Ideal gas equation and Related gas laws Questions in English

(B) From the ideal gas equation,$pV = nRT = \frac{w}{M} RT$.
For the first graph ($p$ vs $T$ at constant $V$): $p = (\frac{wR}{MV}) T$. The slope $m = \frac{wR}{MV}$. Since $V$ is constant,$m \propto w$. As the slope of $m_2$ is greater than $m_1$,$m_2 > m_1$.
For the second graph ($p$ vs $T$ at constant $V$): Similarly,the slope $m$ is proportional to the mass $w$ of the gas. Thus,$m_2 > m_1$.

(A-D) The correct matches are as follows:
$(1)$ Boyle's Law: $p \propto \frac{1}{V}$ $(C)$
$(2)$ Charles's Law: $V \propto T$ or $V = k_2T$ $(E)$
$(3)$ Gay-Lussac's Law: $p \propto T$ $(A)$
$(4)$ Avogadro's Law: $V \propto n$ $(B)$
$(5)$ Ideal Gas Equation: $pV = nRT$ $(D)$
Therefore,the correct sequence is $(1-C, 2-E, 3-A, 4-B, 5-D)$.

(A) The graph of $p$ vs $T$ at constant volume is called an isochore.
$(B)$ The graph of $p$ vs $V$ at constant temperature is called an isotherm.
$(C)$ The graph of $V$ vs $T$ at constant pressure is called an isobar.

(A) Boyle's Law: $p \propto \frac{1}{V}$ at constant $T$ and $n$.
$(B)$ Charles's Law: $V \propto T$ at constant $p$ and $n$.
$(C)$ Dalton's Law: $P_{total} = p_1 + p_2 + p_3 + \dots$ at constant $T$ and $V$.
$(D)$ Avogadro's Law: $V \propto n$ at constant $T$ and $p$.

(A-2, B-3, C-1) For an ideal gas,$PV = nRT$:
$(A)$ $P$ vs $V$ at constant $T$ gives a rectangular hyperbola. However,if we plot $P$ vs $1/V$,we get a straight line passing through the origin. Thus,$(A-2)$.
$(B)$ $V$ vs $T$ at constant $P$ gives a straight line passing through the origin $(V \propto T)$. Thus,$(B-3)$.
$(C)$ $P$ vs $T$ at constant $V$ gives a straight line passing through the origin $(P \propto T)$. Thus,$(C-1)$.
Therefore,the correct match is $(A-2, B-3, C-1)$.

(A) The ideal gas equation is $PV = nRT$.
Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$.
Rearranging gives $PM = \frac{m}{V}RT = dRT$,where $d$ is density.
Thus,$d = \frac{PM}{RT}$.
For a constant pressure $P$,$d \propto \frac{1}{T}$.
Graph $(I)$ shows $d$ vs $T$ as an inverse relationship,which is correct.
Graph $(II)$ shows $d$ vs $T$ as a direct linear relationship,which is incorrect.
Graph $(III)$ shows $d$ vs $\frac{1}{T}$ as a direct linear relationship,which is correct.
Graph $(IV)$ shows $d$ vs $P$ as a direct linear relationship (at constant $T$),which is correct.
Therefore,graph $(II)$ is not correct.

(C) According to the ideal gas equation,$PV = nRT$. Since $n$,$R$,and $T$ are constant,$PV = \text{constant}$,so $P_1V_1 = P_2V_2$.
Volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Given $r_1 = 3 \ cm$,$P_1 = 48 \times 10^{-3} \ bar$,$r_2 = 12 \ cm$.
$P_1 \times \frac{4}{3} \pi (r_1)^3 = P_2 \times \frac{4}{3} \pi (r_2)^3$.
$P_2 = P_1 \times (\frac{r_1}{r_2})^3 = 48 \times 10^{-3} \times (\frac{3}{12})^3$.
$P_2 = 48 \times 10^{-3} \times (\frac{1}{4})^3 = 48 \times 10^{-3} \times \frac{1}{64}$.
$P_2 = \frac{48}{64} \times 10^{-3} = 0.75 \times 10^{-3} \ bar$.
$P_2 = 750 \times 10^{-6} \ bar$.

