Ideal gas equation and Related gas laws Questions in English

(D) According to Boyle's Law,$PV = K$,where $K$ is a constant at a constant temperature.
Rearranging for $P$,we get $P = K \cdot V^{-1}$.
Differentiating $P$ with respect to $V$ at constant temperature $T$:
$(dP / dV)_T = d(K \cdot V^{-1}) / dV$.
Using the power rule,$(dP / dV)_T = K \cdot (-1) \cdot V^{-2} = -K / V^2$.

(D) hot air balloon works on the principle that the volume of a gas is directly proportional to its absolute temperature at constant pressure,which is known as $Charles's \ Law$ ($V \propto T$ at constant $P$).
As the air inside the balloon is heated,its volume increases,making the air less dense than the surrounding air,which causes the balloon to rise.

(A) According to Boyle's Law,at constant temperature,the pressure of a fixed amount of gas is inversely proportional to its volume $(P \propto 1/V)$.
Since density $(d)$ is mass $(m)$ divided by volume $(V)$,and mass is constant for a fixed amount of gas,we have $V = m/d$.
Substituting this into the Boyle's Law expression,we get $P \propto d$.
Therefore,at sea level,the pressure is higher,which leads to a higher density of air compared to higher altitudes.

(A) At constant pressure,according to Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$
Given: $V_1 = 1000 \ mL$,$T_1 = 25 + 273 = 298 \ K$,$T_2 = 35 + 273 = 308 \ K$
Substituting the values: $\frac{1000}{298} = \frac{V_2}{308}$
$V_2 = \frac{1000 \times 308}{298} \approx 1033.56 \ mL$
Volume of air escaped = $V_2 - V_1 = 1033.56 - 1000 = 33.56 \ mL \approx 33 \ mL$

(D) According to Boyle's Law,for a mixture of gases at constant temperature,the resultant pressure $P$ is given by the formula:
$P = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2}$
Given:
$P_1 = 100 \, kPa, V_1 = 1 \, L$
$P_2 = 320 \, kPa, V_2 = 3 \, L$
Substituting the values:
$P = \frac{(100 \times 1) + (320 \times 3)}{1 + 3}$
$P = \frac{100 + 960}{4}$
$P = \frac{1060}{4}$
$P = 265 \, kPa$

(A) Given:
$T_1 = 75^{\circ}C = 348 \, K$
$V_1 = V$
$V_2 = V - 0.15V = 0.85V$
$P_1 = P$
$P_2 = 2P$
Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Substituting the values:
$\frac{P \times V}{348} = \frac{2P \times 0.85V}{T_2}$
$\frac{1}{348} = \frac{1.7}{T_2}$
$T_2 = 348 \times 1.7 = 591.6 \, K$
Converting to Celsius:
$T_2(^{\circ}C) = 591.6 - 273.15 = 318.45^{\circ}C \approx 319^{\circ}C$

(B) According to Gay-Lussac's Law,for a fixed amount of gas at constant volume,$P \propto T$ or $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given: $P_1 = 12 \ atm$,$T_1 = 27 \ ^oC = 27 + 273 = 300 \ K$.
Maximum pressure the cylinder can withstand,$P_2 = 15 \ atm$.
Substituting the values: $\frac{12}{300} = \frac{15}{T_2}$.
$T_2 = \frac{15 \times 300}{12} = 375 \ K$.
Converting back to Celsius: $T_2( ^oC) = 375 - 273 = 102 \ ^oC$.
Thus,the cylinder will not burst up to $102 \ ^oC$.

(B) Given:
$P_1 = 1\, atm$,$V_1 = 100\, cc$,$T_1 = 0\,^{\circ}C = 273\, K$
$P_2 = 1.5\, atm$
$T_2 = T_1 + \frac{T_1}{3} = 273 + \frac{273}{3} = 273 \times \frac{4}{3} = 364\, K$
Using the ideal gas equation: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{1 \times 100 \times (273 \times \frac{4}{3})}{1.5 \times 273}$
$V_2 = \frac{100 \times 4}{1.5 \times 3} = \frac{400}{4.5} = 88.88\, cc \approx 88.9\, cc$

(C) Using the ideal gas equation: $PV = nRT$
Given: $P = 10 \ atm$,$V = 0.082 \ L$,$T = 400 \ K$,$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$
$n = \frac{PV}{RT} = \frac{10 \times 0.082}{0.082 \times 400} = \frac{10}{400} = 0.025 \ mol$
Number of molecules = $n \times N_A = 0.025 \times 6.022 \times 10^{23} \approx 1.5 \times 10^{22}$ molecules.

