Ideal gas equation and Related gas laws Questions in English

(A) Given: Mass of $O_2$ $(w)$ = $2 \ g$,Molar mass of $O_2$ $(M)$ = $32 \ g/mol$,Temperature $(T)$ = $27 + 273 = 300 \ K$,Pressure $(P)$ = $760 \ mm \ Hg = 1 \ atm$.
Using the ideal gas equation: $PV = nRT$.
Number of moles $(n)$ = $\frac{w}{M} = \frac{2}{32} = 0.0625 \ mol$.
Substituting values into the equation: $V = \frac{nRT}{P} = \frac{0.0625 \times 0.0821 \times 300}{1}$.
$V = 1.539 \ L \approx 1.5 \ L$.

(A) The temperature $0 \, ^\circ C$ is equivalent to $273 \, K$ $(T = 0 + 273 = 273 \, K)$.
Since the pressure and temperature remain constant,according to the gas laws,the volume of the gas will remain unchanged.
Therefore,the volume at $273 \, K$ is $V \, mL$.

(D) The total volume of the two flasks when connected is $V_{total} = 1 \ L + 3 \ L = 4 \ L$.
Applying Boyle's Law $(P_1V_1 = P_2V_2)$ for each gas individually:
For $N_2$: $100 \ kPa \times 1 \ L = P_{N_2} \times 4 \ L$,which gives $P_{N_2} = 25 \ kPa$.
For $O_2$: $320 \ kPa \times 3 \ L = P_{O_2} \times 4 \ L$,which gives $P_{O_2} = 240 \ kPa$.
According to Dalton's Law of partial pressures,the total pressure $P_{total} = P_{N_2} + P_{O_2} = 25 \ kPa + 240 \ kPa = 265 \ kPa$.

(C) Boyle's law states that for a fixed amount of an ideal gas kept at a fixed temperature,pressure and volume are inversely proportional.
Mathematically,this is expressed as $V \propto \frac{1}{P}$ at constant $T$ and $n$.

(D) According to Boyle's law,for a given mass of an ideal gas at constant temperature,$V \propto \frac{1}{P}$.
This implies $V = \frac{k}{P}$,where $k$ is a constant.
Therefore,$PV = k$,meaning the product of pressure and volume always remains constant.

(A) According to Boyle's Law,at constant temperature $(T)$,the pressure $(P)$ and volume $(V)$ are related as $P_1 V_1 = P_2 V_2$.
Given: $P_1 = 1 \ atm$,$V_1 = 20 \ cm^3$,$V_2 = 50 \ cm^3$.
Substituting the values: $1 \ atm \times 20 \ cm^3 = P_2 \times 50 \ cm^3$.
Therefore,$P_2 = \frac{20}{50} \times 1 \ atm = 0.4 \ atm$.
Comparing this with the given options,the correct expression is $20 \times \frac{1}{50}$.

(A) The correct answer is $A$. According to Boyle's Law,$PV = \text{constant}$ only at a constant temperature. As the temperature changes,the value of this constant also changes. Therefore,the product $PV$ is dependent on temperature.

(B) According to Boyle's law,for a fixed amount of an ideal gas at constant temperature,the product of pressure $(P)$ and volume $(V)$ is constant,i.e.,$PV = k$ (where $k$ is a constant).
This means that the value of $PV$ does not change with a change in pressure $(P)$ or volume $(V)$.
Therefore,a graph of $PV$ versus $P$ or $PV$ versus $V$ will be a horizontal straight line parallel to the x-axis.
Both options $(b)$ and $(c)$ represent this relationship. However,in standard textbook representations for Boyle's law,the $PV$ vs $P$ plot is the most common illustration.

(A) According to $Charles's \ Law$,at constant pressure,the volume $(V)$ of a fixed mass of an ideal gas is directly proportional to its absolute temperature $(T)$ in $Kelvin$. $V \propto T$.

(C) Charles's law states that at constant pressure,the volume of a fixed mass of a gas is directly proportional to its absolute temperature $(T)$.
Mathematically,this is expressed as $V \propto T$ or $\frac{V}{T} = k$ (where $k$ is a constant).
Therefore,the correct expression is $V \propto T$.

