Ideal gas equation and Related gas laws Questions in English

(200 ML) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
At $STP$,$T_1 = 273 \, K$ and $P_1 = 1 \, bar$.
Given $V_1 = 204.75 \, mL$.
For the final state: $T_2 = 127 + 273 = 400 \, K$ and $P_2 = 1.5 \, bar$.
Substituting the values: $\frac{1 \times 204.75}{273} = \frac{1.5 \times V_2}{400}$.
Solving for $V_2$: $V_2 = \frac{204.75 \times 400}{273 \times 1.5} = 200 \, mL$.

(N/A) $1$. Using the ideal gas law $PV = nRT$,calculate the number of moles $(n)$: $n = \frac{PV}{RT} = \frac{1.5 \times 0.2}{0.0831 \times 400} = 0.009025\, mol$.
$2$. At $STP$ $(273.15\, K, 1\, bar)$,the volume of $1\, mol$ of gas is $22.7\, L$. Thus,$V_{STP} = n \times 22.7 = 0.009025 \times 22.7 = 0.2048\, L = 204.8\, mL$.
$3$. Weight of $N_2$ $(M = 28\, g/mol)$: $w = n \times M = 0.009025 \times 28 = 0.2527\, g$.
$4$. Number of molecules: $N = n \times N_A = 0.009025 \times 6.022 \times 10^{23} = 5.435 \times 10^{21}$ molecules.

(N/A) According to the ideal gas equation,$pV = nRT$.
$\therefore \frac{p}{RT} = \frac{n}{V}$ (Eq.-$i$).
We know that the number of moles $(n) = \frac{\text{mass }(m)}{\text{molar mass }(M)}$ (Eq.-$ii$).
Substituting the value of $n$ from (Eq.-$ii$) into (Eq.-$i$):
$\frac{p}{RT} = \left(\frac{m}{V}\right) \frac{1}{M}$.
Since density $(d) = \frac{\text{mass }(m)}{\text{volume }(V)}$,we can substitute $d$ into the equation:
$\frac{p}{RT} = \frac{d}{M}$.
Rearranging the equation to solve for $d$ or $M$:
$d = \frac{pM}{RT}$ or $M = \frac{dRT}{p}$.
This shows that the molar mass $(M)$ is directly proportional to the density $(d)$ of the gas at constant pressure and temperature.

(A) Given:
Density at sea level $(d_1)$ = $1.5 \ mg \ L^{-1}$
Pressure at sea level $(p_1)$ = $1.0 \ bar$
Pressure at Mount Abu $(p_2)$ = $0.5 \ bar$
Using the formula: $\frac{d_1}{d_2} = \frac{p_1}{p_2}$
$\frac{1.5}{d_2} = \frac{1.0}{0.5}$
$d_2 = \frac{1.5 \times 0.5}{1.0} = 0.75 \ mg \ L^{-1}$
Therefore,the density of the gas on Mount Abu is $0.75 \ mg \ L^{-1}$.

(B) The ideal gas equation is given by $PV = nRT$.
Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M} RT$.
Rearranging for pressure $P$,we get $P = \frac{m}{V} \times \frac{RT}{M} = \frac{dRT}{M}$.
Here,density $d = 0.9 \ g \ L^{-1}$,$R = 8.34 \times 10^{-2} \ bar \ L \ K^{-1} \ mol^{-1}$,$T = 350 \ K$,and the molar mass of neon $M = 20.18 \ g \ mol^{-1}$.
Substituting the values: $P = \frac{0.9 \times 8.34 \times 10^{-2} \times 350}{20.18} \approx 1.309 \ bar$.

(N/A) $1$. Calculate moles of each gas:
$n(H_2) = \frac{10\, g}{2\, g/mol} = 5\, mol$
$n(CO_2) = \frac{22\, g}{44\, g/mol} = 0.5\, mol$
$2$. Convert temperature to Kelvin:
$T = 27 + 273 = 300\, K$
$3$. Calculate partial pressures using $p = \frac{nRT}{V}$:
$p(H_2) = \frac{5 \times 0.08314 \times 300}{2} = 62.355\, bar$
$p(CO_2) = \frac{0.5 \times 0.08314 \times 300}{2} = 6.2355\, bar$
$4$. Calculate total pressure:
$P_{total} = p(H_2) + p(CO_2) = 62.355 + 6.2355 = 68.5905\, bar$