(B) According to Boyle's Law,for a fixed amount of gas at a constant temperature,the product of pressure and volume is constant: $P_{1}V_{1} = P_{2}V_{2}$.
Given:
$P_{1} = 1 \; bar$
$V_{1} = 600 \; dm^{3}$
$V_{2} = 150 \; dm^{3}$
Substituting the values into the equation:
$(1 \; bar) \times (600 \; dm^{3}) = P_{2} \times (150 \; dm^{3})$
$P_{2} = \frac{600 \; bar \cdot dm^{3}}{150 \; dm^{3}}$
$P_{2} = 4 \; bar$.

(D) First,calculate the number of moles of each gas:
$n_{CH_4} = \frac{6.4 \ g}{16 \ g \ mol^{-1}} = 0.4 \ mol$
$n_{CO_2} = \frac{8.8 \ g}{44 \ g \ mol^{-1}} = 0.2 \ mol$
Total moles,$n = 0.4 + 0.2 = 0.6 \ mol$
Convert volume to $m^3$:
$V = 10 \ L = 10 \times 10^{-3} \ m^3 = 0.01 \ m^3$
Convert temperature to Kelvin:
$T = 27 + 273 = 300 \ K$
Using the ideal gas equation $PV = nRT$:
$P = \frac{nRT}{V} = \frac{0.6 \ mol \times 8.314 \ J \ mol^{-1} K^{-1} \times 300 \ K}{0.01 \ m^3}$
$P = 149652 \ Pa = 149.652 \ kPa$
Rounding to the nearest integer,we get $150 \ kPa$.

(D) Assuming the volume of the tyre remains constant, according to Gay-Lussac's Law, $P \propto T$ (where $T$ is in Kelvin).
Given: $P_1 = 35 \ psi$, $T_1 = 27 + 273 = 300 \ K$, $P_2 = 40 \ psi$.
Using the formula $\frac{P_2}{P_1} = \frac{T_2}{T_1}$:
$\frac{40}{35} = \frac{T_2}{300}$
$T_2 = \frac{40 \times 300}{35} = \frac{12000}{35} \approx 342.86 \ K$.
Converting back to Celsius: $T_2(^{\circ} C) = 342.86 - 273 = 69.86^{\circ} C$.
Rounding to the nearest integer, we get $70^{\circ} C$.

(A) The molar mass of acetylene $(C_2H_2)$ is $26 \, g \, mol^{-1}$.
Number of moles $(n)$ $= \frac{4.75 \, g}{26 \, g \, mol^{-1}} \approx 0.1827 \, mol$.
Temperature $(T)$ $= 50 + 273 = 323 \, K$.
Pressure $(P)$ $= \frac{740}{760} \, atm \approx 0.9737 \, atm$.
Using the ideal gas equation $PV = nRT$,we get $V = \frac{nRT}{P}$.
$V = \frac{0.1827 \times 0.0826 \times 323}{0.9737} \approx 5.0059 \, L$.
Rounding off to the nearest integer,we get $5 \, L$.

(A) For an ideal gas,the ideal gas equation is $PV = nRT$.
At constant temperature $(T)$ and for a fixed amount of gas $(n)$,the product $nRT$ is a constant.
Therefore,$PV = \text{constant}$.
This means that the value of $PV$ does not change with a change in pressure $(P)$.
Thus,the plot of $PV$ versus $P$ is a straight line parallel to the $P$-axis.

(D) According to the ideal gas equation,$PV = nRT$.
For a fixed amount of gas,$P = \frac{nRT}{V}$.
At a constant volume $V$,$P \propto T$.
This means that for a given volume,the pressure is higher at a higher temperature.
Therefore,if $T_1 > T_2 > T_3$,then the corresponding pressures at a constant volume will be $P_1 > P_2 > P_3$.
Looking at the graphs,the curve for the highest temperature $(600 \ K)$ must be the uppermost curve,followed by $400 \ K$ and then $200 \ K$ at the bottom.
This matches the representation where the curve for $600 \ K$ is above the curve for $400 \ K$,which is above the curve for $200 \ K$.

(C) The total pressure is calculated using the ideal gas equation: $PV = n_{total}RT$.
First,calculate the number of moles of each gas:
$n_{O_2} = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$
$n_{H_2} = \frac{2 \ g}{2 \ g/mol} = 1.0 \ mol$
Total moles $n_{total} = 0.125 + 1.0 = 1.125 \ mol$.
Using $PV = nRT$ with $V = 1 \ L$,$R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$,and $T = 273 \ K$:
$P \times 1 = 1.125 \times 0.082 \times 273$
$P = 25.18 \ atm$.