(D) The use of hot air balloons is based on $Charles's \ law$,which states that the volume of a fixed mass of gas is directly proportional to its absolute temperature at constant pressure $(V \propto T)$.
As the air inside the balloon is heated,its volume increases,causing its density to decrease relative to the surrounding cooler air,which creates the buoyancy required for the balloon to rise.

(A) Given:
$T_1 = 273 \ ^oC = (273 + 273) \ K = 546 \ K$
$T_2 = 0 \ ^oC = (273 + 0) \ K = 273 \ K$
$P_1 = 1 \ atm$
Since the volume is constant,we apply Gay-Lussac's law:
$\frac{P_1}{T_1} = \frac{P_2}{T_2}$
$P_2 = \frac{P_1 \times T_2}{T_1} = \frac{1 \times 273}{546} = 0.5 \ atm$

(A) According to Charles's Law,the relationship between volume and temperature at constant pressure is given by: $V_t = V_0(1 + \alpha_v t)$.
Here,$V_t$ is the volume at $t\,^{\circ}C$,$V_0$ is the volume at $0\,^{\circ}C$,and $\alpha_v$ is the coefficient of volume expansion,which is equal to $\frac{1}{273.15}\,^{\circ}C^{-1}$.
For every $1\,^{\circ}C$ rise in temperature,the change in volume $\Delta V$ is given by $\Delta V = V_0 \times \alpha_v \times 1 = \frac{V_0}{273.15}$.
Thus,the volume increases by a definite fraction $\frac{1}{273.15}$ of its volume at $0\,^{\circ}C$.

(A) According to Charles's Law,at constant pressure,the volume of a fixed mass of gas is directly proportional to its absolute temperature: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 0.2 \ L$,$T_1 = 0 \ ^oC = 273 \ K$,$T_2 = 273 \ ^oC = 546 \ K$.
Substituting the values: $V_2 = V_1 \times \frac{T_2}{T_1} = 0.2 \ L \times \frac{546 \ K}{273 \ K} = 0.2 \ L \times 2 = 0.4 \ L$.

(A) According to Charles's Law,at constant pressure,$V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given: $V_1 = 400 \ cm^{3}$,$T_1 = 27 + 273 = 300 \ K$,$T_2 = -3 + 273 = 270 \ K$.
Calculating the final volume: $V_2 = \frac{T_2 \times V_1}{T_1} = \frac{270 \ K \times 400 \ cm^{3}}{300 \ K} = 360 \ cm^{3}$.
The contraction in volume is the difference between the initial and final volume: $\Delta V = V_1 - V_2 = 400 \ cm^{3} - 360 \ cm^{3} = 40 \ cm^{3}$.

(C) From the ideal gas equation,$PV = nRT$,we get $P = (\frac{nR}{V})T$.
Comparing this with the equation of a straight line $y = mx$,where $y = P$ and $x = T$,the slope $m = \frac{nR}{V}$.
Since $n$ and $R$ are constants,the slope is inversely proportional to the volume,i.e.,$m \propto \frac{1}{V}$.
Given $V_1 > V_2$,it follows that the slope for $V_2$ $(m_2)$ must be greater than the slope for $V_1$ $(m_1)$,i.e.,$m_2 > m_1$.
Therefore,the curve for $V_2$ has a greater slope than that for $V_1$.

(A) Initially,the total number of moles $n$ in both vessels of volume $V$ is given by:
$n = n_1 + n_2 = \frac{P_1 V}{R T_1} + \frac{P_1 V}{R T_1} = \frac{2 P_1 V}{R T_1}$
When the temperatures are changed to $T_1$ and $T_2$ respectively,the new pressure $P$ is the same in both vessels. The total number of moles remains constant:
$n = \frac{P V}{R T_1} + \frac{P V}{R T_2} = \frac{P V}{R} \left( \frac{1}{T_1} + \frac{1}{T_2} \right) = \frac{P V}{R} \left( \frac{T_1 + T_2}{T_1 T_2} \right)$
Equating the initial and final moles:
$\frac{2 P_1 V}{R T_1} = \frac{P V}{R} \left( \frac{T_1 + T_2}{T_1 T_2} \right)$
$P = \frac{2 P_1 T_2}{T_1 + T_2}$

(C) The ideal gas equation is given by $PV = nRT$.
Rearranging for the gas constant $R$,we get $R = \frac{PV}{nT}$.
Substituting the units for pressure $(P)$,volume $(V)$,amount of substance $(n)$,and temperature $(T)$,we have:
$R = \frac{atm \cdot litre}{mole \cdot K} = litre \cdot atm \cdot K^{-1} \cdot mole^{-1}$.