(A) Given: $V = 1 \, dm^3 = 1 \, L$,$T = -23 \, ^\circ C = 250 \, K$,$P = 7.6 \times 10^{-5} \, torr = \frac{7.6 \times 10^{-5}}{760} \, atm = 10^{-7} \, atm$.
Using the ideal gas equation $PV = nRT$,we find the number of moles $n$:
$n = \frac{PV}{RT} = \frac{10^{-7} \times 1}{0.0821 \times 250} \approx 4.87 \times 10^{-9} \, mol$.
The number of molecules $N = n \times N_A$:
$N = 4.87 \times 10^{-9} \times 6.022 \times 10^{23} \approx 2.93 \times 10^{15}$ molecules.

(A) According to the ideal gas law,$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
Given $V_1 = V_2 = x \ L$,the equation simplifies to $\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Substituting the initial values,we get $\frac{y}{z} = \frac{P_2}{T_2}$.
For the volume to remain $x \ L$,the ratio of pressure to temperature must remain constant,i.e.,$\frac{P_2}{T_2} = \frac{y}{z}$.
Checking option $A$: $\frac{2y}{2z} = \frac{y}{z}$.
Thus,at $2y \ atm$ and $2z \ K$,the volume remains $x \ L$.

(D) Given: $P_1 = P, V_1 = V, T_1 = 27 + 273 = 300 \ K, n_1 = n$
$P_2 = P, T_2 = -173 + 273 = 100 \ K, n_2 = 2n$
Using the ideal gas equation $PV = nRT$,since $P$ is constant:
$\frac{V_1}{V_2} = \frac{n_1 T_1}{n_2 T_2}$
$\frac{V}{V_2} = \frac{n \times 300}{2n \times 100}$
$\frac{V}{V_2} = \frac{300}{200} = \frac{3}{2}$
$V_2 = \frac{2V}{3} \ L$

(C) The molar volume $V_m$ is calculated using the ideal gas equation $PV = nRT$.
For $n = 1 \, mol$,$P = 1 \, atm$,$T = 27 + 273 = 300 \, K$,and $R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.
$V_m = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 300}{1} = 24.63 \, L$.
Rounding to the nearest provided option,the answer is $24.6 \, L$.

(D) The decomposition reaction is: $CH_3OH_{(g)} \to CO_{(g)} + 2H_{2(g)}$
Initial moles: $CH_3OH = 0.01 \, mol$,$CO = 0$,$H_2 = 0$
At equilibrium (complete decomposition): $CH_3OH = 0 \, mol$,$CO = 0.01 \, mol$,$H_2 = 0.02 \, mol$
Total moles of gas $n = 0.01 + 0.02 = 0.03 \, mol$
Using the ideal gas law $PV = nRT$:
$P = \frac{nRT}{V}$
Given $R = 0.082 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,$T = 600 \, K$,$V = 0.082 \, L$
$P = \frac{0.03 \times 0.082 \times 600}{0.082}$
$P = 0.03 \times 600 = 18 \, atm$

(C) Given: $P = 76 \ cm \ of \ Hg = 1 \ atm$,$T = 273 \ K$,$V = 1 \ L$,$w = 1.786 \ g$.
Using the ideal gas equation: $PV = nRT = (w/M)RT$.
$M = (wRT) / (PV) = (1.786 \times 0.0821 \times 273) / (1 \times 1) \approx 40 \ g/mol$.
However,at $STP$ $(273 \ K, 1 \ atm)$,$1 \ mole$ of an ideal gas occupies $22.4 \ L$.
Thus,$M = (1.786 \ g / 1 \ L) \times 22.4 \ L/mol \approx 40 \ g/mol$.
Checking molar masses: $2-\text{methylpropane} (C_4H_{10}) = 58 \ g/mol$,$\text{Propane} (C_3H_8) = 44 \ g/mol$,$\text{Propene} (C_3H_6) = 42 \ g/mol$,$\text{Phosgene} (COCl_2) = 12 + 16 + 71 = 99 \ g/mol$.
Re-evaluating the calculation: $1.786 \times 22.4 = 40.0064 \ g/mol$. The closest value to $40 \ g/mol$ is $42 \ g/mol$ (Propene). Given the options,$42 \ g/mol$ is the most appropriate choice.