(C) Given:
Volume $V = 4.00 \times 10^{3} \ m^{3} = 4.00 \times 10^{3} \times 10^{3} \ L = 4.00 \times 10^{6} \ L$
Pressure $P = 1.0 \ atm$
Temperature $T = 300 \ K$
Gas constant $R = 0.083 \ L \ atm \ K^{-1} \ mol^{-1}$
Using the ideal gas equation $PV = nRT$,where $n = \frac{mass (m)}{Molar \ mass (M)}$:
$n = \frac{PV}{RT} = \frac{1.0 \times 4.00 \times 10^{6}}{0.083 \times 300} \ mol$
$n = \frac{4.00 \times 10^{6}}{24.9} \ mol \approx 1.6064 \times 10^{5} \ mol$
Molar mass of $CH_{4} = 12 + 4 \times 1 = 16 \ g \ mol^{-1}$
Mass $m = n \times M = 1.6064 \times 10^{5} \times 16 \ g$
$m = 25.7024 \times 10^{5} \ g$
Comparing with $X \times 10^{5} \ g$,we get $X = 25.7024$.
Rounding to the nearest integer,$X = 26$.

(C) According to Gay-Lussac's Law,for a fixed volume of gas,pressure is directly proportional to temperature: $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$.
Given: $P_{1} = 300 \ kPa = 300 \times 10^{3} \ Pa$,$T_{1} = 27 + 273 = 300 \ K$,$P_{2} = 1.2 \times 10^{6} \ Pa$.
Substituting the values: $\frac{300 \times 10^{3}}{300} = \frac{1.2 \times 10^{6}}{T_{2}}$.
$1000 = \frac{1.2 \times 10^{6}}{T_{2}} \Rightarrow T_{2} = \frac{1.2 \times 10^{6}}{1000} = 1200 \ K$.
Converting to Celsius: $T(^{\circ} C) = 1200 - 273 = 927^{\circ} C$.

(C) Given:
Volume $V = 10.0 \, L$
Mass of $O_2$ $(W)$ = $64 \, g$
Molar mass of $O_2$ $(M)$ = $32 \, g \, mol^{-1}$
Temperature $T = 27^{\circ} C = 27 + 273 = 300 \, K$
Gas constant $R = 0.0831 \, L \, bar \, K^{-1} \, mol^{-1}$
Step $1$: Calculate the number of moles $(n)$:
$n = \frac{W}{M} = \frac{64 \, g}{32 \, g \, mol^{-1}} = 2 \, mol$
Step $2$: Use the ideal gas equation $PV = nRT$ to find pressure $(P)$:
$P = \frac{nRT}{V}$
$P = \frac{2 \, mol \times 0.0831 \, L \, bar \, K^{-1} \, mol^{-1} \times 300 \, K}{10.0 \, L}$
$P = \frac{49.86}{10} \, bar = 4.986 \, bar \approx 4.9 \, bar$

(D) Given: Ideal gas $A$ and $H_2$ gas are at the same pressure $(P)$ and volume $(V)$.
From the ideal gas equation,$PV = nRT$,we have $n = \frac{PV}{RT}$.
Since $P$ and $V$ are the same for both gases,$n_A T_A = n_{H_2} T_{H_2}$.
Here,$n_A = \frac{3.0}{M_A}$ and $n_{H_2} = \frac{0.2}{2.0} = 0.1 \ mol$.
Substituting the values: $\frac{3.0}{M_A} \times 300 = 0.1 \times 200$.
$\frac{900}{M_A} = 20$.
$M_A = \frac{900}{20} = 45 \ g \ mol^{-1}$.

(A) Since the tank is rigid, the volume $V$ remains constant. According to Gay-Lussac's Law, for a fixed amount of gas at constant volume, the pressure is directly proportional to the absolute temperature: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given:
$P_1 = 30 \ atm$
$T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$
$T_2 = 45^{\circ} C = 45 + 273 = 318 \ K$
Substituting the values into the equation:
$\frac{30}{300} = \frac{P_2}{318}$
$0.1 = \frac{P_2}{318}$
$P_2 = 0.1 \times 318 = 31.8 \ atm$.
Rounding to the nearest integer, we get $P_2 = 32 \ atm$.

