Critical state and Liquefaction of gases Questions in English

(C) The point at which all three phases of water (ice,liquid water,and water vapour) coexist in equilibrium is known as the triple point of water.
This occurs at a temperature of $0.01\,^\circ C$ $(273.16\, K)$ and a pressure of $4.58\, mm \, Hg$ $(611.657\, Pa)$.

(A) Liquefaction of a gas occurs when intermolecular forces become significant enough to hold molecules together in a liquid state.
Low temperature decreases the kinetic energy of gaseous molecules,bringing them closer together.
High pressure forces the molecules into a smaller volume,further decreasing the intermolecular distance.
Therefore,the combination of low temperature and high pressure is the most effective condition for liquefaction.

(C) The critical temperature $(T_c)$ of a gas is the temperature above which it cannot be liquefied,no matter how much pressure is applied.
Above this temperature,the substance exists only as a gas and cannot be converted into a liquid state by compression.

(C) The temperature below which a gas can be liquefied by the application of pressure alone is called the critical temperature $(T_c)$.
Therefore,a gas can be liquefied only when its temperature is below its critical temperature.

(D) The critical temperature $(T_c)$ is defined as the temperature above which a gas cannot be liquefied,regardless of the pressure applied.
At this temperature,the density of the liquid phase becomes equal to the density of the vapour phase,meaning the meniscus between the two phases disappears.
Consequently,the surface tension of the system becomes zero at the critical temperature.
Therefore,all the given statements are correct.

(A) Liquefaction of a gas occurs when the intermolecular forces become strong enough to hold the molecules together in a liquid state.
Decreasing the temperature reduces the kinetic energy of the gas molecules,allowing them to come closer.
Increasing the pressure forces the molecules into a smaller volume,further decreasing the intermolecular distance.
Therefore,$Low \ temperature$ and $High \ pressure$ are the most favorable conditions for liquefaction.

(C) The liquefaction of gases is achieved by reducing the temperature (cooling) and increasing the pressure.
By cooling,the kinetic energy of gas molecules decreases,and by increasing pressure,the molecules are brought closer together,leading to the formation of liquid state.
Therefore,both cooling and increasing pressure are required.

(D) The critical temperature $(T_c)$ is given by the formula: $T_c = \frac{8a}{27Rb}$.
Since $T_c \propto \frac{a}{b}$,we calculate the ratio $\frac{a}{b}$ for each gas:
For $W$: $\frac{4.0}{0.027} \approx 148.15$
For $X$: $\frac{8.0}{0.030} \approx 266.67$
For $Y$: $\frac{6.0}{0.032} = 187.5$
For $Z$: $\frac{12.0}{0.027} \approx 444.44$
Since gas $Z$ has the highest ratio of $\frac{a}{b}$,it has the highest critical temperature.

(B) At the $Critical \ point$,the density of the liquid and vapor phases becomes identical,and the interface between them disappears. Consequently,the surface tension becomes zero.

(D) An ideal gas cannot be liquefied because it assumes the absence of intermolecular forces of attraction.
Since liquefaction requires the presence of intermolecular forces to bring gas molecules together,an ideal gas,by definition,does not possess these forces and therefore cannot be liquefied under any conditions of $P$ and $T$.

(B) The temperature above which a gas cannot be liquefied by pressure alone is called the $Critical \ temperature$ $(T_c)$. Below this temperature,the gas can be liquefied by applying sufficient pressure.

(C) gas can be liquefied by applying pressure when its temperature is below its critical temperature $(T_c)$.
Above the critical temperature,a gas cannot be liquefied by pressure alone.

(A) The critical temperature $(T_c)$ is the temperature above which a gas cannot be liquefied by pressure alone. Therefore,it is possible to compress and liquefy a gas at temperatures below $T_c$. Statement $A$ is incorrect because it claims it is impossible to compress a gas below $T_c$,whereas it is actually the condition required for liquefaction.

(D) In an ideal gas,there are no intermolecular forces of attraction present between the molecules.
Liquefaction of a gas requires the presence of intermolecular forces to bring molecules together to form a liquid phase.
Since ideal gases lack these forces,they cannot be liquefied.

(B) At low temperature and high pressure,the intermolecular forces of attraction increase,which facilitates the liquefaction of gases. $Liquefaction$ of a gas is typically carried out at a temperature below its $Critical \ Temperature$.

