NCERT Exemplar Class 11 Chemistry Solutions Chapter 5 States of Matter

Question 1. A person living in Shimla observed that cooking food without using pressure cooker takes more time. The reason for this observation is that at high altitude:
(i) pressure increases
(ii) temperature decreases
(iii) pressure decreases
(iv) temperature increases
Answer:

The answer is option (iii), pressure decreases.
Explanation: Due to low pressure at high altitudes, boiling takes place at low temperature.

Question 2. Which of the following property of water can be used to explain the spherical the shape of rain droplets?
(i) viscosity
(ii) surface tension
(iii) critical phenomena
(iv) pressure
Answer:

The answer is the option (ii) surface tension.
Explanation: We know that for a given volume, a sphere has a lower state of energy and minimum surface area. Hence, the raindrops are spherical.

Question 3. A plot of volume (V ) versus temperature (T ) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in Fig. 5.1. Which of the following order of pressure is correct for this gas?

(i) p₁ > p₂ > p₃ > p₄
(ii) p₁ = p₂ = p₃ = p₄
(iii) p₁ < p₂ < p₃ < p₄
(iv) p₁ < p₂ = p₃ < p₄
Answer:

The answer is the option (iii) $p_{1} < p_{2} < p_{3} < p_{4}$
Explanation: ……… (PV = constant, at constant temperature)
Hence, if $V_{1} = V_{2} = V_{3} = V_{4} ,$
Then, $P_{1} = P_{2} = P_{3} = P_{4} ,$

Question 4. The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles but their magnitude depends upon
(i) charge of interacting particles
(ii) mass of interacting particles
(iii) polarizability of interacting particles
(iv) strength of permanent dipoles in the particles.
Answer:

The answer is the option (iii) polarizability of interacting particles
Explanation: Because, if the magnitude of interaction energy will be greater, then the polarizability of the interacting particles will also be greater.

Question 5. Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess ‘partial charges’. The partial charge is
(i) more than unit electronic charge
(ii) equal to unit electronic charge
(iii) less than unit electronic charge
(iv) double the unit electronic charge
Answer:

The answer is the option (iii) less than unit electronic charge
Explanation: The unit charge is always greater than partial charges.

Question 6. The pressure of a 1:4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen?
(i) $0.8 \times 10^{5} atm$
(ii) $0.008\; Nm^{-2}$
(iii) $8\times 10^{4}Nm^{-2}$
(iv) $0.25\; atm$

Answer:

We know that,
The partial pressure of O₂ = Mole fraction of O₂ × total pressure of the mixture
$=\frac{4}{5}\times 1\; atm=0.8\; atm$
$=0.8\times 10^{5}Nm^{-2}=8\times 10^{4}Nm^{-2}$
Hence opiton (iii) is correct

Question 7. As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
(i) increases
(ii) decreases
(iii) remains same
(iv) becomes half
Answer:

The answer is the option (i) increases
Explanation: Kinetic energy of molecules α temperature
Therefore, according to Gay Lussac's law, at constant volume, temperature as well as pressure both increases.

Question 8. Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases.

From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first
(i) $H_{2},He,O_{2},N_{2}$
(ii) $He,O_{2},H_{2},N_{2}$
(iii) $N_{2},O_{2},He,H_{2}$
(iv) $O_{2},N_{2},H_{2},He$
Answer:

Ans. The answer is the option (iv) $O_{2},N_{2},H_{2},He$
Explanation: Gas is easily liquified at a higher critical temperature.

Question 9. What is SI unit of viscosity coefficient $(\eta )$?
(i) Pascal
(ii) Nsm^–2
(iii) Km^–2 s
(iv) N m^–2
Answer:

The answer is the option (ii) Nsm^-2

Question 10. Atmospheric pressures recorded in different cities are as follows:

Consider the above data and mark the place at which liquid will boil first.
(i) Shimla
(ii) Bangalore
(iii) Delhi
(iv) Mumbai
Answer:

The answer is the option (i) Shimla
Explanation: Boiling point of liquid α atmospheric pressure. Shimla has the lowest atmospheric pressure; hence, the liquid will boil first in Shimla.

Question 11. Which curve in Fig. 5.2 represents the curve of ideal gas?

(i) B only
(ii) C and D only
(iii) E and F only
(iv) A and B only
Answer:

The answer is the option (i) B only
Explanation: Since we know that PV is constant at all temperatures for an ideal gas, hence the curve B.

Question 12. Increase in kinetic energy can overcome intermolecular forces of attraction. How will the viscosity of liquid be affected by the increase in temperature?
(i) Increase
(ii) No effect
(iii) Decrease
(iv) No regular pattern will be followed
Answer:

The answer is the option (iii) Decrease
Explanation: Since kinetic energy increases with temperature, it can overcome intermolecular forces, and liquid starts flowing; hence the viscosity decreases.