(D) Mass of liquid $= 135.0 - 40.0 = 95.0 \ g$
Volume of the vessel $= \frac{\text{mass of liquid}}{\text{density of liquid}} = \frac{95.0 \ g}{0.95 \ g \ mL^{-1}} = 100 \ mL = 0.1 \ L$
Mass of ideal gas $= 40.5 - 40.0 = 0.5 \ g$
Using the ideal gas equation $PV = nRT$,where $n = \frac{w}{M}$:
$PV = \frac{w}{M} RT$
$0.82 \ atm \times 0.1 \ L = \frac{0.5 \ g}{M} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 250 \ K$
$0.082 = \frac{0.5 \times 0.082 \times 250}{M}$
$M = \frac{0.5 \times 0.082 \times 250}{0.082} = 0.5 \times 250 = 125 \ g \ mol^{-1}$

(B) For an ideal gas,the ideal gas equation is $pV = nRT$.
Since $n = \frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have $pV = \frac{m}{M} RT$.
Rearranging this gives $p = \left( \frac{RT}{M} \right) \left( \frac{m}{V} \right)$.
Since density $d = \frac{m}{V}$,the equation becomes $p = \left( \frac{RT}{M} \right) d$.
This is in the form of a straight line $y = mx$,where the slope is $\frac{RT}{M}$.
Since the slope is directly proportional to temperature $(T)$,a higher temperature will result in a steeper slope.
Given $T_3 > T_2 > T_1$,the slope order will be $T_3 > T_2 > T_1$.
Therefore,the plot with the steepest line for $T_3$ and the least steep for $T_1$ is correct.

(B) The ideal gas equation is $PV = nRT$,where $n = \frac{m}{M}$.
Given: $P = 1.5 \, bar$,$V = 416 \, L$,$m = 100 \, g$,$T = 27 + 273 = 300 \, K$,and $R = 0.083 \, L \, bar \, K^{-1} \, mol^{-1}$.
Substituting the values: $1.5 \times 416 = \frac{100}{M} \times 0.083 \times 300$.
$624 = \frac{2490}{M}$.
$M = \frac{2490}{624} \approx 3.99 \, g \, mol^{-1}$.
Rounding to the nearest integer,we get $M = 4 \, g \, mol^{-1}$.

(D) The number of moles of water $(n)$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.90 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$.
The temperature $(T)$ is $27 + 273 = 300 \ K$.
The pressure $(P)$ in $atm$ is $\frac{32.0 \ Torr}{760 \ Torr \ atm^{-1}} = \frac{32}{760} \ atm$.
Using the ideal gas equation $PV = nRT$,we find the volume $(V)$:
$V = \frac{nRT}{P} = \frac{0.05 \times 0.082 \times 300}{32 / 760} = \frac{0.05 \times 0.082 \times 300 \times 760}{32} = 29.21 \ L$.
The nearest integer is $29$.

(B) Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$ and $d = \frac{m}{V}$,we get $P = \frac{dRT}{M}$.
Rearranging for molar mass: $M = \frac{dRT}{P}$.
Given values:
$d = 0.46 \, g \, L^{-1}$
$R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1}$
$T = 257 + 273 = 530 \, K$
$P = \frac{100}{760} \, atm$
Substituting these values:
$M = \frac{0.46 \times 0.082 \times 530 \times 760}{100}$
$M = 151.93 \, g \, mol^{-1}$.
Rounding to the nearest integer,we get $152$.

(C) $1$. Calculate the number of moles of propane $(C_{3}H_{8})$:
Molar mass of $C_{3}H_{8} = (3 \times 12) + (8 \times 1) = 44 \, g \, mol^{-1}$.
Moles $(n)$ = $\frac{11 \, g}{44 \, g \, mol^{-1}} = 0.25 \, mol$.
$2$. Use the ideal gas equation $PV = nRT$:
Given $P = 2 \, MPa = 2 \times 10^{6} \, Pa$,$V = 2 \, dm^{3} = 2 \times 10^{-3} \, m^{3}$,$n = 0.25 \, mol$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$.
$2 \times 10^{6} \times 2 \times 10^{-3} = 0.25 \times 8.3 \times T$.
$4000 = 2.075 \times T$.
$T = \frac{4000}{2.075} \approx 1927.71 \, K$.
$3$. Convert temperature to Celsius:
$T(^{\circ}C) = T(K) - 273.15 = 1927.71 - 273.15 = 1654.56 \, ^{\circ}C$.
Rounding to the nearest integer,we get $1655 \, ^{\circ}C$.