(D) For a van der Waals gas,the critical constants are related to the parameters $a$ and $b$ as follows:
$V_C = 3b \implies b = \frac{V_C}{3} = \frac{0.123}{3} = 0.041 \ L \ mol^{-1}$. Thus,option $B$ is incorrect.
The volume of a single molecule is related to the excluded volume $b$ by $b = 4 \times N_A \times V_{molecule}$. Therefore,$V_{molecule} = \frac{b}{4 \times N_A} = \frac{0.123}{3 \times 4 \times 6.023 \times 10^{23}} \ L$. Converting to $cc$ $(1 \ L = 1000 \ cc)$,we get $V_{molecule} = \frac{0.123 \times 1000}{12 \times 6.023 \times 10^{23}} \ cc$. Thus,option $A$ is incorrect.
The Boyle temperature $T_B$ is given by $T_B = \frac{a}{Rb}$. Since $T_C = \frac{8a}{27Rb}$,we have $T_B = \frac{27}{8} T_C = 3.375 \times 180 \ K = 607.5 \ K$. Since $T_B > T_C$,option $C$ is incorrect.
By definition,a gas cannot be liquefied above its critical temperature $T_C$. Since $T_C = 180 \ K$,the gas cannot be liquefied at $200 \ K$. Thus,option $D$ is correct.

(D) The ease of liquefaction of a gas is directly proportional to its critical temperature $(T_C)$.
Higher the critical temperature,easier it is to liquefy the gas.
Given critical temperatures:
$T_C(O_2) = 154.3 \ K$
$T_C(H_2) = 33.2 \ K$
$T_C(He) = 5.3 \ K$
Comparing the values: $154.3 > 33.2 > 5.3$.
Therefore,the order of liquefaction (from easiest to hardest) is $O_2 > H_2 > He$.

(B) Option $B$ is incorrect. When a liquid is in equilibrium with its vapor in a closed vessel,increasing the temperature increases the kinetic energy of the molecules,causing more liquid to evaporate. This leads to an increase in the number of vapor molecules per unit volume,thereby increasing the density of the vapor,not decreasing it.

(D) gas can be liquefied only if its temperature is below its critical temperature $(T_C)$.
If the temperature of the gas is above the critical temperature $(T > T_C)$,it cannot be liquefied by the application of pressure alone,no matter how high the pressure is.
Therefore,under the condition $T > T_C, P > P_C$,the gas remains in the gaseous state.

(B) Gases can be liquefied at low temperature and high pressure.
This is because,at low temperature,the kinetic energy of the molecules of the gases is minimum,and high pressure increases the intermolecular forces of attraction between the molecules.

(A) The noble gas $Xe$ is non-polar and consists of individual atoms.
For non-polar species,the only intermolecular forces present are London dispersion forces,also known as instantaneous dipole-induced dipole forces.
These weak forces are responsible for the liquefaction of $Xe$ gas at low temperatures.

(B) gas can be liquefied if its temperature is below its critical temperature $(T < T_c)$.
Once the temperature is below $T_c$,the gas can be liquefied by applying sufficient pressure $(P > P_c)$.
Option $(b)$ represents the condition where $T < T_c$ and $P > P_c$,which is the standard condition for liquefaction of a gas.

(C) The ease of liquefaction of a gas is directly proportional to its critical temperature $(T_c)$.
Higher the critical temperature,stronger are the intermolecular forces of attraction,and easier it is to liquefy the gas.
Comparing the given critical temperatures:
$R (154.3 \ K) > S (126 \ K) > P (33.2 \ K) > Q (5.3 \ K)$.
Therefore,the order of liquefaction (from easiest to hardest) is $R, S, P, Q$.

(D) The triple point of a substance is the temperature and pressure at which the three phases of that substance (solid,liquid,and gas) coexist in thermodynamic equilibrium. For water,this occurs at a temperature of $273.16 \ K$ $(0.01 \ ^\circ C)$ and a pressure of $611.657 \ Pa$ $(0.006037 \ atm)$.

(D) An ideal gas is defined as a gas that obeys the gas laws at all temperatures and pressures and does not possess intermolecular forces of attraction.
Liquefaction of a gas requires the presence of intermolecular forces to bring molecules together to form a liquid phase.
Since an ideal gas lacks these forces,it cannot be liquefied under any conditions of temperature $(T)$ or pressure $(P)$.

