Critical state and Liquefaction of gases Questions in English

(N/A) Below the critical temperature in a closed vessel,the surface of separation (meniscus) between the liquid and its vapour is clearly visible.
As the temperature approaches the critical point,the density of the liquid decreases due to expansion,and the density of the vapour increases due to compression.
At the critical temperature,the densities of the liquid and the vapour become equal,and the surface of separation disappears. The liquid and gaseous states are no longer distinguishable.
The fluid,which is now a homogeneous mixture,is called a supercritical fluid. Any fluid above its critical temperature and pressure is referred to as a supercritical fluid.
Supercritical fluids are used for the extraction of organic substances. For example,$CO_{2}$ above $31.1^{\circ}C$ and above $73.8 \ bar$ pressure has a density of about $1 \ g/cm^{3}$. It is used to dissolve caffeine from coffee beans as a safer alternative to chlorofluorocarbons.

(N/A) It is possible to change a gas into a liquid or a liquid into a gas by a process such that there is a single phase present at all times.
For example,in the given figure,we can move from point $A$ vertically upwards by increasing the temperature above the critical temperature $(31.1^{\circ}C)$.
Then,we can compress the gas at this constant temperature along the isotherm. The pressure will increase.
Finally,we can move vertically downwards by lowering the temperature. As we cross the region above the critical point $E$,we obtain the liquid phase.
Thus,at no stage during this process do we pass through the two-phase region. This process is carried out above the critical temperature $T_c$,where the substance always remains in a single phase. This is known as the continuity of state between the gaseous and the liquid state.

(C) The liquefaction of gases depends on both pressure and temperature. $1$. Increasing pressure brings gas molecules closer together. $2$. Decreasing temperature reduces the kinetic energy of gas molecules,allowing intermolecular forces to hold them together in the liquid state. Thus,liquefaction is most effective at high pressure and low temperature.

(D) The critical constants are defined as follows:
$(i)$ Critical temperature $(T_C)$
$(ii)$ Critical pressure $(P_C)$
$(iii)$ Critical volume $(V_C)$
Therefore,all the mentioned parameters are critical constants.

(N/A) The critical temperature of a gas is the temperature above which it cannot be liquefied,no matter how much pressure is applied.
For $CO_2$,the critical temperature is $30.98^{\circ}C$ (at $73.0 \ atm$ pressure).

(C) The critical temperature of $CO_2$ is $30.98 \, ^\circ C$.
Above this temperature,$CO_2$ exists only as a gas and cannot be liquefied by pressure alone.
Below this temperature,$CO_2$ can exist as a liquid or a gas-liquid equilibrium mixture depending on the pressure applied.

(N/A) The critical temperature $(T_C)$ of $CO_2$ is $31.1^{\circ}C$ (standard value).
At $30.98^{\circ}C$ (which is approximately $T_C$),$CO_2$ exists in equilibrium between gas and liquid phases.
At $13.1^{\circ}C$ $(T < T_C)$,$CO_2$ exists as a liquid at high pressure.
At $31.5^{\circ}C$ and $50^{\circ}C$ $(T > T_C)$,$CO_2$ exists as a gas regardless of the pressure applied.

(C) The temperature $30.98 \, ^\circ C$ is the critical temperature of $CO_2$.
At this temperature,the (gas + liquid) coexistence region is observed.
As the pressure increases (compression),the gas gradually liquefies.
As the pressure decreases (expansion),the liquid turns back into gas.
Beyond the critical point,it exists as a supercritical fluid.

(N/A) substance is called a vapor if it exists in the gaseous state below its critical temperature and can be liquefied by applying pressure. $CO_2$ is referred to as $CO_2$ vapor when it is at a temperature below its critical temperature of $31.1 \ ^\circ C$.

(A) Liquefaction of gases is easier for gases with higher critical temperatures because they can be liquefied at higher temperatures.
Since the critical temperature of $O_2$ $(154.3 \ K)$ is higher than that of $N_2$ $(126.0 \ K)$,$O_2$ will liquefy first during the cooling process of air.

(A) substance is called a $vapor$ if its temperature is below its critical temperature $(T_{c})$.
It can be easily liquefied by applying pressure.
$A$ substance is called a $gas$ if its temperature is above its critical temperature $(T_{c})$.
It cannot be liquefied by applying pressure alone,regardless of how much pressure is applied.

(C) At the critical temperature $(T_c)$,the distinction between the liquid and gaseous phases disappears. Since surface tension is a property arising from the intermolecular forces at the interface between a liquid and its vapor,it becomes $0$ at the critical temperature.

(A) Critical temperature $(T_{C})$ is defined as the temperature above which a gas cannot be liquefied,no matter how much pressure is applied.
At this temperature,the density of the liquid and its vapor become identical,and the boundary between the liquid and the gas phase disappears.

(D)
All given statements are correct.
$(i)$ At temperatures above the critical temperature $(T_c)$,such as $T_1$,the gas cannot be liquefied by applying pressure alone.
$(ii)$ At $T_2$,point $B$ represents the onset of condensation where the gas begins to liquefy.
$(iii)$ $T_c$ (critical temperature) is defined as the highest temperature at which a gas can be liquefied by pressure.
$(iv)$ At point $A$,the substance is entirely in the liquid state; further compression leads to a sharp increase in pressure as liquids are nearly incompressible.