(D) The universal gas constant $R$ has different values depending on the units used for pressure and volume.
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$ ($SI$ units).
Since $1 \ J = 10^7 \ erg$ and $1 \ J = 1 \ N \ m = 10^5 \ dyne \times 10^2 \ cm = 10^7 \ dyne \ cm$,options $A$,$B$,and $C$ are equivalent.
Option $D$ $(8.31 \ atm \ K^{-1} \ mol^{-1})$ is incorrect because the unit of $R$ in $atm$ is $0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,not $8.31$.

(A) The universal gas constant $R$ has different values depending on the units used:
$1$. In $SI$ units,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
$2$. In terms of calories,since $1 \, cal \approx 4.184 \, J$,we have $R = \frac{8.314}{4.184} \approx 1.987 \, cal \, K^{-1} \, mol^{-1}$.
$3$. In terms of pressure-volume units,$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$.
Comparing these with the given options,option $A$ is the correct value.

(C) The ideal gas equation is given by $PV = nRT$.
Rearranging for $R$,we get $R = \frac{PV}{nT}$.
Since $PV$ represents work done $(W)$,$n$ represents the number of moles,and $T$ represents the absolute temperature in Kelvin,the units of $R$ are $\text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$.
Therefore,$R$ represents the work done per mole per degree of absolute temperature.

(C) The ideal gas equation is given by $PV = nRT$.
In this equation:
$P$ represents the pressure of the gas.
$V$ represents the volume occupied by $n$ moles of the gas.
$n$ represents the number of moles of the gas.
$R$ is the universal gas constant.
$T$ is the absolute temperature.
Therefore,$V$ is the volume occupied by $n$ moles of the gas at pressure $P$ and temperature $T$. Thus,the statement '$n$ moles of the gas have a volume $V$' is correct.

(B) The universal gas constant $R$ is defined by the ideal gas equation $PV = nRT$.
Its value depends on the units used for pressure,volume,and temperature.
In units of liters,atmospheres,Kelvin,and moles,the value is approximately $0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.

(C) The ideal gas equation is given by $PV = nRT$.
Rearranging for $R$,we get $R = \frac{PV}{nT}$.
In $S.I.$ units,pressure $P$ is measured in $Pascal$ ($Pa$ or $N \, m^{-2}$),volume $V$ in $m^3$,and temperature $T$ in $K$.
Substituting these units: $R = \frac{(N \, m^{-2}) \times (m^3)}{mol \times K} = N \, m \, K^{-1} \, mol^{-1} = J \, K^{-1} \, mol^{-1}$.
The value of the universal gas constant $R$ in $S.I.$ units is $8.314 \, J \, K^{-1} \, mol^{-1}$.

(C) The ideal gas equation $PV = nRT$ describes the state of an ideal gas.
An isothermal process occurs at a constant temperature $(T = \text{constant})$,and an adiabatic process is one where there is no heat exchange $(q = 0)$.
For an ideal gas,the equation of state $PV = nRT$ holds true for any process as long as the gas behaves ideally.
Therefore,the equation is obeyed in both isothermal and adiabatic processes.

(C) The ideal gas equation is given by $PV = nRT$.
Here,$P$ is pressure,$V$ is volume,$n$ is the number of moles,$R$ is the gas constant,and $T$ is the temperature.
We need to find the number of moles per litre,which is the molar concentration $\frac{n}{V}$.
Rearranging the ideal gas equation:
$\frac{n}{V} = \frac{P}{RT}$
Therefore,the correct option is $C$.