(D) According to the Ideal Gas Law,$PV = nRT = (\frac{W}{M})RT$.
Since $P, V, T$ are constant,$n_1 = n_2$,which implies $\frac{W_1}{M_1} = \frac{W_2}{M_2}$.
Given $W_1 = 0.6 \ g$ (for $N_2$,$M_1 = 28 \ g \ mol^{-1}$) and $W_2 = 0.73 \ g$.
Substituting the values: $\frac{0.6}{28} = \frac{0.73}{M_2}$.
$M_2 = \frac{0.73 \times 28}{0.6} = \frac{20.44}{0.6} \approx 34.06 \ g \ mol^{-1}$.
The molar mass of $PH_3$ is $31 + 3(1) = 34 \ g \ mol^{-1}$.
Therefore,the gas is $PH_3$.

(C) For an open flask,pressure $(P)$ and volume $(V)$ remain constant.
According to the ideal gas law,$PV = nRT$,which implies $n \propto \frac{1}{T}$ at constant $P$ and $V$.
Given: $T_1 = 27^{\circ}C = 300 \, K$.
Let the initial number of moles be $n_1 = n$.
After heating,one-third of the air is expelled,so the remaining moles are $n_2 = n - \frac{n}{3} = \frac{2n}{3}$.
Using the relation $\frac{n_1}{n_2} = \frac{T_2}{T_1}$:
$\frac{n}{2n/3} = \frac{T_2}{300}$
$\frac{3}{2} = \frac{T_2}{300}$
$T_2 = \frac{3}{2} \times 300 = 450 \, K$.
Converting to Celsius: $T_2 = 450 - 273 = 177^{\circ}C$.

(B) Using the ideal gas equation $PV = nRT$,the number of moles $n$ is given by $n = \frac{PV}{RT}$.
For container $A$: $P_A = 2P$,$V_A = 2V$,$T_A = 2T$. Thus,$n_A = \frac{(2P)(2V)}{R(2T)} = 2 \frac{PV}{RT}$.
For container $B$: $P_B = P$,$V_B = V$,$T_B = T$. Thus,$n_B = \frac{PV}{RT}$.
The ratio of the number of molecules (which is proportional to the number of moles) is $\frac{n_A}{n_B} = \frac{2(PV/RT)}{(PV/RT)} = 2 : 1$.

(B) Given: $W_1 = 3.7 \ g$,$T_1 = 25 + 273 = 298 \ K$,$W_2 = 0.184 \ g$,$M_2 = 2 \ g/mol$ $(H_2)$,$T_2 = 17 + 273 = 290 \ K$.
Since pressure $P$ and volume $V$ are the same,from the ideal gas equation $PV = nRT$,we have $n_1 T_1 = n_2 T_2$.
Substituting $n = \frac{W}{M}$,we get $\frac{W_1}{M_1} T_1 = \frac{W_2}{M_2} T_2$.
$\frac{3.7}{M_1} \times 298 = \frac{0.184}{2} \times 290$.
$M_1 = \frac{3.7 \times 298 \times 2}{0.184 \times 290}$.
$M_1 = \frac{2205.2}{53.36} \approx 41.3 \ g/mol$.

(B) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.
Substituting this,$PV = \frac{m}{M} RT$.
Rearranging for density $(d = \frac{m}{V})$,we get $d = \frac{PM}{RT}$.
For a given gas,$M$ is constant,so $d \propto \frac{P}{T}$.
To maximize density $(d)$,we need to maximize pressure $(P)$ and minimize temperature $(T)$.
Comparing the options:
$(A)$ $\frac{P}{T} = \frac{1}{100} = 0.01$
$(B)$ $\frac{P}{T} = \frac{2}{100} = 0.02$
$(C)$ $\frac{P}{T} = \frac{1}{200} = 0.005$
$(D)$ $\frac{P}{T} = \frac{3}{200} = 0.015$
Comparing the values,option $(B)$ has the highest ratio of $\frac{P}{T}$.