(B) The $Critical \ temperature$ $(T_c)$ of a gas is defined as the temperature above which a gas cannot be liquefied,no matter how much pressure is applied. Conversely,it is the temperature below which a gas can be liquefied by the application of pressure alone. Therefore,the temperature below which a gas cannot exist in the gaseous state (as it would condense) is related to the $Critical \ temperature$.

(C) The critical temperature $(T_c)$ of a gas is defined as the temperature above which a gas cannot be liquefied,no matter how much pressure is applied. $ \newline $ Therefore,for a gas to be liquefied,it must be cooled to a temperature below its critical temperature $(T < T_c)$. $ \newline $ Thus,the correct option is $C$.

(A) The critical temperature $(T_C)$ is defined as the temperature above which a gas cannot be liquefied,no matter how much pressure is applied.
At temperatures below $T_C$,the gas can be liquefied by applying pressure because the intermolecular forces are strong enough to hold the molecules together upon compression.
Therefore,the statement that it is not possible to compress a gas at a temperature lower than $T_C$ is incorrect,as compression is actually the method used to liquefy gases below $T_C$.

(D) An ideal gas is defined as a gas where there are no intermolecular forces of attraction between the molecules.
Liquefaction of a gas occurs when intermolecular forces become strong enough to hold the molecules together in the liquid state.
Since an ideal gas assumes zero intermolecular forces,it is impossible to overcome the kinetic energy of the molecules to bring them into a liquid state,regardless of the pressure or temperature applied.

(D) The ease of liquefaction of a gas is directly proportional to its critical temperature $(T_c)$. Higher critical temperature indicates stronger intermolecular forces,making it easier to liquefy the gas.
The given critical temperatures are:
$O_2: 154.3 \ K$
$N_2: 126 \ K$
$H_2: 33.2 \ K$
$He: 5.3 \ K$
Therefore,the decreasing order of liquefaction is $O_2 > N_2 > H_2 > He$.

(D) The critical temperature $(T_c)$ of a gas is directly proportional to the strength of the intermolecular forces of attraction.
Stronger intermolecular forces (such as dipole-dipole interactions or higher van der Waals forces) lead to a higher critical temperature.
Among the given gases,$SO_{2(g)}$ is a polar molecule with significant dipole-dipole interactions and a higher molecular weight compared to $He$,$Ne$,and $N_2$.
Therefore,$SO_{2(g)}$ has the highest critical temperature.

(A) The critical temperature $(T_C)$ of a gas is defined as the temperature above which a gas cannot be liquefied,regardless of the pressure applied.
It is given by the relation $T_C = \frac{8a}{27Rb}$,which implies $T_C \propto a$,where $a$ is the van der Waals constant representing the magnitude of intermolecular forces of attraction.
$A$ higher critical temperature indicates stronger intermolecular forces,making the gas easier to liquefy.
Comparing the given values: $A = 25\,^{\circ}C$,$B = 10\,^{\circ}C$,$C = -80\,^{\circ}C$,and $D = 15\,^{\circ}C$.
Since gas $A$ has the highest critical temperature $(25\,^{\circ}C)$,it will be liquefied most easily.

(B) Critical temperature $(T_C)$ is the maximum temperature at which a gas can be liquefied by the application of pressure alone.
Below the critical temperature,the thermal motion of the molecules is slow enough for the intermolecular forces to come into play,leading to the condensation of the gas.
At the critical temperature,the liquid and gaseous phases become indistinguishable.
Ideal gases do not have a characteristic critical temperature because they lack intermolecular forces of attraction.
Therefore,the correct statement is that below the critical temperature,thermal motion of the molecules is slow enough for the intermolecular forces to come into play leading to condensation of the gas.

(D) The critical temperature $(T_c)$ of a gas depends on the magnitude of intermolecular forces of attraction.
$CO_2$ has the highest critical temperature among the given options,which is $304.2 \ K$,due to stronger van der Waals forces compared to $H_2$,$He$,and $N_2$.

(A) The critical temperature $(T_c)$ of a gas is the temperature above which it cannot be liquefied,no matter how much pressure is applied.
This is because,above $T_c$,the kinetic energy of the gas molecules is so high that the intermolecular forces of attraction are insufficient to hold the molecules together in the liquid state.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) At the critical temperature $(T_c)$,the distinction between the liquid and gaseous phases disappears.
This is because the density of the liquid phase becomes equal to the density of the gaseous phase at this temperature.
Consequently,the liquid passes into the gaseous state imperceptibly and continuously,meaning there is no phase boundary.
Both the Assertion and the Reason are correct,and the Reason correctly explains why the transition occurs in this manner.

