
Therefore,$\frac{m_A}{m_B} = \frac{P_A V_A}{P_B V_B} \times \frac{T_B}{T_A}$
Given: $P_A = 4P_B$,$V_A = 4V_B$,$T_A = 4T_B$,and $m_B = x \ g$.
Substituting these values: $\frac{m_A}{x} = \frac{(4P_B)(4V_B)}{P_B V_B} \times \frac{T_B}{4T_B}$
$\frac{m_A}{x} = 16 \times \frac{1}{4} = 4$
$m_A = 4x \ g$

(A) Step $1$: Calculate the number of moles of each gas.
$n_{CH_4} = \frac{4 \ g}{16 \ g/mol} = 0.25 \ mol$
$n_{CO_2} = \frac{4.4 \ g}{44 \ g/mol} = 0.1 \ mol$
Step $2$: Calculate the total number of moles $(n_T)$.
$n_T = 0.25 \ mol + 0.1 \ mol = 0.35 \ mol$
Step $3$: Use the ideal gas equation $PV = nRT$ to find the pressure.
Given: $V = 1 \ L$,$T = 27 + 273 = 300 \ K$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
$P = \frac{n_T RT}{V} = \frac{0.35 \times 0.082 \times 300}{1} = 8.61 \ atm \approx 8.6 \ atm$.

(A) According to the ideal gas equation,$PV = nRT$,which can be rearranged as $V = (\frac{nR}{P})T$.
This equation represents a straight line passing through the origin $(V = mT)$,where the slope $(m)$ is equal to $\frac{nR}{P}$.
Since the slope is inversely proportional to pressure $(P)$,a line with a steeper slope corresponds to a lower pressure.
In the given plot,the slope of the line for $P_1$ is the highest,followed by $P_2$,and then $P_3$ (which has the lowest slope).
Therefore,the order of pressure is $P_3 > P_2 > P_1$,or $P_1 < P_2 < P_3$.

(C) The ideal gas equation is given by $PM = dRT$.
Here,$P = 1 \ atm$,$d = 1.24 \ g/L$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,and $T = 273 \ K$.
Rearranging the formula to solve for molar mass $(M)$:
$M = \frac{dRT}{P} = \frac{1.24 \times 0.0821 \times 273}{1} \approx 28 \ g/mol$.
The molar mass of $CO$ is $12 + 16 = 28 \ g/mol$.
Therefore,the gas is $CO$.

(B) According to Charle's law, at constant pressure, $V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 3 \,L$, $T_1 = 35 + 273 = 308 \,K$.
$V_2 = 2.5 \,L$, $T_2 = T$.
Substituting the values: $\frac{3}{308} = \frac{2.5}{T_2}$.
$T_2 = \frac{2.5 \times 308}{3} = 256.67 \,K$.
Converting to Celsius: $T(^{\circ}C) = 256.67 - 273 = -16.33^{\circ} C$.
Rounding to the nearest option, $T \approx -16^{\circ} C$.

(D) Using the ideal gas equation: $PV = nRT = \frac{w}{M} RT$
Where $M$ is the molar mass of the gas.
Given: $w = 4.4 \ g$,$T = 0^{\circ} C = 273 \ K$,$R = 0.082 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,$P = 0.82 \ atm$,$V = 2.73 \ L$.
Substituting the values: $0.82 \times 2.73 = \frac{4.4}{M} \times 0.082 \times 273$
$M = \frac{4.4 \times 0.082 \times 273}{0.82 \times 2.73}$
$M = \frac{4.4 \times 22.386}{2.2386} = 44 \ g/mol$.
The molar mass of $CO_2$ is $12 + 2 \times 16 = 44 \ g/mol$.
Therefore,the gas is $CO_2$.

(B) Using the ideal gas equation $PV = nRT$.
Given: $n = 1 \ mol$,$V = 12 \ L = 12 \times 10^{-3} \ m^3$,$T = 297 + 273 = 570 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$P = \frac{nRT}{V} = \frac{1 \times 8.314 \times 570}{12 \times 10^{-3}} \ Pa$.
$P = \frac{4738.98}{0.012} \ Pa = 394915 \ Pa \approx 395 \ kPa$.

(C) According to Boyle's Law, at constant temperature, $P_1V_1 = P_2V_2$.
Given: $P_1 = 2 \,atm$, $V_1 = V$, $V_2 = \frac{V}{4}$.
Substituting the values: $2 \times V = P_2 \times \frac{V}{4}$.
$P_2 = 2 \times 4 = 8 \,atm$.
The final pressure is $8 \,atm$.
The increase in pressure $= P_2 - P_1 = 8 \,atm - 2 \,atm = 6 \,atm$.