(D) From the ideal gas equation,$PM = dRT$. Since temperature $T$ is constant,$P = \frac{dRT}{M}$,which implies $P \propto \frac{d}{M}$.
Given:
For $G_1$: $M_1 = M$,$d_1 = 2d$
For $G_2$: $M_2 = 3M$,$d_2 = d$
Ratio of pressures:
$\frac{P_1}{P_2} = \frac{d_1}{M_1} \times \frac{M_2}{d_2} = \frac{2d}{M} \times \frac{3M}{d} = 6$
Therefore,the ratio of pressure exerted by $G_2$ to that of $G_1$ is:
$\frac{P_2}{P_1} = \frac{1}{6}$

(C) Given: $T_1 = 27 + 273 = 300\,K$,$d_1 = d$,$d_2 = 0.75d = \frac{3}{4}d$,$P_1 = P_2 = P$.
From the ideal gas equation $PV = nRT$,we know $PV = (\frac{m}{M})RT$,which rearranges to $PM = dRT$ (where $d = \frac{m}{V}$).
Since $P$,$M$,and $R$ are constant,$d_1 T_1 = d_2 T_2$.
Substituting the values: $d \times 300 = 0.75d \times T_2$.
$T_2 = \frac{300}{0.75} = \frac{300}{3/4} = 300 \times \frac{4}{3} = 400\,K$.

(C) The ideal gas equation is given by $PM = dRT$,where $P$ is pressure,$M$ is molar mass,$d$ is density,$R$ is the gas constant,and $T$ is temperature.
Given: $P = 2.0\, atm$,$T = 27 + 273 = 300\, K$,$M$ (for $CH_4$) $= 16\, g\, mol^{-1}$,and $R = 0.0821\, L\, atm\, K^{-1}\, mol^{-1}$.
Substituting the values: $d = \frac{PM}{RT} = \frac{2.0 \times 16}{0.0821 \times 300}$.
$d = \frac{32}{24.63} \approx 1.30\, g\, L^{-1}$.

(A) From the ideal gas equation,$PV = nRT = (m/M)RT$.
Since density $d = m/V$,we have $PM = dRT$,which implies $d = PM / RT$.
Therefore,$d \propto P / T$.
This gives the relation $\frac{d_2}{d_1} = \frac{P_2 / T_2}{P_1 / T_1} = \frac{P_2 T_1}{P_1 T_2}$.
Thus,$d_2 = d_1 [T_1 P_2 / T_2 P_1]$.

(C) From the ideal gas equation,$PV = nRT$,we know $PV = (m/M)RT$,where $m$ is mass and $M$ is molar mass.
Rearranging for density $d = m/V$,we get $d = PM/RT$.
Thus,$P = dRT/M$.
Given $d_x = 2d_y$ and $M_y = 3M_x$.
The ratio of pressures is $\frac{P_x}{P_y} = \frac{d_x R T / M_x}{d_y R T / M_y} = \frac{d_x}{d_y} \times \frac{M_y}{M_x}$.
Substituting the given values: $\frac{P_x}{P_y} = (2) \times (3) = 6$.

(A) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.
Substituting $n$,we get $PV = \frac{m}{M}RT$.
Rearranging for density $(d = \frac{m}{V})$,we get $P = \frac{dRT}{M}$.
Since the molar masses $(M)$ are equal,the relationship is $P \propto d \times T$.
Given the ratio of densities $\frac{d_1}{d_2} = \frac{1}{2}$ and the ratio of temperatures $\frac{T_1}{T_2} = \frac{2}{1}$.
The ratio of pressures is $\frac{P_1}{P_2} = \frac{d_1}{d_2} \times \frac{T_1}{T_2} = \frac{1}{2} \times \frac{2}{1} = \frac{1}{1}$.
Therefore,the ratio of their pressures is $1 : 1$.

(A) The density $(d)$ of a gas is given by the ideal gas equation: $d = \frac{PM}{RT}$.
Here,$P = 3\, atm$,$M$ (molar mass of $NH_3$) = $17\, g\, mol^{-1}$,$R = 0.0821\, L\, atm\, K^{-1}\, mol^{-1}$,and $T = 27 + 273 = 300\, K$.
Substituting the values:
$d = \frac{3 \times 17}{0.0821 \times 300} = \frac{51}{24.63} \approx 2.07\, g\, L^{-1}$.

