Answer

(c) Pressure at the top of a mountain [or at high altitude] is low. This suggests that boiling takes place at lower temperature, due to which things take more time to boil.

However, in a pressure cooker, pressure is increased and hence, boiling point increases. Thus, things comes to boil in a pressure cooker in a less period of time.

• (a) Pressure increases: This is incorrect because at high altitudes, atmospheric pressure actually decreases, not increases. Increased pressure would lead to a higher boiling point, which is not the case at high altitudes.

• (b) Temperature decreases: While it is true that temperatures can be lower at high altitudes, this is not the primary reason why cooking takes longer. The main factor is the decreased atmospheric pressure, which lowers the boiling point of water.

• (d) Temperature increases: This is incorrect because temperatures generally decrease with altitude. Even if temperatures were to increase, it would not explain the longer cooking times, which are primarily due to the lower boiling point caused by decreased pressure.

Answer

(b) The spherical shape of rain droplets is due to surface tension. The lowest energy state of a liquid will be when the surface area is minimum. Surface tension tries to decrease the surface area of the liquid to the minimum. The rain droplets are spherical because for a given volume, a sphere has minimum surface area.

• Viscosity: Viscosity is a measure of a fluid’s resistance to flow. It does not directly influence the shape of droplets but rather affects how quickly they can move or spread. Therefore, it is not responsible for the spherical shape of rain droplets.

• Critical phenomena: Critical phenomena refer to the behavior of physical systems during phase transitions, such as the transition from liquid to gas. This concept does not explain the shape of droplets but rather the conditions under which different phases coexist. Hence, it is not related to the spherical shape of rain droplets.

• Pressure: Pressure is the force exerted per unit area. While pressure can influence the formation and behavior of droplets, it does not determine their shape. The spherical shape is specifically due to surface tension minimizing the surface area, not pressure.

Thinking Process

The question is based upon Boyle’s law. It states that at constant temperature, the volume of a given proportional to its pressure. i.e., $V \propto \frac{1}{p}$ or, $p V=$ constant

Answer

(c) At a particular temperature, $p V=$ constant

$$ \begin{aligned} Thus, \quad p_{1} V_{1} & =p_{2} V_{2}=p_{3} V_{3}=p_{4} V_{4} \\ As \quad V_{1} & >V_{2}>V_{3}>V_{4} \\ Therefore, \quad p_{1} & <p_{2}<p_{3}<p_{4} \end{aligned} $$

• Option (a) $p_{1}>p_{2}>p_{3}>p_{4}$: This option is incorrect because it suggests that the pressure decreases as the volume increases. However, according to the ideal gas law (PV = constant at constant temperature), if the volume increases, the pressure should decrease. Therefore, the order should be the opposite of what is stated in this option.

• Option (b) $p_{1}=p_{2}=p_{3}=p_{4}$: This option is incorrect because it implies that the pressure is the same for all volumes. If the pressure were the same, the volume would also be the same for a given temperature, which contradicts the given plot where different lines represent different pressures.

• Option (d) $p_{1}<p_{2}=p_{3}<p_{4}$: This option is incorrect because it suggests that two of the pressures are equal (i.e., $p_{2}$ and $p_{3}$). However, the plot shows distinct lines for each pressure, indicating that all pressures are different. Therefore, this option does not correctly represent the relationship between pressure and volume as depicted in the plot.

Answer

(c) London dispersion forces operate only over very short distance. The energy of interaction varies as $\frac{1}{\text { (distance between two interacting particles)}^{6}}$.

Large or more complex are the molecules, greater is the magnitude of London forces. This is obviously due to the fact that the large electron clouds are easily distorted or polarised.

Hence, greater the polarisability of the interacting particles, greater is the magnitude of the interaction energy.

• (a) Charge of interacting particles: London dispersion forces are a type of van der Waals force that arise due to temporary dipoles induced in atoms or molecules. These forces do not depend on the charge of the interacting particles, as they occur even between neutral atoms and molecules.

• (b) Mass of interacting particles: While the mass of the particles can influence the overall strength of the London dispersion forces indirectly (since larger mass often correlates with larger size and more electrons), the interaction energy itself is not directly dependent on the mass. It is more directly related to the polarisability of the electron cloud.

• (d) Strength of permanent dipoles in the particles: London dispersion forces are specifically due to temporary dipoles and not permanent dipoles. The strength of permanent dipoles is relevant to other types of intermolecular forces, such as dipole-dipole interactions, but not to London dispersion forces.