(A) Using the ideal gas equation: $PV = nRT$
Given: $n = 2 \ mol$,$T = 546 \ K$,$V = 44.8 \ L$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
$P = \frac{nRT}{V} = \frac{2 \times 0.0821 \times 546}{44.8} \approx 2 \ atm$.
Therefore,the correct option is $A$.

(A) Using the ideal gas equation $PV = nRT$,where $P = 1 \, atm$,$V = 22.4 \, L$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 30 + 273 = 303 \, K$.
$n = \frac{PV}{RT} = \frac{1 \times 22.4}{0.0821 \times 303}$.
$n = \frac{22.4}{24.8763} \approx 0.9004 \, mol$.
Thus,the number of moles is approximately $0.9 \, mol$.

(B) The ideal gas equation is given by $PV = nRT$.
Given: $n = 0.5 \ mole$,$P = 1 \ atm$,$T = 273 \ K$,and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $V = \frac{nRT}{P} = \frac{0.5 \times 0.0821 \times 273}{1} \approx 11.2 \ L$.
Thus,the correct option is $(B)$.

(B) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given:
$P_1 = P$,$V_1 = 100\, cc$,$T_1 = 0 + 273 = 273\, K$.
Conditions:
$P_2 = 1.5 P = \frac{3}{2} P$.
$T_2 = T_1 + \frac{1}{3} T_1 = \frac{4}{3} T_1 = \frac{4}{3} \times 273\, K$.
Calculation:
$V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 100 \times \frac{P}{\frac{3}{2} P} \times \frac{\frac{4}{3} \times 273}{273}$.
$V_2 = 100 \times \frac{2}{3} \times \frac{4}{3} = 100 \times \frac{8}{9} = \frac{800}{9} \approx 88.89\, cc$.

(B) According to the combined gas law,for a fixed amount of gas,the relationship between pressure $(P)$,volume $(V)$,and temperature $(T)$ is given by $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Rearranging this equation,we get $\frac{P_1 V_1}{P_2 V_2} = \frac{T_1}{T_2}$.
Therefore,the correct option is $B$.

(C) Given: $d_A = 2d_B$ and $M_A = 0.5M_B$ (or $M_B = 2M_A$).
Using the ideal gas equation: $PV = nRT = \frac{m}{M}RT$.
Rearranging for pressure: $P = \frac{m}{V} \cdot \frac{RT}{M} = \frac{dRT}{M}$.
Since temperature $T$ is constant,the ratio of pressures is $\frac{P_A}{P_B} = \frac{d_A}{d_B} \times \frac{M_B}{M_A}$.
Substituting the given values: $\frac{P_A}{P_B} = \frac{2d_B}{d_B} \times \frac{2M_A}{M_A} = 2 \times 2 = 4$.

(C) Step $1$: Calculate the number of moles of each gas.
$n(O_2) = \frac{16 \, g}{32 \, g/mol} = 0.5 \, mol$
$n(H_2) = \frac{3 \, g}{2 \, g/mol} = 1.5 \, mol$
Step $2$: Calculate the total number of moles.
$n_{total} = 0.5 \, mol + 1.5 \, mol = 2.0 \, mol$
Step $3$: Use the ideal gas equation $PV = nRT$.
Given $P = 760 \, mm \, Hg = 1 \, atm$,$T = 0 \, ^oC = 273 \, K$,and $R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
$V = \frac{nRT}{P} = \frac{2.0 \, mol \times 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 273 \, K}{1 \, atm} \approx 44.8 \, L$.
Since $1 \, L = 1000 \, mL$,$44.8 \, L = 44800 \, mL$.

(C) Given: $P_1 = 760 \ mm \ Hg$,$V_1 = 273 \ mL$,$T_1 = 273 \ K$ (at $N.T.P.$).
Final conditions: $P_2 = 600 \ mm \ Hg$,$T_2 = 273 + 273 = 546 \ K$.
Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
$V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{760 \times 273 \times 546}{273 \times 600} = \frac{760 \times 546}{600} = 691.6 \ mL$.