(A) $1$. Calculate the number of moles for each gas:
$n(O_2) = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$
$n(H_2) = \frac{2 \ g}{2 \ g/mol} = 1.0 \ mol$
$2$. Total moles $(n_{total})$ = $0.125 + 1.0 = 1.125 \ mol$
$3$. Use the ideal gas equation $PV = nRT$:
$P = \frac{n_{total} \times R \times T}{V}$
$P = \frac{1.125 \ mol \times 0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} \times 273 \ K}{1 \ L}$
$P = 1.125 \times 22.4133 = 25.215 \ atm$

(B) The kinetic energy $(KE)$ of an ideal gas is given by the formula $KE = \frac{3}{2} nRT$.
Given that the kinetic energy of $He$ is equal to the kinetic energy of $Ar$:
$KE_{He} = KE_{Ar}$
$\frac{3}{2} n_{He} R T_{He} = \frac{3}{2} n_{Ar} R T_{Ar}$
$n_{He} \times T_{He} = n_{Ar} \times T_{Ar}$
Substituting the given values:
$0.3 \times T_{He} = 0.4 \times 400$
$0.3 \times T_{He} = 160$
$T_{He} = \frac{160}{0.3} = 533.33 \ K$.
Since the closest option provided is $533.33 \ K$ is not present,let us re-evaluate the question context. If the question implies average kinetic energy per mole,then $KE = \frac{3}{2} RT$. However,for total kinetic energy,the calculation holds. Given the options,$533.33 \ K$ is the correct mathematical result.

(B) The compressibility factor $Z$ is defined as the ratio of the molar volume of a real gas $(V_{real})$ to the molar volume of an ideal gas $(V_{ideal})$ at the same temperature and pressure.
$Z = \frac{PV}{nRT}$
For an ideal gas,the equation of state is $PV = nRT$.
Therefore,for an ideal gas,$Z = \frac{nRT}{nRT} = 1$.
Thus,the value of the compressibility factor for an ideal gas is $1$.

(C) For an ideal gas,the equation is $PV = nRT$. At $STP$,$P = 1 \ atm$ and $T = 273.15 \ K$. The volume of $2 \ mol$ of an ideal gas at $STP$ is $V = n \times 22.4 \ L/mol = 2 \times 22.4 = 44.8 \ L$. Since the given volume is $50 \ L$,which is greater than $44.8 \ L$,the gas occupies more space than expected for an ideal gas at $STP$. Under these conditions (low pressure and high volume),the gas behaves very closely to an ideal gas. However,in the context of standard textbook problems,gases at $STP$ are generally considered to exhibit ideal gas behavior.

(D) The compressibility factor $Z$ is defined as $Z = \frac{PV}{nRT}$.
At very low pressure,real gases behave like ideal gases.
For an ideal gas,the compressibility factor $Z$ is equal to $1$.
Therefore,at very low pressure,the compressibility factor of $CO_2$ gas approaches $1$.

(C) From the ideal gas equation,$PV = nRT$,we know that $P = \frac{dRT}{M}$,where $d$ is density and $M$ is molecular weight.
Given: $d_A = 2d_B$ and $M_B = 2M_A$,at the same temperature $T$.
Taking the ratio of pressures: $\frac{P_A}{P_B} = \frac{d_A}{M_A} \times \frac{M_B}{d_B}$.
Substituting the given values: $\frac{P_A}{P_B} = \frac{2d_B}{M_A} \times \frac{2M_A}{d_B} = 2 \times 2 = 4$.
Therefore,the ratio $\frac{P_A}{P_B} = 4:1$.

(B) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$ and $d = \frac{m}{V}$.
Thus,$P = \frac{dRT}{M}$.
For gases $A$ and $B$ at the same temperature $T$,the ratio of pressures is given by:
$\frac{P_A}{P_B} = \frac{d_A}{d_B} \times \frac{M_B}{M_A}$.
Given: $d_A = 4 d_B$ and $M_A = 0.5 M_B$ (or $M_B = 2 M_A$).
Substituting these values:
$\frac{P_A}{P_B} = \frac{4 d_B}{d_B} \times \frac{2 M_A}{M_A} = 4 \times 2 = 8$.