(A) According to the ideal gas equation,$pV = nRT$.
For $1 \ mole$ of gas,$n = 1$,so $pV = RT$ or $p = \frac{RT}{V}$.
Taking $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
For point $X$: $V = 50 \ L, T = 200 \ K$.
$p_X = \frac{0.0821 \times 200}{50} = 0.3284 \ atm \approx 0.328 \ atm$.
For point $Y$: $V = 50 \ L, T = 500 \ K$.
$p_Y = \frac{0.0821 \times 500}{50} = 0.821 \ atm \approx 0.820 \ atm$.
For point $Z$: $V = 20 \ L, T = 200 \ K$.
$p_Z = \frac{0.0821 \times 200}{20} = 0.821 \ atm \approx 0.820 \ atm$.
Thus,the pressures at $X, Y, Z$ are $0.328 \ atm, 0.820 \ atm, 0.820 \ atm$ respectively.

(B) The density of the gas is $\rho = 5 \ mg \ cm^{-3} = 5 \ g \ L^{-1}$.
Using the ideal gas equation $pV = nRT$,where $n = \frac{w}{M}$,we have $p = \frac{wRT}{MV} = \frac{\rho RT}{M}$.
Rearranging for the molar mass $M$ of the vapor cluster: $M = \frac{\rho RT}{p}$.
Given $\rho = 5 \ g \ L^{-1}$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,and $p = 1 \ atm$:
$M = \frac{5 \times 0.0821 \times 300}{1} = 123.15 \ g \ mol^{-1}$.
The molar mass of a single acetic acid molecule $(CH_3COOH)$ is $60 \ g \ mol^{-1}$.
The number of molecules in the cluster is $n = \frac{M_{cluster}}{M_{monomer}} = \frac{123.15}{60} \approx 2.05$.
Thus,the number of molecules is closest to $2$.

(D) Using the ideal gas equation $pV = nRT$,where $p = 2 \, atm$,$V = 2.24 \, L$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 298 \, K$.
$n = \frac{pV}{RT} = \frac{2 \times 2.24}{0.0821 \times 298} \approx 0.183 \, mol$.
The molar mass of $N_2$ is $28 \, g/mol$.
The maximum mass of $N_2$ is $m = n \times M = 0.183 \times 28 \approx 5.124 \, g$.
Since the container explodes at $2 \, atm$,the safe amount must be slightly less than $5.124 \, g$. Among the given options,$4.2 \, g$ is the closest value that ensures safety.

(B) According to the ideal gas equation,$p V = n R T$.
Rearranging for the volume as a function of temperature,we get $V = (\frac{n R}{p}) T$.
The slope of the plot of $V$ versus $T$ is given by $\text{slope} = \frac{n R}{p}$.
Given: $\text{slope} = 0.328 \, L \, K^{-1}$,$n = 2 \, mol$,and $R = 0.0821 \, L \, atm \, mol^{-1} \, K^{-1}$.
Substituting the values: $0.328 = \frac{2 \times 0.0821}{p}$.
Solving for $p$: $p = \frac{2 \times 0.0821}{0.328} = \frac{0.1642}{0.328} = 0.5 \, atm$.

(C) According to the ideal gas equation,$pV = nRT$.
Since the vessel is closed,the volume $(V)$ and the number of moles $(n)$ remain constant.
Therefore,$p \propto T$,which implies $\frac{p_1}{T_1} = \frac{p_2}{T_2}$.
Given: $p_1 = 1 \, atm$,$T_1 = 27^{\circ} C = 27 + 273 = 300 \, K$,and $T_2 = 327^{\circ} C = 327 + 273 = 600 \, K$.
Substituting the values: $\frac{1}{300} = \frac{p_2}{600}$.
$p_2 = \frac{600}{300} = 2 \, atm$.