(A) The total number of moles of the gas mixture is $n_{total} = n_{Cl_2} + n_{N_2} + n_{O_2} = 4 + 4 + 2 = 10 \,mol$.
The temperature in Kelvin is $T = 27 + 273 = 300 \,K$.
The volume of the vessel is $V = 5 \,L$.
Using the ideal gas equation $PV = nRT$,the total pressure $P$ is given by $P = \frac{n_{total}RT}{V}$.
Substituting the values: $P = \frac{10 \,mol \times 8.34 \times 10^{-2} \,bar \,L \,mol^{-1} \,K^{-1} \times 300 \,K}{5 \,L}$.
$P = \frac{10 \times 0.0834 \times 300}{5} = \frac{250.2}{5} = 50.04 \,bar$.
Given the options,the closest value is $49.88 \,bar$.

(A) $1$. Initial moles of $H_2$ $(n_1)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{4 \, g}{2 \, g/mol} = 2 \, mol$.
$2$. Final moles of $H_2$ $(n_2)$ remaining in the vessel can be calculated using the ideal gas equation $PV = nRT$:
$n_2 = \frac{PV}{RT} = \frac{50 \, bar \times 0.5 \, L}{8.314 \times 10^{-2} \, L \, bar \, mol^{-1} \, K^{-1} \times 298 \, K} \approx 1.0086 \, mol$.
$3$. Moles of $H_2$ escaped = $n_1 - n_2 = 2 - 1.0086 = 0.9914 \, mol$.
$4$. Number of molecules escaped = $\text{moles} \times N_A = 0.9914 \times 6.022 \times 10^{23} \approx 5.97 \times 10^{23}$ molecules.

(676.6 MM HG) Given:
$p_1 = 760 \, mm \, Hg$
$V_1 = 600 \, mL$
$T_1 = 25 + 273 = 298 \, K$
$V_2 = 640 \, mL$
$T_2 = 10 + 273 = 283 \, K$
Using the combined gas law: $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$
$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}$
$p_2 = \frac{760 \times 600 \times 283}{298 \times 640}$
$p_2 = \frac{129048000}{190720} \approx 676.6 \, mm \, Hg$

(D) Step $1$: Calculate the initial pressure $(P_1)$ using the Ideal Gas Equation $PV = nRT$.
$n_1 = \frac{212\, g}{32\, g/mol} = 6.625\, mol$.
$T = 21 + 273.15 = 294.15\, K$.
$P_1 = \frac{n_1RT}{V} = \frac{6.625 \times 0.083 \times 294.15}{34} = 4.76\, bar$.
Step $2$: Calculate the new amount of gas $(n_2)$ at $P_2 = 1.24\, bar$ using $PV = nRT$ (assuming $V$ and $T$ are constant).
$n_2 = \frac{P_2V}{RT} = \frac{1.24 \times 34}{0.083 \times 294.15} = 1.727\, mol$.
Step $3$: Convert moles to mass.
$Mass = n_2 \times Molar\, mass = 1.727 \times 32 = 55.26\, g$.
(Note: The provided solution $156.74\, g$ appears to be incorrect based on the calculation; the correct remaining mass is $55.26\, g$).

(A) According to Charles's Law,at constant pressure,$V \propto T$ (where $T$ is in Kelvin).
Initial temperature $T_1 = 17 + 273 = 290 \ K$.
Initial volume $V_1 = V$.
Final volume $V_2 = V/2$.
Using the relation $\frac{V_1}{T_1} = \frac{V_2}{T_2}$:
$\frac{V}{290} = \frac{V/2}{T_2}$
$T_2 = \frac{290}{2} = 145 \ K$.
Converting back to Celsius: $145 - 273 = -128^{\circ}C$.

(B) According to Gay-Lussac's Law,for a constant volume,$\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given: $P_1 = 2 \ atm$,$T_1 = 0 \ ^{\circ}C = 273 \ K$,$P_2 = 2.5 \ atm$.
Substituting the values: $\frac{2}{273} = \frac{2.5}{T_2}$.
$T_2 = \frac{2.5 \times 273}{2} = 341.25 \ K$.
Converting to Celsius: $T(^{\circ}C) = 341.25 - 273 = 68.25 \ ^{\circ}C$.