(A) According to Boyle's law,$P_1 V_1 = P_2 V_2$ at constant temperature.
At the bottom of the lake,the total pressure is $P_1 = P_{atm} + P_{water} = \rho g \ell + \rho g h = \rho g(h + \ell)$.
The volume of the bubble is $V_1 = \frac{4}{3} \pi r^3$.
At the surface,the pressure is $P_2 = P_{atm} = \rho g \ell$.
The radius becomes $3r$,so the volume is $V_2 = \frac{4}{3} \pi (3r)^3 = 27 \times \frac{4}{3} \pi r^3$.
Substituting these into the equation: $\rho g(h + \ell) \times \frac{4}{3} \pi r^3 = \rho g \ell \times 27 \times \frac{4}{3} \pi r^3$.
Simplifying,we get $h + \ell = 27 \ell$.
Therefore,$h = 26 \ell$.

(B) For an ideal gas,the equation of state is $PV = nRT$.
For $1 \text{ mole}$ of gas,$P = RT/V$.
Differentiating $P$ with respect to $V$ at constant temperature $T$:
$[\partial P/\partial V]_T = \partial(RT/V)/\partial V = RT \times (\partial(V^{-1})/\partial V) = RT \times (-V^{-2}) = -RT/V^2$.

(B) 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) The compressibility factor $(Z)$ is a measure of the deviation of real gases from ideal behavior.
$Z = \frac{PV}{nRT}$
For an ideal gas,the equation of state is $PV = nRT$.
Substituting this into the expression for $Z$:
$Z = \frac{nRT}{nRT} = 1$
Therefore,the compressibility factor for an ideal gas is $1$.

(B) According to Gay-Lussac's Law,for a fixed volume of gas,$\frac{P_1}{T_1} = \frac{P_2}{T_2}$.
Given: $P_1 = 12 \text{ atm}$,$T_1 = 27 + 273 = 300 \text{ K}$,$P_2 = 14.9 \text{ atm}$.
Substituting the values: $\frac{12}{300} = \frac{14.9}{T_2}$.
$T_2 = \frac{14.9 \times 300}{12} = 372.5 \text{ K}$.
Converting to Celsius: $T_2 = 372.5 - 273 = 99.5 \,^oC$.

(A) According to the ideal gas equation,$PV = nRT = (m/M)RT$,where $m$ is the mass and $M$ is the molar mass of the gas.
Rearranging the equation,we get $P = (m/V) \times (RT/M)$.
Since density $\rho = m/V$,we can substitute this into the equation: $P = \rho \times (RT/M)$.
At a constant temperature $T$,$R$ and $M$ are constants,so $P \propto \rho$.

(A) According to the ideal gas equation,$PV = nRT$,which can be rearranged as $V = (\frac{nR}{P})T$.
This represents a straight line equation $y = mx$,where the slope $m = \frac{nR}{P}$.
Since the slope is inversely proportional to pressure $(m \propto \frac{1}{P})$,the line with the smallest slope corresponds to the highest pressure.
Looking at the graph,the slope of the line for $p_1$ is the smallest,followed by $p_3$,and then $p_2$.
Therefore,the correct order of pressures is $p_1 > p_3 > p_2$.

(C) Using the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.
Given:
$P = 2 \, atm$
$V = 100 \, L$
$T = 20 + 273 = 293 \, K$
$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$
$M$ (molar mass of $H_2S$) $= 2(1) + 32 = 34 \, g/mol$.
Substituting the values:
$m = \frac{MPV}{RT} = \frac{34 \times 2 \times 100}{0.0821 \times 293} \approx 282.68 \, g$.