Answer

(c) Dipole-dipole forces act between the molecules possessing permanent dipole and ends of dipoles possess ‘partial charges’. Partial charges present on ends of a dipole are always less than the unit electronic charge.

e.g.,

• (a) more than unit electronic charge: This option is incorrect because partial charges on the ends of a dipole are always less than the unit electronic charge. A unit electronic charge refers to the charge of a single electron or proton, which is a fundamental constant. Partial charges arise due to the unequal sharing of electrons between atoms in a molecule and are always a fraction of the unit electronic charge.

• (b) equal to unit electronic charge: This option is incorrect because partial charges are not equal to the unit electronic charge. The unit electronic charge is a fixed value (approximately 1.602 x 10^-19 coulombs), while partial charges are smaller and result from the distribution of electron density in a molecule. They represent a fraction of the unit electronic charge.

• (d) double the unit electronic charge: This option is incorrect because partial charges cannot exceed the unit electronic charge, let alone be double it. Partial charges are always less than the unit electronic charge, as they represent the slight imbalance in electron distribution between atoms in a molecule, not a full transfer of electrons.

Answer

(c) Given that, pressure of a $1: 4$ mixture of $H_{2}$ and $O_{2}$ enclosed in a vessel is one atmosphere. This suggests that the moles ratio of $H_{2}$ and $O_{2}$ is $1: 4$. Thus, partial pressure of dioxygen $\left(O_{2}\right)$ is given as

$\therefore$ Partial pressure of $O_{2}=$ Mole fraction of $O_{2} \times$ total pressure of mixture

$$ \begin{aligned} & =\frac{4}{1+4} \times 1=\frac{4}{5} \times 1 \mathrm{~atm} \\ & =0.8 \mathrm{~atm}=0.8 \times 10^{5} \mathrm{Nm}^{-2}=8 \times 10^{4} \mathrm{Nm}^{-2} \end{aligned} $$

• Option (a) $0.8 \times 10^{5} \mathrm{~atm}$: This option is incorrect because it incorrectly represents the unit of pressure. The correct partial pressure of dioxygen is $0.8 \mathrm{~atm}$, not $0.8 \times 10^{5} \mathrm{~atm}$.

• Option (b) $0.008 \mathrm{Nm}^{-2}$: This option is incorrect because it significantly underestimates the partial pressure of dioxygen. The correct partial pressure is $8 \times 10^{4} \mathrm{Nm}^{-2}$, which is much higher than $0.008 \mathrm{Nm}^{-2}$.

• Option (d) $0.25 \mathrm{~atm}$: This option is incorrect because it incorrectly calculates the partial pressure of dioxygen. Given the mole ratio of $1:4$ for $H_{2}$ and $O_{2}$, the correct partial pressure of dioxygen is $0.8 \mathrm{~atm}$, not $0.25 \mathrm{~atm}$.

Thinking Process

This problem is based on Gay-Lussac’s law which states that “volume remaining constant, the pressure of a given mass of a gas is directly proportional to its temperature in degrees Kelvin”.

Answer

(a) As the temperature increases, average kinetic energy of molecules increases. From Gay-Lussac’s law, at constant volume, as the temperature is increased, pressure increases.

• (b) Decreases: This option is incorrect because, according to Gay-Lussac’s law, at constant volume, an increase in temperature results in an increase in pressure, not a decrease.

• (c) Remains same: This option is incorrect because, according to Gay-Lussac’s law, at constant volume, an increase in temperature leads to an increase in pressure, so the pressure cannot remain the same.

• (d) Becomes half: This option is incorrect because, according to Gay-Lussac’s law, at constant volume, an increase in temperature causes an increase in pressure, not a decrease to half.

Answer

(d) Higher the critical temperature, more easily is the gas liquified. Hence, order of liquefaction starting with the gas liquefying first will be $O_{2}, N_{2}, H_{2}$, $He$.

Note Critical temperature of a gas may be defined as that temperature above which it cannot be liquified howsoever high pressure may be applied on the gas.

• Option (a) $H_{2}, He, O_{2}, N_{2}$:

  • Incorrect because $H_{2}$ and $He$ have the lowest critical temperatures (33.2 K and 5.3 K respectively), meaning they are the hardest to liquefy. They should not be listed first in the order of liquefaction.

• Option (b) $He, O_{2}, H_{2}, N_{2}$:

  • Incorrect because $He$ has the lowest critical temperature (5.3 K), making it the hardest to liquefy. It should be listed last, not first.