(A) At point $(a)$,$CO_2$ exists as a gas.
At point $(b)$,the liquefaction of $CO_2$ starts.
At point $(c)$,the liquefaction ends.
Between points $(b)$ and $(c)$,liquid and gaseous $CO_2$ coexist.
As the volume decreases from $(b)$ to $(c)$,the amount of gas decreases and the amount of liquid increases.
Evaluating the statements:
$A.$ Correct: $CO_2$ remains as a gas up to point $(b)$.
$B.$ Incorrect: Liquid $CO_2$ starts appearing at point $(b)$,not $(c)$.
$C.$ Correct: Liquid and gaseous $CO_2$ coexist between points $(b)$ and $(c)$.
$D.$ Incorrect: As the volume decreases from $(b)$ to $(c)$,the amount of liquid increases,not decreases.
Therefore,there are $2$ correct statements ($A$ and $C$).

(D) The ease of liquefaction of a gas is directly proportional to its critical temperature $(T_C)$.
As the temperature is lowered from $500 \ K$,the gas with the highest critical temperature will reach its liquefaction point first.
Comparing the given critical temperatures: $NH_3$ $(405.5 \ K)$ > $CO_2$ $(304.1 \ K)$ > $N_2$ $(126 \ K)$ > $He$ $(5.3 \ K)$.
Since $NH_3$ has the highest critical temperature,it will liquefy first.

(B) According to Andrews' experiments on carbon dioxide,the gas obeys Boyle's law only at temperatures above its critical temperature.
For carbon dioxide,the critical temperature $(T_c)$ is $31.1^{\circ}C$.
At temperatures below $31.1^{\circ}C$,the gas can be liquefied by pressure,and it deviates significantly from ideal gas behavior.
Therefore,the minimum temperature at which $CO_2$ behaves like an ideal gas (obeying Boyle's law) is its critical temperature,which is $31.1^{\circ}C$.

(B) According to Andrews' experiments on $CO_2$ isotherms,the critical pressure $(P_C)$ of carbon dioxide is $73 \ atm$.
At this critical pressure,the temperature at which the gas starts to condense (or the temperature at which the gas becomes indistinguishable from its liquid phase) is known as the critical temperature $(T_C)$.
For $CO_2$,the critical temperature $(T_C)$ is $30.98^{\circ} C$.

(D) The ease of liquefaction of a gas depends on its critical temperature $(T_{c})$.
Gases with higher critical temperatures are easier to liquefy because they can be liquefied at higher temperatures.
The critical temperatures of the given gases are approximately: $SO_2$ $(430 \ K)$,$Cl_2$ $(417 \ K)$,$NH_3$ $(405 \ K)$,and $O_2$ $(154.6 \ K)$.
Since $O_2$ has the lowest critical temperature,it is the most difficult to liquefy among the given options.

(A) The ease of liquefaction of a gas depends on the magnitude of intermolecular forces of attraction.
$SO_2$,$Cl_2$,and $NH_3$ are polar molecules,which possess stronger dipole-dipole interactions.
$O_2$ is a non-polar molecule,which only possesses weak London dispersion forces.
Therefore,$O_2$ has the lowest critical temperature and is the most difficult to liquefy.

(A) The critical temperature of water is approximately $647 \ K$ $(647.096 \ K)$.

(C) Gases with higher critical temperatures and stronger intermolecular forces of attraction are easier to liquefy.
Among the given options,$Cl_{2}$ (chlorine) has a relatively high critical temperature $(417 \ K)$ and strong van der Waals forces,making it easy to liquefy.
In contrast,gases like $O_{2}$,$N_{2}$,and $He$ have very low critical temperatures and are considered permanent gases,which are difficult to liquefy at room temperature.

(D) Critical temperature $(T_c)$ is defined as the temperature above which a gas cannot be liquefied,regardless of the pressure applied.
It is not the lowest temperature at which liquefaction occurs; rather,it is the maximum temperature at which a gas can exist as a liquid.
Therefore,statement $(D)$ is incorrect.

(A) Based on the phase diagram,the liquid phase is bounded by the triple point at the lower temperature limit and the critical point at the upper temperature limit. Below the triple point,the substance exists as a solid or vapor. Above the critical temperature $(T_c)$,the substance exists as a supercritical fluid,where the distinction between liquid and gas disappears. Therefore,a liquid can exist only between the triple point and the critical point.

(D) The ease of liquefaction of a gas is directly proportional to its critical temperature $(T_c)$.
$C$ has the highest critical temperature,which is $405.5 \ K$.
Therefore,gas $C$ will be the easiest to liquefy and will get liquefied first upon cooling.

(C) Given,$T_c = \frac{8 a}{27 R b}$,$p_c = \frac{a}{27 b^2}$,$V_c = 3 b$.
The compressibility factor $(Z)$ is defined as $Z = \frac{p V}{R T}$.
At the critical state,$Z = \frac{p_c V_c}{R T_c}$.
Substituting the given values:
$Z = \frac{(\frac{a}{27 b^2}) \times (3 b)}{R \times (\frac{8 a}{27 R b})}$.
$Z = \frac{\frac{3 a}{27 b}}{\frac{8 a}{27 b}} = \frac{3}{8}$.
Thus,the correct option is $(C)$.

(A, C, D) Above the critical temperature $(T_c)$,a gas cannot be liquefied by the application of pressure alone.
At this temperature,the distinction between the liquid and gas phases disappears,meaning the liquid phase cannot be distinguished from the gas phase.
Furthermore,the density of the substance changes continuously with changes in pressure $(P)$ or volume $(V)$ as it exists as a supercritical fluid.
Therefore,statements $A$,$C$,and $D$ are applicable.

(B) gas can be liquefied at a temperature $T$ only when it is below its critical temperature,$T_{c}$.
At temperatures below $T_{c}$,the gas can be liquefied by applying sufficient pressure,specifically when the pressure $p$ is greater than the critical pressure,$p_{c}$.
Therefore,the condition for liquefaction is $T < T_{c}$ and $p > p_{c}$.