Question 13. How does the surface tension of a liquid vary with increase in temperature?
(i) Remains same
(ii) Decreases
(iii) Increases
(iv) No regular pattern is followed
Answer:

The answer is the option (ii) Decreases
Explanation: We know that, Surface tension $\alpha$ 1/temperature & kinetic energy α temperature, hence the intermolecular attraction decreases.

Question 15. Which of the following figures does not represent 1 mole of dioxygen gas at STP?
(i) 16 grams of gas
(ii) 22.7 litres of gas
(iii) 6.022 × 10²³ dioxygen molecules
(iv) 11.2 litres of gas
Answer:

The answer is the option (i) 16 grams of gas, (iv) 11.2 litres of gas
Explanation: 1 mole of O₂ has 6.023 × 10²³ molecules of O₂, & at STP it occupies 22.4-litre volume and has a molar mass of 32g.

Question 16. Under which of the following two conditions applied together, a gas deviates most from the ideal behaviour?
(i) Low pressure
(ii) High pressure
(iii) Low temperature
(iv) High temperature
Answer:

The answer is the option (ii) High pressure, (iii) Low temperature
Explanation: =Gases deviate from PV=nRT under high pressure and low temperature, these are real gases, and it deviates from ideal behaviour.

Question 17. Which of the following changes decrease the vapour pressure of water kept in a sealed vessel?
(i) Decreasing the quantity of water
(ii) Adding salt to water
(iii) Decreasing the volume of the vessel to one-half
(iv) Decreasing the temperature of water
Answer:

The answer is the option (ii) Adding salt to the water, (iv) Decreasing the temperature of the water
Explanation: Adding salt to water decreases the surface area for a water molecule to vaporize, and hence vapour pressure of water also decreases.
Now, vapour pressure α temperature, hence, option (iv).

Question 19. Physical properties of ice, water and steam are very different. What is the chemical composition of water in all the three states.
Answer:

The molecular arrangement of water is different in the solid, liquid and gaseous state; however, the chemical composition is the same in all the states.

Question 20. The behaviour of matter in different states is governed by various physical laws. According to you what are the factors that determine the state of matter?
Answer:

The factors that determine the different states of matter are volume, mass, temperature and pressure.

Question 21. Use the information and data given below to answer the questions (a) to (c):

• Stronger intermolecular forces result in higher boiling point.

• Strength of London forces increases with the number of electrons in the molecule.

• Boiling point of HF, HCl, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively.

(a) Which type of intermolecular forces are present in the molecules HF, HCl, HBr and HI?
(b) Looking at the trend of boiling points of HCl, HBr and HI, explain out of dipole-dipole interaction and London interaction, which one is predominant here.
(c) Why is boiling point of hydrogen fluoride highest while that of hydrogen chloride lowest?

Answer:

(a) Dipole-dipole are present in HCl, HBr, HI and HF; but, in HF along with these Intermolecular hydrogen bonding is also present.
(b)Dipole moment decreases from HCl to HI due to decrease in electronegativity, whereas, from HCl to HI its boiling point increases, hence we can say that London forces are predominant here.
(c) HF has hydrogen bonding as well as the highest electronegativity; hence it has the highest dipole moment as well as the highest boiling point.

Question 22. What will be the molar volume of nitrogen and argon at 273.15K and 1 atm?
Answer:

The conditions given above are of STP where the volume occupies by q mole of gas is 22.4 L. Hence, the volume of N₂ and argon, here, will be 22.4 L.

Question 23. A gas that follows Boyle’s law, Charles law and Avogadro’s law are called an ideal gas. Under what conditions a real gas would behave ideally?
Answer:

A real gas would behave ideal gas under low pressure and high temperature.

Question 24. Two different gases ‘A’ and ‘B’ are filled in separate containers of equal capacity under the same conditions of temperature and pressure. On increasing the pressure slightly the gas ‘A’ liquefies but gas B does not liquefy even on applying high pressure until it is cooled. Explain this phenomenon.
Answer:

This is based on the phenomenon of critical temperature; here, Gas 'A' liquefies easily, hence, it is below the critical temperature, whereas Gas' B' doesn't liquefy even on applying high pressure; hence it is above the critical temperature.

Question 25. Value of universal gas constant (R) is the same for all gases. What is its physical significance?
Answer:

The eq. of universal gas constant (R) is R = PV/nRT. Therefore, the unit of R will depend on P, V & T viz, Pa m³ K^-1 mol^-1 or Jmol^-1K^-1. Since Joule is the unit of work; therefore, R is the work done by gas mole per kelvin.