(A) The ideal gas equation is given by $PV = nRT$.
Since $n = \frac{m}{M}$ and density $d = \frac{m}{V}$,we can write $P = \frac{dRT}{M}$.
Given: $d = 8 \, kg \, m^{-3}$,$T = -40 + 273.15 = 233.15 \, K$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$,and $M = 0.0365 \, kg \, mol^{-1}$.
Substituting the values: $P = \frac{8 \times 8.314 \times 233.15}{0.0365} \, Pa$.
$P = \frac{15511.45}{0.0365} \, Pa \approx 4.25 \times 10^{5} \, Pa$.

(B) The density $d$ of an ideal gas is given by the formula $d = \frac{PM}{RT}$.
At $STP$,$P_1 = 1 \, atm = 760 \, torr$,$T_1 = 273.15 \, K$,and $d_1 = 1.43 \, g \, L^{-1}$.
At the new condition,$P_2 = 800 \, torr$,$T_2 = 17 + 273.15 = 290.15 \, K$.
Since $M$ and $R$ are constant,we have $\frac{d_1 T_1}{P_1} = \frac{d_2 T_2}{P_2}$.
Rearranging for $d_2$: $d_2 = d_1 \times \frac{P_2}{P_1} \times \frac{T_1}{T_2}$.
Substituting the values: $d_2 = 1.43 \times \frac{800}{760} \times \frac{273.15}{290.15}$.
$d_2 = 1.43 \times 1.0526 \times 0.9414 \approx 1.417 \, g \, L^{-1}$.

(A) The payload is the difference between the mass of air displaced by the balloon and the mass of $He$ gas inside the balloon.
Using the ideal gas equation $PV = nRT = (m/M)RT$,the mass $m = (PVM)/(RT)$.
Given: $P = 1 \, atm$,$V = 10^6 \, L$,$T = 27 + 273 = 300 \, K$,$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$.
Mass of air displaced $(m_{air})$ = $(1 \times 10^6 \times 28.84) / (0.0821 \times 300) \approx 1170.36 \, kg$.
Mass of $He$ inside $(m_{He})$ = $(1 \times 10^6 \times 4) / (0.0821 \times 300) \approx 162.32 \, kg$.
Payload = $m_{air} - m_{He} = 1170.36 - 162.32 = 1008.04 \, kg \approx 1010 \, kg$.

(C) Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Given: $P_1 = 5 \, bar$,$V_1 = V$,$T_1 = 0 + 273 = 273 \, K$
$T_2 = 546 + 273 = 819 \, K$,$V_2 = \frac{V}{3}$
Substituting the values: $\frac{5 \times V}{273} = \frac{P_2 \times (V/3)}{819}$
$P_2 = \frac{5 \times 819 \times 3}{273} = 5 \times 3 \times 3 = 45 \, bar$

(A) According to the ideal gas law,$PV = nRT$. At standard temperature and pressure $(STP)$,defined as $273.15 \ K$ and $1 \ atm$,the molar volume of an ideal gas is approximately $22.4 \ L \ mol^{-1}$.
Since both nitrogen $(N_2)$ and argon $(Ar)$ behave nearly as ideal gases under these conditions,their molar volumes are approximately equal to $22.4 \ L \ mol^{-1}$ each.

(N/A) The unit of $R$ depends upon the units in which $p$,$V$,and $T$ are measured,as $R = \frac{p \cdot V}{n \cdot T}$.
If pressure is measured in $Pa$ (Pascal),volume per mole is measured in $m^3 \cdot mol^{-1}$,and temperature is measured in $K$ (Kelvin),then the units of $R$ are $Pa \cdot m^3 \cdot K^{-1} \cdot mol^{-1}$,which is equivalent to $J \cdot mol^{-1} \cdot K^{-1}$.
Since $J$ (Joule) defines the unit of work done,$R$ represents the work done by the gas per mole per Kelvin.

(B) According to Boyle's law,$p \propto 1/V$ at constant temperature.
$(i)$ The graph of $p$ vs $V$ is a rectangular hyperbola,which is a curve.
$(ii)$ The graph of $p$ vs $1/V$ is a straight line passing through the origin because $p = k(1/V)$,where $k$ is a constant.