(B) For the first graph ($p$ vs $1/V$): According to the ideal gas equation $pV = nRT$,we have $p = (nRT) \times (1/V)$. The slope of the line is $nRT$. Since the slope for $T_2$ is greater than the slope for $T_1$,it follows that $T_2 > T_1$.
For the second graph ($V$ vs $T$): According to Charles' law $V = (nR/p) \times T$. The slope of the line is $nR/p$. Since the slope for $p_2$ is greater than the slope for $p_1$,and the slope is inversely proportional to pressure,it follows that $p_1 > p_2$.
Therefore,the correct observation is $T_2 > T_1$ and $p_1 > p_2$.

(C) Boyle's Law states that for a given mass of an ideal gas at constant temperature,the pressure $(p)$ is inversely proportional to its volume $(V)$.
Mathematically,this is expressed as:
$p \propto \frac{1}{V}$
$p \cdot V = k$ (where $k$ is a constant)
This relationship represents a rectangular hyperbola when plotting pressure $(p)$ on the $y$-axis against volume $(V)$ on the $x$-axis.
Therefore,the graph that correctly represents Boyle's Law is the one showing a hyperbolic curve where $p$ decreases as $V$ increases.

(B) According to Charles' law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$ at constant pressure.
Given: $V_1 = 420 \ cm^3$,$T_1 = 27 + 273 = 300 \ K$.
The temperature is reduced by $20^{\circ} C$,so $T_2 = 300 \ K - 20 \ K = 280 \ K$.
Using the formula $V_2 = \frac{V_1 \times T_2}{T_1}$:
$V_2 = \frac{420 \ cm^3 \times 280 \ K}{300 \ K} = 392 \ cm^3$.

(A) The ideal gas equation is $PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$,which rearranges to $P = \frac{m}{V} \times \frac{RT}{M} = \frac{dRT}{M}$.
For equal amounts of the same gas,the molar mass $M$ is constant,so $P \propto d \times T$.
Therefore,the ratio of pressures is $\frac{P_1}{P_2} = \frac{d_1}{d_2} \times \frac{T_1}{T_2}$.
Given $\frac{d_1}{d_2} = \frac{1}{2}$ and $\frac{T_1}{T_2} = \frac{2}{1}$.
Substituting these values: $\frac{P_1}{P_2} = \frac{1}{2} \times \frac{2}{1} = \frac{1}{1}$.
Thus,the ratio of their pressures is $1:1$.

(D) Given:
Pressure of $N_2$ gas $(P) = 2.3 \ atm$
Volume $(V) = 2.6 \ cm^3 = 2.6 \times 10^{-3} \ L$
Temperature $(T) = 27^{\circ} C + 273 = 300 \ K$
Gas constant $(R) = 0.0821 \ L \ atm \ mol^{-1} \ K^{-1}$
Using the ideal gas equation: $PV = nRT$
$n = \frac{PV}{RT}$
$n = \frac{2.3 \ atm \times 2.6 \times 10^{-3} \ L}{0.0821 \ L \ atm \ mol^{-1} \ K^{-1} \times 300 \ K}$
$n = \frac{5.98 \times 10^{-3}}{24.63} \ mol$
$n \approx 2.42 \times 10^{-4} \ mol$
Rounding to the nearest provided option,the value is $2 \times 10^{-4} \ mol$.

(B) Boyle's law states that at constant temperature $(T)$ and amount of gas $(n)$,the pressure $(p)$ of a fixed amount of gas is inversely proportional to its volume $(V)$:
$p \propto \frac{1}{V} \implies pV = \text{constant} \quad (k)$
$1$. Plot $(ii)$: Shows $p$ vs $V$ at different temperatures. Since $pV = nRT$,$p = \frac{nRT}{V}$. For a given $V$,$p$ increases as $T$ increases. Thus,for $T_3 > T_2 > T_1$,the curve for $T_3$ should be above $T_2$,which is above $T_1$. The plot $(ii)$ shows $T_1 < T_2 < T_3$ with curves correctly ordered,making it a correct representation of Boyle's law isotherms.
$2$. Plot $(iv)$: Shows $pV$ vs $p$. Since $pV = nRT$,for a fixed $T$,$pV$ is constant. As $T$ increases,the value of the constant $nRT$ increases. Thus,$pV$ should be higher for higher temperatures. The plot $(iv)$ shows $T_1 < T_2 < T_3$ with $pV$ values increasing accordingly,which is correct.
Plots $(i)$ and $(iii)$ do not correctly represent the relationship between the variables at different temperatures based on the ideal gas equation.
Therefore,plots $(ii)$ and $(iv)$ are correct.