(D) Liquid $CO_2$ is used as a solvent in dry cleaning because it is environmentally friendly and effective at dissolving grease and stains when combined with a suitable detergent. This process is known as supercritical fluid extraction or liquid $CO_2$ cleaning.

(A) The critical temperature $(T_c)$ is the temperature above which a gas cannot be liquefied,no matter how much pressure is applied.
When cooling a gas from $500 \ K$,the gas that reaches its critical temperature first will start to liquefy.
Since the critical temperature of ammonia $(405.5 \ K)$ is higher than that of carbon dioxide $(304.10 \ K)$,ammonia will reach its critical temperature first during the cooling process.
Therefore,ammonia will liquefy first.

(A) The critical temperature of a gas is a measure of the strength of intermolecular forces of attraction.
Higher critical temperature indicates that the gas can be liquefied more easily,which implies stronger intermolecular forces of attraction.
Since the critical temperature of $CO_2$ $(31.1^{\circ} C)$ is higher than that of $CH_4$ $(-81.9^{\circ} C)$,$CO_2$ has stronger intermolecular forces of attraction.

(N/A) The first complete data on the pressure-volume-temperature relations of a substance in both gaseous and liquid states was obtained by Thomas Andrews for $CO_2$.
From the analysis of the Andrews isotherms:
$(i)$ At high temperatures,isotherms resemble those of an ideal gas,and the gas cannot be liquefied even at very high pressures.
$(ii)$ As the temperature is lowered,the shape of the curve changes,and the data shows a significant deviation from ideal behavior.
$(iii)$ Explanation of $T_C$ (Critical Temperature),$P_C$ (Critical Pressure),and $V_C$ (Critical Volume): At $30.98^{\circ}C$,$CO_2$ remains a gas up to $73 \ atm$ pressure (Point $E$ in the figure). At $73 \ atm$ pressure,liquid $CO_2$ appears for the first time. This point $E$ represents the critical state where the distinction between liquid and gas disappears.

(N/A) Thomas Andrews obtained the first complete data on the pressure-volume-temperature relationships of a substance in both gaseous and liquid states using $CO_2$.
Key observations from the Andrews isotherms:
$(i)$ At high temperatures (e.g.,$50^{\circ}C$),the isotherms resemble those of an ideal gas,and the gas cannot be liquefied regardless of the pressure applied.
$(ii)$ As the temperature is lowered,the shape of the curve changes,showing significant deviation from ideal behavior. $A$ horizontal portion appears,representing the coexistence of liquid and gas phases.
$(iii)$ Critical Constants $(T_C, P_C, V_C)$: Point $E$ represents the critical state. At $30.98^{\circ}C$ $(T_C)$,$CO_2$ remains a gas up to $73 \ atm$ $(P_C)$. At this pressure,liquid $CO_2$ appears for the first time. The volume at this point is the critical volume $(V_C)$.

(N/A) Liquefaction of Real Gases: All real gases,upon isothermal compression,exhibit behavior similar to $CO_2$. To liquefy a gas,it must be cooled below its critical temperature $(T_c)$. The critical temperature is defined as the highest temperature at which a gas can be liquefied by pressure alone.
Liquefaction of Permanent Gases: Permanent gases (gases that show continuous positive deviation in the compressibility factor $Z$) require both significant cooling and high compression. Compression brings molecules into close proximity,while cooling reduces their kinetic energy. Consequently,intermolecular forces become strong enough to hold the molecules together,leading to liquefaction.

(N/A) Liquefaction of a gas can occur without passing through the two-phase (liquid + gas) region if the process is carried out above the critical temperature. This is demonstrated using the $CO_2$ isotherm diagram:
Step-$i$ (Point $A$ to Point $F$): We move from point $A$ to $F$ by increasing the temperature at constant volume,which also increases the pressure.
Step-$ii$ (Point $F$ to Point $G$): We compress the gas at a constant temperature along the isotherm (above the critical temperature,e.g.,$31.1^{\circ}C$). The pressure increases as the volume decreases.
Step-$iii$ (Point $G$ to Point $D$): We move vertically down from $G$ to $D$ by lowering the temperature. As we cross the critical isotherm,the substance transitions into the liquid state without ever entering the heterogeneous two-phase region.
In this series of changes,the substance remains in a single phase throughout the process.