• Option (c) $N_{2}, O_{2}, He, H_{2}$:

  • Incorrect because $N_{2}$ has a critical temperature of 126 K, which is lower than $O_{2}$ (154.3 K). Therefore, $N_{2}$ should not be listed before $O_{2}$ in the order of liquefaction.

Answer

(b) The SI unit of viscosity coefficient $(\eta)$ is $\mathrm{Nm}^{-2} \mathrm{~s}$ or $\mathrm{Nsm}^{-2}$.

$\begin{array}{rlrl} & \text { As we know that, } & f & =\eta A \frac{d v}{d x} \\ \text { where, } & f =\text { force } \\ \eta & =\text { viscosity coefficient } \\ \frac{d v}{d x} & =\text { velocity gradient }\end{array}$

Substitute SI units of $f=N, d x=m, A=m^{2}$ and $v=m s^{-1}$ in above equation, to get,

$$ \eta=\frac{\mathrm{N} \times \mathrm{m}}{\mathrm{m}^{2} \times \mathrm{ms}^{-1}}=\mathrm{Nm}^{-2} \mathrm{~s} $$

Hence, the si unit of $\eta=$ is $n s m^{-2}$

• (a) Pascal: Pascal (Pa) is the SI unit of pressure, not viscosity. Viscosity is a measure of a fluid’s resistance to flow, whereas pressure is the force exerted per unit area.

• (c) $\mathrm{km}^{-2} \mathrm{~s}$: This unit is incorrect because it combines kilometers (a unit of length) with seconds in a way that does not correspond to the physical dimensions of viscosity. Viscosity involves force, area, and velocity gradient, not just length and time.

• (d) $\mathrm{Nm}^{-2}$: Newton per square meter (Nm$^{-2}$) is the unit of pressure (Pascal), not viscosity. Viscosity requires an additional time dimension (seconds) to account for the rate of deformation of the fluid.

Thinking Process

(i) A liquid boils when vapour pressure above it becomes equal to the atmospheric pressure. Lower the atmospheric pressure, lower is the boiling point.

(ii) Arrange the four cities according to the increase of atmospheric pressures and pick out the lowest one.

Answer

(a) Shimla has the lowest atmospheric pressure among all the four cities. Thus, at Shimla liquid will boil first because lower the atmospheric pressure, lower is the boiling point.

• Bangalore: Bangalore has a higher atmospheric pressure ($1.2 \times 10^{5} , \text{N/m}^2$) compared to Shimla. Higher atmospheric pressure increases the boiling point of a liquid, so it will not boil first in Bangalore.

• Delhi: Delhi has a slightly higher atmospheric pressure ($1.02 \times 10^{5} , \text{N/m}^2$) compared to Shimla. This means the boiling point of a liquid will be slightly higher in Delhi than in Shimla, so it will not boil first in Delhi.

• Mumbai: Mumbai has the highest atmospheric pressure ($1.21 \times 10^{5} , \text{N/m}^2$) among all the cities listed. Higher atmospheric pressure results in a higher boiling point for liquids, so it will not boil first in Mumbai.

Thinking Process

(i) The graph given in the question depend upon the Boyle’s law.

(ii) At constant temperature, the graph between $\mathrm{pV}$ vs $\mathrm{p}$ at all pressures will be straight line parallel to $x$-axis for ideal gas.

(iii) At constant temperature $pV$ vs $p$ plot for real gases is not straight line.

Answer

(a) For ideal gas $p V=$ constant at all pressures. Therefore, only $B$ represents ideal gas.

• Option (b) $C$ and $D$: These curves do not represent an ideal gas because they do not show a constant $pV$ product at all pressures. For an ideal gas, the $pV$ product should remain constant, but these curves likely show variations in $pV$ with pressure, indicating non-ideal behavior.

• Option (c) $E$ and $F$: Similar to option (b), these curves do not represent an ideal gas because they do not maintain a constant $pV$ product at all pressures. The variations in $pV$ with pressure suggest deviations from ideal gas behavior.

• Option (d) $A$ and $B$: While curve $B$ represents an ideal gas with a constant $pV$ product, curve $A$ does not. Curve $A$ likely shows a $pV$ product that changes with pressure, indicating non-ideal gas behavior.

Answer

(c) With increase of temperature, the kinetic energy of the molecules of liquid increases which can overcome the intermolecular forces. Hence, the liquid starts flowing.

In other words the viscosity of a liquid decreases with increase in temperature. It has been found that the decrease is about $2 $% per degree rise in temperature.