(B) According to Boyle's law,$PV = K$ (where $K$ is a constant).
The value of the constant $K$ depends on the temperature $T$.
As the temperature increases,the value of $K$ increases,causing the isotherm to shift away from the origin.
Therefore,for different temperatures $T_3 > T_2 > T_1$,the slopes of the isotherms increase in the same order.

(A) According to Boyle's Law,at constant temperature and amount of gas,$P_1V_1 = P_2V_2$.
Given that the pressure is halved,$P_2 = P_1 / 2$.
Substituting this into the equation: $P_1V_1 = (P_1 / 2) \times V_2$.
Therefore,$V_2 = 2V_1$.
Thus,the volume becomes double the initial volume.

(A) Boyle's Law quantitatively proves that,"Gases are highly compressible."
This is because when a given mass of gas is compressed,the same number of molecules occupies a smaller space,meaning that gases become denser at higher pressures.

(A) According to the ideal gas equation,$PV = nRT$ and $d = \frac{PM}{RT}$,where $d$ is density,$P$ is pressure,$M$ is molar mass,$R$ is the gas constant,and $T$ is temperature.
At constant temperature $(T)$ and constant molar mass $(M)$,the density $(d)$ is directly proportional to the pressure $(P)$: $d \propto P$.
Therefore,increasing the pressure of the gas will result in an increase in its density.

(A) According to Boyle's Law,for a fixed amount of gas at a constant temperature,$P_1V_1 = P_2V_2$.
Given:
$P_1 = 2.0 \times 10^4 \, Pa$
$V_1 = 112.0 \times 10^{-3} \, m^3$
$P_2 = 4.0 \times 10^4 \, Pa$
Substituting the values:
$V_2 = \frac{P_1V_1}{P_2} = \frac{2.0 \times 10^4 \, Pa \times 112.0 \times 10^{-3} \, m^3}{4.0 \times 10^4 \, Pa}$
$V_2 = 0.5 \times 112.0 \times 10^{-3} \, m^3 = 56.0 \times 10^{-3} \, m^3$.

(N/A) Boyle's Law states that at constant temperature,the pressure of a fixed amount of gas is inversely proportional to its volume ($p \propto 1/V$ or $pV = k$).
$(i)$ Plot of $V$ vs $p$: Shows an isothermal curve (hyperbola) representing $V \propto 1/p$.
(ii) Plot of $V$ vs $1/p$: Shows a straight line passing through the origin,confirming $V = k(1/p)$.
(iii) Plot of $p$ vs $1/V$: Shows a straight line passing through the origin,confirming $p = k(1/V)$.
(iv) Plot of $\log_{10} p$ vs $\log_{10} V$: Since $pV = k$,taking log on both sides gives $\log p + \log V = \log k$,or $\log p = -\log V + \log k$. This is a straight line with a slope of $-1$.
$(v)$ Plot of $p$ vs $V$ at different temperatures: Shows isotherms where $p$ decreases as $V$ increases. Higher temperature curves are further from the origin $(T_3 > T_2 > T_1)$.
(vi) Plot of $p$ vs $1/V$ at different temperatures: Shows straight lines passing through the origin. The slope $(k = nRT/V)$ increases with temperature,so $T_3 > T_2 > T_1$ (Note: The image label $T_3 < T_2 < T_1$ in the provided diagram is physically incorrect for the slope of $p$ vs $1/V$ at constant $n$).

(B) The plot of volume $(V)$ versus temperature $(T)$ at constant pressure for a fixed amount of gas is a straight line passing through the origin.
$\therefore V \propto T$ (at constant $p$ and $n$)
This represents Charles's Law,which states that the volume of a fixed mass of gas is directly proportional to its absolute temperature at constant pressure.

(N/A) The combined gas law is given by $\frac{p_{1}V_{1}}{T_{1}} = \frac{p_{2}V_{2}}{T_{2}}$.
The ideal gas equation is given by $pV = nRT$,where $p$ is pressure,$V$ is volume,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature.

(A) The ideal gas equation is also known as the $Equation \ of \ state$.
This is because the equation $pV = nRT$ describes the relationship between the four variables $p, V, n,$ and $T$ that define the state of a gas.