(D) The ideal gas equation is $pV = nRT$.
$(A)$ At constant $T$,$p = \frac{nRT}{V}$,so $p \propto \frac{1}{V}$. This is a rectangular hyperbola,which is correct for an ideal gas.
$(B)$ The graph of $p$ versus $1/V$ at constant $p$ is incorrect because $p$ cannot be constant while $1/V$ varies in an ideal gas equation unless $T$ also changes proportionally. Furthermore,for an ideal gas at constant $T$,$p$ is directly proportional to $1/V$,not constant.
$(C)$ The graph of $pV$ versus $T$ should be a straight line passing through the origin with a slope of $nR$,not a constant horizontal line. Thus,this graph is also incorrect.
Therefore,both graphs $B$ and $C$ do not represent the behaviour of an ideal gas.

(C) According to Boyle's law,the pressure $(p)$ of a given mass of an ideal gas is inversely proportional to its volume $(V)$ at a constant temperature.
$\therefore p \propto \frac{1}{V}$ or $pV = \text{constant}$.
This relationship represents a rectangular hyperbola when $p$ is plotted against $V$.
Therefore,the graph showing an inverse relationship between $p$ and $V$ is the correct representation of Boyle's law.

(D) For an ideal gas,the internal energy $U$ is a function of temperature only,i.e.,$U = f(T)$.
Since the process is isothermal,the temperature remains constant $(T_2 = T_1 = 300 \ K)$.
Therefore,the change in internal energy $\Delta U = nC_v(T_2 - T_1) = 0$.

(A) Moles of dioxygen gas $(O_2)$ $= \frac{3.2 \ g}{32 \ g \ mol^{-1}} = 0.1 \ mol$.
Volume of dioxygen gas at $STP$ $(V_1)$ $= 0.1 \ mol \times 22.4 \ L \ mol^{-1} = 2.24 \ L$.
According to Boyle's Law,at constant temperature,$P_1V_1 = P_2V_2$.
Given $P_2 = \frac{1}{3}P_1$,so $P_1 \times 2.24 = (\frac{1}{3}P_1) \times V_2$.
$V_2 = 2.24 \times 3 = 6.72 \ L$.

(B) Using the ideal gas equation $PV = nRT$,since $T$ is constant,$n = \frac{PV}{RT}$.
For $CH_4$: $n_1 = \frac{2 \times 2}{RT} = \frac{4}{RT}$.
For $CO_2$: $n_2 = \frac{4 \times 3}{RT} = \frac{12}{RT}$.
For $Ne$: $n_3 = \frac{3 \times 4}{RT} = \frac{12}{RT}$.
Total moles $n_T = n_1 + n_2 + n_3 = \frac{4 + 12 + 12}{RT} = \frac{28}{RT}$.
Total volume $V_T = 2 + 3 + 4 = 9 \ L$.
Final pressure $P_T = \frac{n_T RT}{V_T} = \frac{28}{RT} \times \frac{RT}{9} = \frac{28}{9} \approx 3.11 \ atm$.
The nearest integer is $3$.

(B) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Let the initial volume $V_1 = 100 \, units$.
Since the volume decreases by $40 \%$,the final volume $V_2 = 100 - 40 = 60 \, units$.
Given $P_1 = 940.3 \, mm \, Hg$.
Substituting the values: $940.3 \times 100 = P_2 \times 60$.
$P_2 = \frac{94030}{60} = 1567.16 \, mm \, Hg$.
The nearest integer is $1567 \, mm \, Hg$.

(C) According to Boyle's law,for a fixed amount of gas at a constant temperature,the pressure is inversely proportional to the volume $(P \propto \frac{1}{V})$.
Using the ideal gas equation,$PV = nRT$,we can rearrange it as:
$P = (nRT) \times (\frac{1}{V})$
This equation is in the form of $y = mx$,where $y = P$,$x = \frac{1}{V}$,and the slope $m = nRT$.
Since the slope $(nRT)$ is directly proportional to the temperature $(T)$,a graph of $P$ versus $\frac{1}{V}$ will yield a straight line passing through the origin for each temperature. As the temperature increases,the slope $(nRT)$ also increases. Therefore,for $T_3 > T_2 > T_1$,the slope of the line for $T_3$ will be the steepest,followed by $T_2$ and then $T_1$.