• (a) Increase: This option is incorrect because an increase in temperature leads to an increase in the kinetic energy of the molecules, which reduces the intermolecular forces of attraction, causing the liquid to flow more easily and thus decreasing its viscosity, not increasing it.

• (b) No effect: This option is incorrect because temperature has a significant effect on the kinetic energy of molecules. As temperature increases, the kinetic energy increases, which reduces the intermolecular forces and decreases the viscosity of the liquid.

• (d) No regular pattern will be followed: This option is incorrect because there is a well-established relationship between temperature and viscosity. As temperature increases, the viscosity of a liquid consistently decreases, following a regular pattern.

Answer

(b) The surface tension of liquids generally decreases with increase of temperature and becomes zero at the critical temperature. The decrease in surface tension with increase of temperature is obviously due to the fact that with increase of temperature, the kinetic energy of the molecules increases and therefore, the intermolecular attraction decreases.

Note Surface tension of a liquid is defined as the force acting at right angles to the surface along one centimetre length of the surface. Units of surface tension are dyne per $\mathrm{cm}$ (or newtons per metre i.e., $\mathrm{Nm}^{-1}$ in the SI systems).

• (a) Remains same: This option is incorrect because surface tension is influenced by temperature. As temperature increases, the kinetic energy of the molecules also increases, leading to a decrease in intermolecular forces and thus a decrease in surface tension. Therefore, surface tension does not remain the same with an increase in temperature.

• (c) Increases: This option is incorrect because an increase in temperature results in higher kinetic energy of the molecules, which reduces the intermolecular forces. Consequently, the surface tension decreases rather than increases with an increase in temperature.

• (d) No regular pattern is followed: This option is incorrect because there is a well-established pattern that surface tension decreases with an increase in temperature. The relationship between temperature and surface tension is predictable and consistent, contrary to the suggestion that no regular pattern is followed.

Answer

$(b, c)$

In gaseous state, molecules are in a state of random motion i.e., it is the state in which molecules are disorderly arranged. Gaseous state have higher entropy than the liquid as well as solid.

• (a) Complete order of molecules: This is incorrect because in the gaseous state, molecules are not in a state of complete order. Instead, they are in a state of complete disorder and move randomly.

• (d) Fixed position of molecules: This is incorrect because in the gaseous state, molecules do not have fixed positions. They move freely and randomly throughout the available space.

Answer

$(a, b, d)$

It is known that, 1 mole of a gas = gram molecular mass = Avogadro’s number of molecules $=22.4 ~L$ at STR

Thus, 1 mole of $O_{2}$ gas $=32 g$ of $O_2=6.02 \times 10^{23}$ molecules $=22.4 ~L$ of $O_{2}$ at STP

• (a) $16 \mathrm{~g}$ of gas: This is incorrect because 1 mole of dioxygen gas ($O_2$) has a molar mass of 32 grams, not 16 grams. Therefore, 16 grams of $O_2$ represents only 0.5 moles of dioxygen gas.

• (b) $22.7 \mathrm{~L}$ of gas: This is incorrect because at STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters, not 22.7 liters. Therefore, 22.7 liters of $O_2$ represents slightly more than 1 mole of dioxygen gas.

• (d) $11.2 \mathrm{~L}$ of gas: This is incorrect because at STP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, 11.2 liters of $O_2$ represents only 0.5 moles of dioxygen gas.

Answer

( $b, c$ )

A gas which obeys the ideal gas equation, $p V=n R T$ under all conditions of temperature and pressure is called an ‘ideal gas’.

However, there is no gas which obeys the ideal gas equation under all conditions of temperature and pressure. Hence, the concept of ideal gas is only theoretical or hypothetical. The gases are found to obey the gas laws fairly well when the pressure is low or the temperature is high.

Such gases are, therefore, known as ‘real gases’. All gases are real gases. Hence, at high pressure and low temneratıre a real gas deviates most from ideal hehaviour.

• At low pressure, gas molecules are far apart, and intermolecular forces are negligible, so the gas behaves more ideally.
• At high temperature, the kinetic energy of gas molecules is high, which overcomes intermolecular forces, leading to more ideal behavior.

Answer

$(b, d)$

When salt is added, less surface area is available for the water molecules to vaporise because salt molecules occupy the position of water molecules on the surface. Hence, vapour pressure of the water decreases.

Moreover, vapour pressure also decreases with decrease in temperature because kinetic energy of the molecules decrease.