(A) Using the ideal gas law $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is mass,$M$ is molar mass).
Since $V$,$R$,and $M$ are constant,we have $\frac{P_1 V}{m_1 T_1} = \frac{P_2 V}{m_2 T_2}$.
Rearranging for $T_2$: $T_2 = T_1 \times \frac{P_2}{P_1} \times \frac{m_1}{m_2}$.
Given: $P_1 = 1 \ atm$,$T_1 = 300 \ K$,$m_1 = 2 \ g$,$P_2 = 0.75 \ atm$,$m_2 = 1 \ g$.
$T_2 = 300 \times \frac{0.75}{1} \times \frac{2}{1} = 300 \times 1.5 = 450 \ K$.

(B) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Given: $P_1 = 1 \, atm$,$V_1 = 12000 \, L$,$T_1 = 27 + 273 = 300 \, K$
$P_2 = 0.5 \, atm$,$T_2 = -23 + 273 = 250 \, K$
$V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1} = \frac{1 \times 12000 \times 250}{0.5 \times 300} \, L$
$V_2 = \frac{12000 \times 250}{150} = 20000 \, L$
Therefore,the correct option is $(B)$.

(C) From the ideal gas equation,$PV = nRT = (\frac{m}{M})RT$.
Since density $d = \frac{m}{V}$,we have $P = \frac{dRT}{M}$,which implies $d = \frac{PM}{RT}$.
At constant pressure $P$ and molar mass $M$,$d \propto \frac{1}{T}$.
Therefore,$\frac{d_1}{d_2} = \frac{T_2}{T_1}$.
Given $T_1 = 27 + 273 = 300\, K$,$d_1 = d$,and $d_2 = 0.75\, d$.
Substituting the values: $\frac{d}{0.75\, d} = \frac{T_2}{300\, K}$.
$T_2 = \frac{300}{0.75} = 400\, K$.

(C) According to the combined gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given: $P_1 = 740 \ mm$,$V_1 = 100 \ mL$,$T_1 = 27 + 273 = 300 \ K$.
Given: $P_2 = 740 \ mm$,$V_2 = 80 \ mL$.
Since pressure is constant $(P_1 = P_2)$,the equation simplifies to Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
$T_2 = \frac{V_2 \times T_1}{V_1} = \frac{80 \ mL \times 300 \ K}{100 \ mL} = 240 \ K$.
Converting to Celsius: $T_2 = 240 - 273 = -33 \ ^oC$.

(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 or particles.

(B) Dalton's law of partial pressure is applicable only to a mixture of non-reacting gases.
Among the given options,$H_2$ and $Cl_2$ react with each other to form $HCl$ gas $(H_2 + Cl_2 \rightarrow 2HCl)$.
Therefore,Dalton's law of partial pressure will not apply to this mixture.

(C) The ideal gas equation is given by $PV = nRT$.
Rearranging for concentration $C = \frac{n}{V}$,we get $P = CRT$.
Given $P = 1 \, atm$,$C = 1 \, mol \, L^{-1}$,and $R = 0.082 \, L \, atm \, mol^{-1} \, K^{-1}$.
Substituting these values: $1 = 1 \times 0.082 \times T$.
Therefore,$T = \frac{1}{0.082} \approx 12.19 \, K$.
Thus,the condition is $T \approx 12 \, K$.

(D) According to the ideal gas equation,$PV = nRT$,where $n = \frac{w}{M}$.
Substituting $n$,we get $PV = \frac{w}{M} RT$,which implies $V = \frac{wRT}{PM}$.
Since $w$,$R$,$T$,and $P$ are constant for all gases,the volume $V$ is inversely proportional to the molar mass $M$ $(V \propto \frac{1}{M})$.
Therefore,the gas with the highest molar mass will have the least volume.
The molar masses are: $HF = 20 \, g/mol$,$HCl = 36.5 \, g/mol$,$HBr = 81 \, g/mol$,and $HI = 128 \, g/mol$.
Since $HI$ has the highest molar mass,it will occupy the least volume.

(D) According to Boyle's law,at constant temperature,$P_1V_1 = P_2V_2$.
Given: $V_1 = 60 \, mL$,$V_2 = 100 \, mL$,$P_2 = 720 \, mm$.
Substituting the values: $P_1 \times 60 = 720 \times 100$.
$P_1 = \frac{720 \times 100}{60} = 1200 \, mm$.