(C) The work done in a cyclic process is equal to the area enclosed by the cycle in the $P-V$ diagram.
For a $V$ vs $P$ graph,the area enclosed is given by the formula for the area of a triangle: $Area = \frac{1}{2} \times \text{base} \times \text{height}$.
Here,the base along the $P$-axis is $(30 - 10) \ kPa = 20 \ kPa = 20 \times 10^3 \ Pa$.
The height along the $V$-axis is $(30 - 10) \ dm^3 = 20 \ dm^3 = 20 \times 10^{-3} \ m^3$.
Since the cycle $A$ $\rightarrow B$ $\rightarrow C$ $\rightarrow A$ is clockwise in a $V$ vs $P$ graph,the work done is positive.
$W = \frac{1}{2} \times (20 \times 10^3 \ Pa) \times (20 \times 10^{-3} \ m^3) = \frac{1}{2} \times 20 \times 20 \ J = 200 \ J$.

(B) For a reversible isothermal expansion,the work done by the system is given by: $\omega = -nRT \ln \left(\frac{V_2}{V_1}\right)$.
Given:
$n = 1 \ mol$
$T = 18 + 273.15 = 291.15 \ K$
$V_1 = 10 \ L$
$V_2 = 10 + 90 = 100 \ L$
$R = 0.08206 \ L \ atm \ mol^{-1} \ K^{-1}$
Substituting the values:
$\omega = -1 \times 0.08206 \times 291.15 \times \ln \left(\frac{100}{10}\right)$
$\omega = -23.887 \times \ln(10)$
$\omega = -23.887 \times 2.303$
$\omega \approx -55.01 \ L \ atm$
The magnitude of work done by the system is $55 \ L \ atm$.

(B) The total pressure inside the vessel is equal to the external pressure,which is $P_{total} = 1 \ atm$.
The unknown compound exerts a vapour pressure of $0.68 \ atm$. Since it is in equilibrium with its vapour,the partial pressure of the unknown compound is $P_{unknown} = 0.68 \ atm$.
The partial pressure of $He$ is $P_{He} = P_{total} - P_{unknown} = 1 \ atm - 0.68 \ atm = 0.32 \ atm$.
Using the ideal gas law for $He$: $PV = nRT$.
$0.32 \ atm \times V = 0.1 \ mol \times 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 273 \ K$.
$0.32 \times V = 2.24133$.
$V = \frac{2.24133}{0.32} \approx 7.004 \ L$.
Thus,the total volume is close to $7 \ L$.

(A) Initial moles in compartment $A$ $(n_A)$: $n_A = \frac{P_A V_A}{R T_A} = \frac{5 \times 1}{R \times 400} = \frac{1}{80R}$
Initial moles in compartment $B$ $(n_B)$: $n_B = \frac{P_B V_B}{R T_B} = \frac{1 \times 3}{R \times 300} = \frac{1}{100R}$
Total moles $(n_{total})$ = $n_A + n_B = \frac{1}{80R} + \frac{1}{100R} = \frac{5+4}{400R} = \frac{9}{400R}$
At equilibrium,the partition is conducting and movable,so both compartments will have the same pressure $(P)$ and temperature $(T)$.
Let $V_A'$ and $V_B'$ be the new volumes. $V_A' + V_B' = 1 + 3 = 4 \ m^3$.
Using the ideal gas law for the total system: $P(V_A' + V_B') = n_{total} R T$
$P(4) = (\frac{9}{400R}) R T \implies P = \frac{9T}{1600}$
For compartment $A$: $P V_A' = n_A R T \implies (\frac{9T}{1600}) V_A' = (\frac{1}{80R}) R T$
$V_A' = \frac{1}{80} \times \frac{1600}{9} = \frac{20}{9} \approx 2.22 \ m^3$.

(C) In a rigid container,the volume of the gas remains constant,which means $\Delta V = 0$.
Since work done $W = -P_{ext} \Delta V$,it follows that $W = 0$.
This type of process is known as an isochoric process.
According to the first law of thermodynamics,$\Delta U = q + W$. Since $W = 0$,$\Delta U = q$,meaning the absorbed heat increases the internal energy of the system.
An isobaric process is one where pressure remains constant,which is not the case here.
Therefore,the statement that it is an isobaric process is incorrect.