• Decreasing the quantity of water: The vapour pressure of a liquid in a sealed container depends on the temperature and the nature of the liquid, not on the quantity of the liquid. Therefore, decreasing the quantity of water does not affect the vapour pressure.

• Decreasing the volume of the vessel to one-half: The vapour pressure of a liquid in a sealed container is independent of the volume of the container. It is determined by the temperature and the nature of the liquid. Hence, decreasing the volume of the vessel does not change the vapour pressure.

Answer

From Avogadro’s law, we know that

Volume of 1 mole of the gas $=$ gram molecular mass $=22.4 \mathrm{~L}$ at STP

Volume occupied by $28 \mathrm{~g} \mathrm{CO}(1 \mathrm{~mol} \mathrm{CO})=22.4 \mathrm{~L}$ at STP

$$ \left(\because \text { Molar mass of } C O=12+16=28 \mathrm{~g} \mathrm{~mol}^{-1}\right) $$

$\therefore \quad$ Volume occupied by $1 \mathrm{~g} \mathrm{CO}=\frac{22.4}{28} \mathrm{~L}$ at STP

Similarly, volume occupied by $1 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}=\frac{22.4}{18} \mathrm{~L}$ at STP

$$ \left(\because \text { Molar mass of } \mathrm{H}_{2} \mathrm{O}=(2 \times 1)+16=18 \mathrm{~g} \mathrm{~mol}^{-1}\right) $$

Volume occupied by $1 \mathrm{~g} \mathrm{CH}_{4}=\frac{22.4}{16} \mathrm{~L}$ at STP

$$ \left(\because \text { Molar mass of } \mathrm{CH}_{4}=12+(4 \times 1)=16 \mathrm{~g} \mathrm{~mol}^{-1}\right) $$

Volume occupied by $1 \mathrm{~g} \mathrm{NO}=\frac{22.4}{30} \mathrm{~L}$ at STP

$$ \left(\because \text { Molar mass of } \mathrm{NO}=14+16=30 \mathrm{~g} \mathrm{~mol}^{-1}\right) $$

Thus, $1 \mathrm{~g} \mathrm{CH}_{4}$ will occupy maximum volume while $1 \mathrm{~g}$ of $\mathrm{NO}$ will occupy minimum volume at STP.

Answer

Physical properties of ice, water and steam are very different because they found in different states. Ice found in solid, water found in liquid and steam found in vapour states. The chemical composition of water in all the three states (ice, water and steam) is same, i.e., $\mathrm{H}_{2} \mathrm{O}$.

Answer

Temperature, pressure, mass and volume are the factors which determine the different states of matter. i.e., solid, liquid and gas.

Answer

From the information and data given in the question, we concluded that

(a) In $\mathrm{HCl}, \mathrm{HBr}$ and $\mathrm{HI}$, dipole-dipole and London forces are present because molecules possess permanent dipole. In HF dipole-dipole, London forces and hydrogen bonding are present.

(b) Electronegativity of chlorine, bromine and iodine decreases in the order are present

$$ \mathrm{Cl}>\mathrm{Br}>\mathrm{I} $$

Therefore, dipole moment should decrease from $\mathrm{HCl}$ to $\mathrm{HI}$ Thus, dipole-dipole interaction should decrease from HCIto HI But boiling point increases on moving from $\mathrm{HCl}$ to $\mathrm{HI}$ This means that London forces are predominant.

This is so because London forces increases as the number of electrons in a molecule increases and in this case number of electrons is increasing from $\mathrm{HCl}$ towards $\mathrm{HI}$

(c) Hydrogen fluoride has highest dipole moment attributes due to highest electronegativity of fluorine as well as presence of hydrogen bonding in HF. Therefore, HF has highest boiling point.

Answer

When temperature and pressure of a gas is $273.15 \mathrm{~K}$ (or $0{ }^{\circ} \mathrm{C}$ ) and $1 \mathrm{~atm}\left(\right.$ or 1 bar or $10^{5}$ pascal), such conditions are called standard temperature and pressure conditions (STP). Under these conditions, the volume occupied by 1 mole of each and every gas is $22.4 \mathrm{~L}$. Hence, the moler volume of $\mathrm{N}_{2}$ and $\mathrm{Ar}$ at $273.15 \mathrm{~K}$ and $1 \mathrm{~atm}$ is $22.4 \mathrm{~L}$.

Answer

At low pressure and high temperature, a real gas behaves as an ideal gas. Almost all gases are real gas.

Answer

The temperature above which a gas cannot be liquefied howsoever high pressure may be applied on the gas is called critical temperature. Since, gas ‘A’ liquifies easily, this suggests gas ’ $A$ ’ is below its critical temperature.