(N/A) The value of $R$ is $0.08314 \ bar \ L \ mol^{-1} \ K^{-1}$.
From the ideal gas equation,$pV = nRT$,we have $R = \frac{pV}{nT}$.
At standard temperature and pressure $(STP)$,$1 \ mol$ of an ideal gas occupies $22.71 \ L$ at $1 \ bar$ pressure and $273.15 \ K$.
Substituting these values: $R = \frac{(1 \ bar)(22.71 \ L)}{(1 \ mol)(273.15 \ K)}$.
$R = 0.083141 \ bar \ L \ mol^{-1} \ K^{-1}$.
Thus,$R \approx 8.314 \times 10^{-2} \ bar \ L \ mol^{-1} \ K^{-1}$.

(N/A) The molar volume of a gas is defined as the volume occupied by $1 \, \text{mol}$ of an ideal gas at $STP$.
At $STP$ ($273.15 \, K$ temperature and $1 \, \text{bar}$ pressure),the volume of $1 \, \text{mol}$ of an ideal gas is $22.710981 \, L \, \text{mol}^{-1}$,which is its molar volume.

(A) The universal gas constant $R$ has a value of $8.314 \ J \ K^{-1} \ mol^{-1}$ in $SI$ units.
It can also be expressed as $8.314 \ Pa \ m^3 \ K^{-1} \ mol^{-1}$ since $1 \ J = 1 \ Pa \ m^3$.

(N/A) $(i)$ $R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1}$
$(ii)$ $R = 8.20 \times 10^{-2} \, L \, atm \, K^{-1} \, mol^{-1}$
The value of $R$ with full precision is:
$(iii)$ $R = 8.20578 \times 10^{-2} \, L \, atm \, K^{-1} \, mol^{-1}$

(N/A) The molar volume of an ideal gas at $STP$ $(0 \, ^{\circ}C, 1 \, bar)$ is $22.71098 \, L \, mol^{-1}$.
Previously,at $STP$ $(0 \, ^{\circ}C, 1 \, atm)$,the molar volume was taken as $22.413996 \, L \, mol^{-1}$.
The molar volume of an ideal gas at $SATP$ $(25 \, ^{\circ}C, 1 \, bar)$ is $24.787 \, L \, mol^{-1}$.

(B) According to the ideal gas equation,$pV = nRT$.
Rearranging for volume,we get $V = \frac{nRT}{p}$.
Since $n$,$R$,$T$,and $p$ are constant,the volume $V$ is the same for all gases regardless of their chemical nature.

(N/A) For a gas,the relationship between its molecular mass $(M)$,volume $(V)$,and density $(d)$ is given by the ideal gas equation $PV = nRT$.
Since $n = \frac{m}{M}$,where $m$ is the mass of the gas,we have $PV = \frac{m}{M}RT$.
Rearranging for density $(d = \frac{m}{V})$,we get $d = \frac{PM}{RT}$.
Thus,at constant temperature and pressure,the density of a gas is directly proportional to its molecular mass $(d \propto M)$.

(N/A) Avogadro's Law: Equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. Mathematically,$V \propto n$ (at constant $T$ and $p$).
Gay-Lussac's Law: At a constant volume,the pressure of a fixed amount of a gas is directly proportional to its absolute temperature. Mathematically,$p \propto T$ (at constant $n$ and $V$).

(B) Charles's Law states that for a fixed mass of an ideal gas,the volume is directly proportional to its absolute temperature at constant pressure.
Mathematically,$V \propto T$ (at constant $P$ and $n$).
This can be expressed as $\frac{V}{T} = k$,where $k$ is a constant.

(A) According to Charles's Law,for a fixed mass of gas at constant pressure,the volume $V_t$ at temperature $t\,^{\circ}C$ is given by $V_t = V_0 \left(1 + \frac{t}{273.15}\right)$.
Given that the temperature is increased by $1\,^{\circ}C$,we substitute $t = 1$ into the equation.
Therefore,the new volume $V_t = V_0 \left(1 + \frac{1}{273.15}\right) = V_0 + \frac{1}{273.15} V_0$.

(A) According to Charles's Law,$V \propto T$ at constant pressure and amount of gas.
For a temperature change of $1\,^oC$,the volume changes by a factor of $\frac{1}{273.15}$ of the volume at $0\,^oC$.
Therefore,the volume increases by $\frac{1}{273.15}$ of its original volume.

(N/A) The value of the gas constant $R$ depends on the units of the variables $p$,$V$,$n$,and $T$ used in the ideal gas equation $pV = nRT$.
The unit of $R$ is derived based on the units of $p$,$V$,$n$,and $T$ in the equation $pV = nRT$.