(B) According to Avogadro's law,under similar conditions of temperature and pressure,equal volumes of gases contain an equal number of moles. Since the containers are identical,the volume is the same,so $n_A = n_B$.
The number of moles is given by $n = \frac{w}{Mw}$,where $w$ is the mass and $Mw$ is the molar mass.
Therefore,$\frac{w_A}{40} = \frac{w_B}{x}$.
Rearranging the terms,we get $\frac{w_A}{w_B} = \frac{40}{x}$.
Given the mass ratio $\frac{w_A}{w_B} = \frac{5}{8}$,we substitute this into the equation:
$\frac{5}{8} = \frac{40}{x}$.
Solving for $x$: $x = \frac{40 \times 8}{5} = 8 \times 8 = 64$.

(A) According to the ideal gas equation,$PV = nRT$,where $P$ is pressure,$V$ is volume,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is temperature.
Since $V$,$n$,and $T$ are constant for all four gases,the pressure $P$ is given by $P = \frac{nRT}{V}$.
Since $n$,$R$,$T$,and $V$ are identical for all gases,the pressure exerted by each gas will be the same.
However,in real gas behavior,the pressure exerted depends on the compressibility factor and intermolecular forces. For ideal gases,the pressure is equal. If the question implies real gases,the pressure exerted by a gas is influenced by the van der Waals equation: $(P + \frac{an^2}{V^2})(V - nb) = nRT$.
For gases with smaller molecular size and weaker intermolecular forces (like $H_2$),the deviation from ideal behavior is minimal,and they exert pressure closest to the ideal value. Given the options and the standard context of such problems,if all are treated as ideal,the pressure is the same. If the question implies which gas behaves most ideally or exerts the highest pressure due to lower attractive forces,$H_2$ is the correct choice.

(C) Gay-Lussac's law states that the pressure of a given mass of a gas is directly proportional to its absolute temperature,provided the volume remains constant.
Mathematically,this is expressed as $P \propto T$ or $\frac{P}{T} = \text{constant}$ at constant volume $(V)$ and fixed mass $(n)$.

(C) Given mass of $CO_2 = 8.8 \times 10^{-2} \ kg = 88 \ g$.
The molar mass of $CO_2$ is $44 \ g \ mol^{-1}$.
The number of moles $n$ is calculated as $n = \frac{\text{mass}}{\text{molar mass}} = \frac{88 \ g}{44 \ g \ mol^{-1}} = 2 \ mol$.
According to the ideal gas equation,$PV = nRT$.
Substituting $n = 2$,we get $PV = 2 \ RT$.

(B) Using the ideal gas equation: $PV = nRT$
Rearranging for $n$: $n = \frac{PV}{RT}$
Given values: $P = 3.18 \times 10^5 \ N \ m^{-2}$,$V = 8.314 \times 10^{-3} \ m^3$,$T = 318 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
Substituting the values: $n = \frac{(3.18 \times 10^5) \times (8.314 \times 10^{-3})}{8.314 \times 318}$
$n = \frac{3.18 \times 10^2}{318} = \frac{318}{318} = 1 \ mole$

(B) According to Boyle's law,for a fixed amount of an ideal gas at constant temperature,the pressure $P$ is inversely proportional to the volume $V$,i.e.,$P \propto \frac{1}{V}$.
This implies that $PV = k$,where $k$ is a constant.
Therefore,if we plot $PV$ on the y-axis and $P$ on the x-axis,the value of $PV$ remains constant regardless of the change in pressure $P$.
This results in a horizontal straight line parallel to the pressure axis.

(B) According to Boyle's law,at constant temperature and amount of gas:
$P_1 V_1 = P_2 V_2$
Given:
$P_1 = 1 \ atm$,$V_1 = 25 \ mL$
$P_2 = 1.25 \ atm$,$V_2 = ?$
Substituting the values:
$1 \ atm \times 25 \ mL = 1.25 \ atm \times V_2$
$V_2 = \frac{1 \ atm \times 25 \ mL}{1.25 \ atm} = 20 \ mL$

(A) According to Boyle's law,at constant temperature,the product of pressure and volume is constant.
$P_1 V_1 = P_2 V_2$
Given:
$P_1 = 105 \ kPa$,$V_1 = 11.2 \ dm^3$
$P_2 = 210 \ kPa$,$V_2 = ?$
Substituting the values:
$V_2 = \frac{P_1 V_1}{P_2} = \frac{105 \ kPa \times 11.2 \ dm^3}{210 \ kPa} = 5.6 \ dm^3$