On the other hand, gas ’ $B$ ’ does not liquefy easily even on applying high pressure. This suggests that gas ’ $\mathrm{B}$ ’ is above its critical temperature.

Answer

Unit of $R$ depends upon those units in which $p, V$ and $T$ are measured as, $R=\frac{p V}{n T}$. If pressure is measured in Pascal, per mole volume is measured in $\mathrm{m}^{3}$ and temperature is measured in Kelvin then units of ’ $R$ ’ are $\mathrm{Pam}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ or $\mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$. Since, joule is the unit of work done, so ’ $R$ ’ is work done by the gas per mole per kelvin.

Answer

This statement is correct only for ideal gases. It is not possible to liquefy an ideal gas because there is no intermolecular forces of attractions between the molecules of an ideal gas.

Answer

In the above given molecules, only hexane $(CH_3(CH_2)_4 CH_3)$ is a non-polar molecule in which only London dispersion forces exist. These forces are very weak while both water and alcohol are polar molecules in which dipole-dipole interactions as well as $H$-bonding exists.

However, $\mathrm{H}$-bonding interactions are much stronger in water than alcohol, therefore, it possesses stronger intermolecular forces of attraction than alcohol and hexane. Hence, the increasing order of surface tension is

hexane $<$ alcohol $<$ water

Greater is the attractive forces between the molecules, greater is the magnitude of surface tension of liquid.

Answer

Whenever a gas is collected over water, it is moist and saturated with water vapours which exert their own pressure. The pressure due to water vapours is called aqueous tension thus, the total pressure of the gas ( $p$ moist gas) is

Thus, $p_{\text {dry gas }}$ is given as

$$ P_{\text {moist gas }}=p_{\text {dry gas }}+\text { aqueous tension } $$

$P_{\text {dry gas }}=P_{\text {moist gas }}-$ aqueous tension Hence, the correction term applied to the total pressure of the gas in order to obtain pressure of dry gas is $p$ moist gas - aqueous tension.

Answer

The energy which arises due to motion of atoms or molecules in a body is known as thermal energy. It is a measure of average kinetic energy of the particles. It increases with increase in temperature.

Answer

$\mathrm{H}-\mathrm{F}$ is a polar covalent molecule in which dipole-dipole interactions exists. Since, it contains an electronegative atom, $\mathrm{F}$ bonded to $\mathrm{H}$ atom, therefore, $\mathrm{H}$-bond exists between its molecules. Due to these $\mathrm{H}$-bonds, $-\mathrm{HF}$ molecules exists in liquid state. Hence, in liquid state, $\mathrm{H}$-bonds as well as dipole dipole interactions exists within $\mathrm{HF}$ molecules.

Answer

Real gases can be liquefied by cooling and compressing the gas. This proves that force of attraction exist among the molecules.

$$ \ldots \stackrel{S t}{H}-F \ldots \underset{H \text {-bonds } \rightarrow}{\stackrel{S t}{H}}-\stackrel{S-}{F} \ldots \stackrel{S T}{H}-F \ldots . H-\stackrel{S t}{H} \ldots $$

Answer

(i) For ideal gas, compressibility factor, $Z=1$.

(ii) Above Boyle’s temperature, real gases show positive deviation. So, $Z>1$

Note Following important points come out from the compressibility factor, $Z=\frac{p V}{n R T}$.

(i) For ideal gas $Z=1$ at all temperatures and pressures because $p V=n R T$.

(ii) At very low pressure all gases shown have $Z \simeq 1$ and behave as ideal gas.

(iii) At high pressure, all the gases have $Z>1$. These are more difficult to compress.

(iv) At intermediate pressures most gases have $Z<1$.

Answer

The given critical temperature and pressure of $CO_{2}$ gas are $30.98^{\circ} C$ and $73 ~atm$ respectively. This suggests that $CO_{2}$ gas cannot be liquified above $30.98^{\circ} C$ and 73 atm howsoever high temperature and pressure may be applied on $CO_{2}$ gas. Hence, $CO_{2}$ gas cannot be liquified at $32 .{ }^{\circ} C$ and $80 ~atm$.