(A) The ideal gas equation is given by $PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$. Rearranging for pressure $P$,we get $P = \frac{m}{V} \times \frac{RT}{M} = \frac{dRT}{M}$.
Given: Density $d = 0.9 \ g \ L^{-1}$,Temperature $T = 350 \ K$,Molar mass of Neon $M = 20.18 \ g \ mol^{-1}$,Gas constant $R = 0.08314 \ L \ bar \ K^{-1} \ mol^{-1}$.
Substituting the values: $P = \frac{0.9 \times 0.08314 \times 350}{20.18} \approx 1.297 \ bar \approx 1.303 \ bar$ (using standard approximations).

(A) From the ideal gas equation,$PV = nRT$.
Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M} RT$.
Rearranging the terms,$\frac{m}{V} = \frac{PM}{RT}$.
Since density $d = \frac{m}{V}$,we get $d = \frac{PM}{RT}$.
Therefore,the ratio $\frac{d}{M} = \frac{P}{RT}$.

(C) The density $(d)$ of an ideal gas is given by the formula $d = \frac{MP}{RT}$.
From this relation,it is clear that density is directly proportional to pressure $(P)$ and inversely proportional to temperature $(T)$.
Therefore,to achieve maximum density,the gas must be at high pressure and low temperature.

(A) From the ideal gas equation,$PV = nRT$,we know $PV = (\frac{m}{M})RT$,which gives $P = (\frac{d}{M})RT$,where $d$ is the density.
Therefore,$P = \frac{dRT}{M}$.
Given $\frac{d_1}{d_2} = \frac{1}{2}$ and $\frac{T_1}{T_2} = \frac{2}{1}$.
Taking the ratio of pressures: $\frac{P_1}{P_2} = (\frac{d_1}{d_2}) \times (\frac{T_1}{T_2}) = \frac{1}{2} \times \frac{2}{1} = \frac{1}{1}$.
Thus,the ratio of their pressures is $1:1$.

(B) The ideal gas equation is given by $PV = nRT$.
From the first law of thermodynamics,work done $W$ by an ideal gas during expansion is related to the pressure-volume product.
Since $PV$ has the dimensions of energy or work,we can express the gas constant $R$ as $R = \frac{PV}{nT}$.
Substituting work $W$ for $PV$,we get $R = \frac{W}{nT}$.
Thus,$R$ represents the work done per mole per unit temperature change.

(N/A) The universal gas constant $R$ is defined by the ideal gas equation $PV = nRT$,which implies $R = \frac{PV}{nT}$. Its value depends on the units used for pressure,volume,and temperature. The following table summarizes the common values of $R$:

Condition	Value and Unit of $R$
$(i)$ Pressure in $atm$,Volume in $L$	$0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$
$(ii)$ Pressure in $bar$,Volume in $L$	$0.08314 \ L \ bar \ K^{-1} \ mol^{-1}$
$(iii)$ $SI$ Units (Pressure in $Pa$,Volume in $m^3$)	$8.314 \ J \ K^{-1} \ mol^{-1}$
$(iv)$ Energy in $calories$	$1.987 \ cal \ K^{-1} \ mol^{-1}$

(N/A) These laws provide precise statements about experimental facts and tell us how a particular system behaves under different conditions.
Gas laws are useful for making predictions,such as how the pressure of a gas increases when it is compressed.

(B) According to Boyle's Law,for an ideal gas at constant temperature,$pV = \text{constant}$.
Therefore,the product $pV$ remains independent of the pressure $p$.
As a result,the graph of $pV$ versus $p$ is a straight line parallel to the $p$-axis.

(B) The compressibility factor $Z$ is defined as $Z = \frac{pV}{nRT}$.
If $pV = nRT$,then $Z = 1$.
This condition is satisfied by ideal gases under all conditions of temperature and pressure.

(B) gas behaves ideally at low pressure and high temperature.
Under these conditions,the intermolecular forces of attraction become negligible,and the volume occupied by the gas molecules becomes negligible compared to the total volume of the gas.
Consequently,the van der Waals equation $(p + \frac{an^2}{V^2})(V - nb) = nRT$ reduces to the ideal gas equation $pV = nRT$ because the pressure correction term $\frac{an^2}{V^2} \approx 0$ and the volume correction term $nb \approx 0$.