(N/A) There is a continuity between the gaseous and liquid states. The term 'fluid' is used for either a liquid or a gas to recognize this continuity. Thus,a liquid can be viewed as a very dense gas.
$(i)$ $A$ liquid and gas can be distinguished only when the fluid is below its critical temperature and $(ii)$ its pressure and volume lie under the dome.
In that situation,liquid and gas are in equilibrium,and a surface separating the two phases is visible. In the absence of this surface,there is no fundamental way of distinguishing between the two states.
At the critical temperature,a liquid passes into the gaseous state imperceptibly and continuously; the surface separating the two phases disappears.
Substance Vapour: $A$ gas below its critical temperature can be liquefied by applying pressure and is called the 'vapour' of the substance. For example,carbon dioxide gas below its critical temperature is called carbon dioxide vapour.
Critical temperature and pressure of some substances:

Substance	$T_{C} / K$	$p_{C} / bar$	$V_{C} / dm^{3} mol^{-1}$
$H_{2}$	$33.2$	$12.97$	$0.0650$
$He$	$5.3$	$2.29$	$0.0577$
$N_{2}$	$126.0$	$33.9$	$0.0900$
$O_{2}$	$154.3$	$50.4$	$0.0744$
$CO_{2}$	$304.10$	$73.9$	$0.0956$
$H_{2}O$	$647.1$	$220.6$	$0.0450$
$NH_{3}$	$405.5$	$113.0$	$0.0723$

(A) The critical temperature $(T_c)$ of a gas is a measure of the strength of intermolecular forces of attraction. Higher $T_c$ indicates stronger intermolecular forces,making the gas easier to liquefy.
For $CO_2$,$T_c = 31.1 \, ^\circ C$,and for $CH_4$,$T_c = -81.9 \, ^\circ C$.
Since $31.1 \, ^\circ C > -81.9 \, ^\circ C$,the intermolecular forces in $CO_2$ are stronger than in $CH_4$.
Therefore,$CO_2$ can be liquefied more easily than $CH_4$ at a given temperature.

(B) $SO_2$ gas can be liquefied at room temperature by applying a pressure of approximately $2 \text{ atm}$.
This is because $SO_2$ has a relatively high critical temperature $(430 \text{ K})$,allowing it to be easily liquefied under moderate pressure.

(N/A) The temperature above which a gas cannot be liquefied,no matter how much pressure is applied,is called the critical temperature $(T_c)$.
Gas $A$ liquefies upon a slight increase in pressure,which indicates that its temperature is already below its critical temperature $(T < T_c)$.
Gas $B$ does not liquefy even under high pressure unless it is cooled,which indicates that its temperature is currently above its critical temperature $(T > T_c)$. Cooling the gas brings its temperature below the critical temperature,allowing it to be liquefied by pressure.

(N/A) This statement is correct only for an ideal gas. It is impossible to liquefy an ideal gas because there are no intermolecular forces of attraction between the molecules of an ideal gas. Liquefaction requires the presence of intermolecular forces to bring molecules closer together to form a liquid phase.

(B) The critical temperature $(T_c)$ of $CO_2$ is $30.98\,^{\circ}C$.
Above the critical temperature,a gas cannot be liquefied by the application of pressure alone,regardless of how high the pressure is.
Since $32\,^{\circ}C > 30.98\,^{\circ}C$,$CO_2$ exists as a supercritical fluid at this temperature and cannot be liquefied by increasing the pressure to $80\,atm$.

(A) $(i)$ At temperature $T_1$,between points $a$ and $b$,$CO_2$ exists in the gaseous state.
$(ii)$ At point $b$,the isotherm becomes horizontal,indicating the start of the phase transition where $CO_2$ begins to liquefy.
$(iii)$ At temperature $T_2$,the $CO_2$ is completely liquefied at point $g$.
$(iv)$ Condensation will not take place at temperature $T_3$ because $T_3$ is greater than the critical temperature $(T_c)$.
$(v)$ The portion of the isotherm at $T_1$ between points $b$ and $c$ represents the equilibrium between liquid and gaseous $CO_2$.