Answer

(i) Molar volume occupied by the gas molecules $\propto$ size of the molecules and van der Waals’ constant ’ $b$ ’ represents molar volume of the gas molecules. Hence, value of ’ $b$ ’ increases in the following order

$$ H_{2}<He<O_{2}<CO_{2} $$

(ii) van der Waals’ constant ’ $a$ ’ is the measure of magnitude of intermolecular attraction. The magnitude of intermolecular attractions increases with increase in size of electron cloud in a molecule. Hence, for the given gases magnitude of ’ $a$ ’ decreases in the following order

$$ CH_{4}>O_{2}>H_{2} $$

Greater the size of electron cloud, greater is the polarisability of the molecule and greater is the dispersion forces or London forces.

Answer

Given that, $p_{\text {ideal }}=p_{\text {real }}+\frac{a n^{2}}{V^{2}}$

(i) $a=\frac{p V^{2}}{n^{2}}$ If units of $p=\mathrm{Nm}^{-2}$,

units of $V=\mathrm{m}^{3}$, units of $n=\mathrm{mol}$

then, units of $a=\frac{\mathrm{Nm}^{-2}\left(\mathrm{~m}^{3}\right)^{2}}{(\mathrm{~mol})^{2}}=\mathrm{Nm}^{4} \mathrm{~mol}^{-2}$

(ii) If units of $p=\operatorname{atm}$, units of $V=\mathrm{dm}^{3}$, units of $n=\mathrm{mol}$

then, units of $a=\frac{p V^{2}}{n^{2}}=\frac{\text { atm. }\left(\mathrm{dm}^{3}\right)^{2}}{(\mathrm{~mol})^{2}}=\mathrm{atm} \mathrm{dm}^{6} \mathrm{~mol}^{-2}$

Answer

Two phenomena that can be explained on the basis of surface tension are as following

(i) Rise or fall of the liquid in a capillary-(capillary action).

(ii) Spherical shape of small liquid drops.

Thinking Process

Stronger the intermolecular forces existing between the liquids, more viscous is the liquid.

Answer

In water and glycerine - Hydrogen bonding and dipole-dipole interactions exists as both are polar molecules.

hexane - Dispersion forces/London forces are present due ts non-polar nature. The order of intermolecular forces among the given liquids is, hexane $<$ water $<$ glycerine.

Since, hexane has weakest intermolecular forces and glycerine the strongest (three $\mathrm{OH}$ groups) intermolecular forces. Therefore, hexane has minimum viscosity and glycerine has maximum viscosity.

Answer

As the temperature of a liquid increases, kinetic energy of the molecules increases which can overcome intermolecular forces. So, the liquid can flow more easily, this results in decrease in viscosity of the liquid.

Answer

(i) In accordance to Boyle’s law, pressure of a gas is inversly proportional to its volume if temperature is kept constant. Thus, the volume of a gas will decrease if the pressure on the gas is increased keeping the temperature constant. e.g., at $200 \mathrm{~K}$ when pressure increases from $p_{1}$ to $p_{2}$, volume of the gas decreases, $V_{2}<V_{1}$.

(ii) In accordance to Iharles law, volume of a gas is directly proportional to its temperature if pressure is kept constant.

Thus, on increasing temperature, the volume of a gas will increase if pressure is kept constant.

At constant $p$ when we increase the temperature from $200 \mathrm{~K}$ to $400 \mathrm{~K}$, the volume of the gas increases, $V_{4}>V_{3}$.

Answer

(i) At low pressure, the real gas shows very small deviation from ideal behaviour because the two curves almost coincide at low pressure.

(ii) At high pressure, the real gas show large deviations from ideal behaviour as the curves are far apart.

(iii) At point ’ $A$ ‘, both the curves intersect each other. At this point real gas behaves as an ideal gas. $p_{1}$ and $V_{1}$ are the pressure and volume which corresponds to this point $A$.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(4)$

A. When molar volume is constant, the $p-T$ graph is called isochore.

B. When temperature is constant, the $p-V$ graph is called isotherm.

C. When pressure is constant, $V-T$ graph is called isobar.

Answer

A. $\rightarrow(5)$

B. $\rightarrow(4)$

C. $\rightarrow(2)$

D. $\rightarrow(1)$

A. Boyle’s law, $p \propto \frac{1}{V}$ at constant $T$ and $n$.

B. Charle’s law, $V \propto T$ at constant $p$ and $n$.

C. Dalton’s law, $p_{\text {Total }}=p_{1}+p_{2}+p_{3}+\ldots$ at constant $T$, V.

D. Avogadro’s law, $V \propto n$ at constant $T$ and $p$.

Answer

A. $\rightarrow(2)$

B. $\rightarrow(3)$

C. $\rightarrow(1)$

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(a) Both assertion and reason are true and reason is the correct explanation of assertion. Intermolecular forces and thermal energy decide the state of matter because the former kept the molecules together but later keep them apart.