(D) Using the ideal gas equation: $PV = nRT$
First,calculate the number of moles $(n)$ of water vapour:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.8 \; g}{18 \; g \; mol^{-1}} = 0.1 \; mol$
Convert the temperature $(T)$ from Celsius to Kelvin:
$T = 374 + 273 = 647 \; K$
Given pressure $(P)$ is $1 \; bar$ and gas constant $(R)$ is $0.083 \; bar \; L \; K^{-1} \; mol^{-1}$.
Now,calculate the volume $(V)$:
$V = \frac{nRT}{P} = \frac{0.1 \; mol \times 0.083 \; bar \; L \; K^{-1} \; mol^{-1} \times 647 \; K}{1 \; bar} = 5.37 \; L$

(11.35 L) According to Boyle's Law,$p_{1} V_{1} = p_{2} V_{2}$.
Given: $p_{1} = 1 \ bar$,$V_{1} = 2.27 \ L$,and $p_{2} = 0.2 \ bar$.
Substituting the values into the equation:
$V_{2} = \frac{p_{1} V_{1}}{p_{2}}$
$V_{2} = \frac{1 \ bar \times 2.27 \ L}{0.2 \ bar} = 11.35 \ L$.
Since the balloon bursts if the pressure exceeds $0.2 \ bar$,the volume of the balloon can be expanded up to $11.35 \ L$.

(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:
$V_{1} = 2 \ L$
$T_{1} = 23.4 + 273.15 = 296.55 \ K$
$T_{2} = 26.1 + 273.15 = 299.25 \ K$
Rearranging the formula for $V_{2}$:
$V_{2} = \frac{V_{1} \times T_{2}}{T_{1}}$
Substituting the values:
$V_{2} = \frac{2 \ L \times 299.25 \ K}{296.55 \ K}$
$V_{2} \approx 2.018 \ L$

(A) 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:
$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$
Rearranging for $p_{2}$:
$p_{2} = \frac{p_{1} V_{1} T_{2}}{T_{1} V_{2}}$
Substituting the values:
$p_{2} = \frac{760 \times 600 \times 283}{298 \times 640}$
$p_{2} = \frac{129048000}{190720} \approx 676.6 \,mm \,Hg$

(B) Given:
Initial pressure,$p_{1} = 1 \, bar$
Initial volume,$V_{1} = 500 \, dm^{3}$
Final volume,$V_{2} = 200 \, dm^{3}$
Since the temperature remains constant,the final pressure $(p_{2})$ can be calculated using Boyle's law.
According to Boyle's law,
$p_{1} V_{1} = p_{2} V_{2}$
$\Rightarrow p_{2} = \frac{p_{1} V_{1}}{V_{2}}$
$= \frac{1 \times 500}{200} \, bar$
$= 2.5 \, bar$
Therefore,the minimum pressure required is $2.5 \, bar$.

(0.8 BAR) Given:
Initial pressure,$p_{1} = 1.2 \, bar$
Initial volume,$V_{1} = 120 \, mL$
Final volume,$V_{2} = 180 \, mL$
Since the temperature remains constant,the final pressure $(p_{2})$ can be calculated using Boyle's law.
According to Boyle's law,$p_{1} V_{1} = p_{2} V_{2}$.
$p_{2} = \frac{p_{1} V_{1}}{V_{2}}$
$p_{2} = \frac{1.2 \times 120}{180} \, bar$
$p_{2} = 0.8 \, bar$
Therefore,the final pressure of the gas would be $0.8 \, bar$.

(N/A) The equation of state is given by,
$pV = nRT$ .......... $(i)$
Where,
$p \rightarrow$ Pressure of gas
$V \rightarrow$ Volume of gas
$n \rightarrow$ Number of moles of gas
$R \rightarrow$ Gas constant
$T \rightarrow$ Temperature of gas
From equation $(i)$,we have,
$\frac{n}{V} = \frac{p}{RT}$
Replacing $n$ with $\frac{m}{M}$,where $m$ is the mass and $M$ is the molar mass,we have
$\frac{m}{MV} = \frac{p}{RT}$ .......... $(ii)$
Since density $d = \frac{m}{V}$,substituting this into equation $(ii)$ gives,
$\frac{d}{M} = \frac{p}{RT}$
$\Rightarrow d = \left(\frac{M}{RT}\right) p$
Since the molar mass $M$ and gas constant $R$ are constants,at a constant temperature $T$,the term $\left(\frac{M}{RT}\right)$ is constant.
Therefore,$d = (\text{constant}) \times p$
$\Rightarrow d \propto p$
Hence,at a given temperature,the density $d$ of a gas is directly proportional to its pressure $p$.