(D) According to Boyle's law,for a fixed amount of gas at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $V_1 = 100 \ mL$,$P_1 = 720 \ mm$,$V_2 = 84 \ mL$.
Substituting the values into the equation: $720 \times 100 = P_2 \times 84$.
$P_2 = \frac{720 \times 100}{84} = \frac{72000}{84} \approx 857.14 \ mm$.
Therefore,the correct option is $D$.

(B) According to the ideal gas law,$PV = nRT$,where $n$ is the number of moles. Since the number of molecules is directly proportional to the number of moles $(n = \frac{N}{N_A})$,we have $n = \frac{PV}{RT}$.
Given for container $A$: $P_A = 2P_B$,$V_A = 2V_B$,and $T_A = 2T_B$.
The ratio of the number of molecules is equal to the ratio of the number of moles:
$\frac{n_A}{n_B} = \frac{P_A V_A}{R T_A} \times \frac{R T_B}{P_B V_B} = \frac{P_A}{P_B} \times \frac{V_A}{V_B} \times \frac{T_B}{T_A}$.
Substituting the given values:
$\frac{n_A}{n_B} = (2) \times (2) \times (\frac{1}{2}) = 2$.
Thus,the ratio is $2:1$.

(A) Given that the initial pressure at $NTP$ is $P_1 = 76 \, cm$ of $Hg$.
Initial volume $V_1 = 5 \, L$.
The total final volume after connecting the two cylinders is $V_2 = 5 \, L + 30 \, L = 35 \, L$.
According to Boyle's law,at constant temperature,$P_1V_1 = P_2V_2$.
Substituting the values: $76 \times 5 = P_2 \times 35$.
$P_2 = \frac{76 \times 5}{35} = \frac{380}{35} \approx 10.857 \, cm$ of $Hg$.
Rounding to the nearest option,the result is $10.8 \, cm$ of $Hg$.

(C) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given:
$P_1 = 620 \, mm$,$V_1 = 300 \, cc$,$T_1 = 27 + 273 = 300 \, K$.
$P_2 = 640 \, mm$,$T_2 = 47 + 273 = 320 \, K$.
Substituting the values:
$\frac{620 \times 300}{300} = \frac{640 \times V_2}{320}$.
$620 = 2 \times V_2$.
$V_2 = \frac{620}{2} = 310 \, cc$.

(C) Using the ideal gas law equation: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given: $V_1 = 4 \ dm^3$,$P_2 = 2P_1$,and $T_2 = 2T_1$.
Substituting the values: $\frac{P_1 \times 4}{T_1} = \frac{2P_1 \times V_2}{2T_1}$.
Simplifying the equation: $4 = V_2$.
Therefore,the volume remains $4 \ dm^3$.

(C) Given: $P_1 = P, V_1 = V, T_1 = T$
$P_2 = P/2, T_2 = 2T$
Using the ideal gas equation: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Substituting the values: $\frac{PV}{T} = \frac{(P/2) \times V_2}{2T}$
$\frac{PV}{T} = \frac{P \times V_2}{4T}$
$V = \frac{V_2}{4}$
$\therefore V_2 = 4V$
Thus,the volume will be four times the initial volume.

(A) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Initial conditions: $P_1 = P$,$V_1 = V$,$T_1 = 75 + 273 = 348 \ K$
Final conditions: $P_2 = 2P$,$V_2 = 0.15V$ (since volume is reduced to $15\%$ of its initial value)
Substituting the values: $\frac{P \times V}{348} = \frac{2P \times 0.15V}{T_2}$
$T_2 = 348 \times 2 \times 0.15 = 104.4 \ K$
Converting to Celsius: $T_2(^oC) = 104.4 - 273 = -168.6 \ ^oC$
Note: Based on the calculation,the provided options do not match the result. However,if the question meant volume is reduced $BY$ $15\%$ (i.e.,$V_2 = 0.85V$),then $T_2 = 348 \times 2 \times 0.85 = 591.6 \ K = 318.6 \ ^oC$,which rounds to $319 \ ^oC$ (Option $A$).