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(b) Both assertion and reason are true but reason is not the correct explanation of assertion. At constant temperature, $p V v s V$ for real gases is not a straight line because real gases have intermolecular forces of attraction.

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(c) Assertion is true but reason is false.

At high altitude, atmospheric pressure is low.

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(a) Both assertion and reason are true and reason is the correct explanation of assertion. Gases do not liquefy above their critical temperature, even an applying high pressure because above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together.

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(a) Both assertion and reason are true and reason is the correct explanation of assertion. At critical temperature, density of liquid and gaseous phase becomes equal. Due to which liquid passes into gaseous state imperceptibly and continuously.

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(d) Assertion is false but reason is true.

Liquid tend to reduce number of molecules or surface tension at there surface, that’s why small liquid drops have spherical shape.

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(i) In gaseous state, $CO_{2}$ will exist between the points $a$ and $b$ at temperature $T_{1}$.

(ii) At point $b$, the plot becomes linear, this shows the phase transition, i.e., liquification of $\mathrm{CO}_{2}$ starts and at point $c$, it gets completely liquified.

(iii) Similarly, at temperature $T_{2}, g$ is the point at which $\mathrm{CO}_{2}$ will be completely liquified.

(iv) Condensation will not take place at $T_{3}$ temperature because $T_{3}>T_{C}$ (critical temperature).

(v) Between $b$ and $c$, liquid and gaseous $\mathrm{CO}_{2}$ are in equilibrium.

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(i) Boiling point of $A=$ approximately $315 \mathrm{~K}$,

$B=$ approximately $345 \mathrm{~K}$.

(ii) In a closed vessel, liquid $C$ will not boil because pressure inside keeps on increasing

(iii) Temperature corresponding to $60 \mathrm{~mm}=313 \mathrm{~K}$.

(iv) A liquid boils when vapour pressure becomes equal to the atmospheric pressure. However at high altitudes i.e., on, hills, water boils at low temp erature due to low atmospheric pressure. But when pressure cooker is used, the vapour pressure of water is increased due to which water boils at even lower temperature within a short period of time

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In a closed vessel, it is essential to know that below the critical point (i.e., critical temperature and critical pressure), the surface of separation between the liquid and its vapour is clearly visible. As we approach towards the critical point, the density of the liquid decreases while that of the vapour increases due to compression.

Below critical point

(Meniscus clearly visible)

Above critical point (Meniscus disappears)

At the critical point, the densities of the liquid and that of the vapour become equal and the surface of separation disappears i.e., the liquid and the gaseous state become not separable. In other words, the meniscus is no longer visible.

The fluid which is now a homogeneous mixture is called supercritical fluid. Hence, any fluid above its critical temperature and pressure is called a supercritical fluid.

These supercritical fluids dissolve many organic substances. They are used for quick separation of a mixture into its components. e.g., $\mathrm{CO}_{2}$ above $31.1^{\circ} \mathrm{C}$ and above 600 bar pressure has a density of about $1 \mathrm{~g} / \mathrm{cm}^{3}$.

It is used to dissolve out caffeine from coffee beans as it is a better substitute than chlorofluorocarbons which are harmful for the environment.

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Sharp glass edges are heated to make them smooth. Because on heating glass melts and the surface of the liquid tends to take the rounded shape at the edges which has minimum surface area. This is called fire polishing of glass.

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When a liquid flows over a fixed surface, the layer of molecules in the immediate contact of surface is stationary. The velocity of the upper layers increases as the distance of layers from the fixed layer increases.

This type of flow in which there is a regular gradation of velocity in passing from one layer to the next is called laminar flow.

In laminar flow, the velocity of molecules is not same in all the layers because every layer offers some resistance or friction to the layer immediately below it.

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It is possible to change a gas into a liquid or a liquid into a gas by a process such that there is always a single phase present.

e.g., In figure given above, we can move from $A$ to vertically by increasing the temperature, then we can reach the point by compressing the gas at constant temperature along the isotherm (isotherm at $31.1^{\circ} \mathrm{C}$ ). The pressure will increase.

Now, we can move vertically down towards $D$ by lowering the temperature. As soon as we cross the point $\mathrm{H}$ on the critical isotherm, we get liquid. Thus, at no stage during the process, we can through two-phase region.

It process is carried out at the critical temperature, substance always remains in one phase. This is called continuity of state between the gaseous and the liquid state.