Question 26. One of the assumptions of the kinetic theory of gases states that “there is no force of attraction between the molecules of a gas.” How far is this statement correct? Is it possible to liquefy an ideal gas? Explain.
Answer:

Since there are no intermolecular forces of attraction between gas molecules of an ideal gas. Hence, the above statement is correct for ideal gases.

Question 27. The magnitude of surface tension of liquid depends on the attractive forces between the molecules. Arrange the following in increasing order of surface tension : water, alcohol $(C_{2}H_{5}OH)$ and hexane $[CH_{3}(CH_{2})_{4}CH_{3}]$
Answer:

The molecules above are of water, alcohol, and hexane. Here water and alcohol are polar and have dipole-dipole interaction and intermolecular hydrogen bonding, whereas hexane is non-polar and has weak London dispersion forces.
H-bonding is stronger in water; hence, surface tension is also more in water than alcohol. Therefore, Hexane < Alcohol < Water.

Question 28. Pressure exerted by saturated water vapour is called aqueous tension. What correction term will you apply to the total pressure to obtain pressure of dry gas?
Answer:

Let us consider that at temperature 'T' pressure of moist gas is 'P'; then, the pressure of dry gas can be obtained by the equation:
$P_{Drygas} = P_{Total }$ - aqueous tension

Question 29. Name the energy which arises due to motion of atoms or molecules in a body. How is this energy affected when the temperature is increased?
Answer:

Thermal energy arises due to the motion of atoms or molecules in a body. It is a measure of kinetic energy. Thermal energy α temperature.

Question 30. Name two intermolecular forces that exist between HF molecules in liquid state.
Answer:

Hydrogen bond and dipole-dipole interactions are the two forces existing in HF molecules when they are in a liquid state.

Question 31. One of the assumptions of kinetic theory of gases is that there is no force of attraction between the molecules of a gas.State and explain the evidence that shows that the assumption is not applicable for real gases.
Answer:

The assumption is not applicable to real gases because they can be liquified by cooling and applying high pressure. Hence, the force of attraction exists between the molecules of real gases.

Question 32. Compressibility factor, Z, of a gas is given as Z = (pV/nRT)
(i) What is the value of Z for an ideal gas?
(ii) For real gas what will be the effect on value of Z above Boyle’s temperature?
Answer:

(i) Z is a compressibility factor. Its value for an ideal gas is Z = 1.
(ii) Real gases show positive deviation above Boyle's temperature, i.e., Z>1.

Question 33. The critical temperature (T_c) and critical pressure ( p_c) of $CO_{2}$ are 30.98°C and 73 atm respectively. Can $CO_{2}$ (g) be liquefied at 32°C and 80 atm pressure?
Answer:

It does not matter that how high temperature or high-pressure is CO₂ at; it cannot be liquified above 30.98^o and 73 atm. Hence, it cannot be liquified at 32^o and 80 atm pressure.

Question 34. For real gases the relation between p, V and T is given by van der Waals equation:
$\left ( p+\frac{an^{2}}{v^{2}} \right )(v-nb)=nRT$
where ‘a’ and ‘b’ are van der Waals constants, ‘nb’ is approximately equal to the total volume of the molecules of a gas. ‘a’ is the measure of magnitude of intermolecular attraction.
(i) Arrange the following gases in the increasing order of ‘b’. Give reason.$O_{2}, CO_{2}, H_{2}, He$
(ii) Arrange the following gases in the decreasing order of magnitude of ‘a’. Give reason.$CH_{4}, O_{2}, H_{2}$

Answer:

(i) We know that volume α size of the molecules. Hence, the increasing order of the value of 'b' will be:
$H_{2}<He<O_{2}<CO_{2}$
(ii) 'a' is the Van der Waal's constant which represents the magnitude of intermolecular attraction. An $\alpha$ size of the electron cloud. Hence, the greater the size of the electron cloud, the greater will be the dispersion forces and polarizability of the molecule. Hence, the gases in decreasing order of the magnitude of 'a' will be: $CH_{4} > O_{2} > H_{2}$

Question 35. The relation between pressure exerted by an ideal gas (pideal) and observed pressure (preal) is given by the equation P_ideal $= P_{real} + (an^{2}/V^{2})$
If pressure is taken in Nm^–2, number of moles in mol and volume in m³, Calculate the unit of ‘a’.
What will be the unit of ‘a’ when pressure is in atmosphere and volume in dm³?
Answer:

It is given that $P_{ideal}=P_{real}+an^{2}/V^{2}$
We know that the unit of p = Nm^-2, unit of V = m³, number of moles(n) = mol
Unit of ‘a’ = Nm^-2(m³)²/(mol)² = Nm⁴mol^-4
If we take the unit of p = atm, V = dm³ and n = mol
Then unit of ‘a’ = pV²/n² = atm(dm³)²/(mol)² = atm dm⁶mol^-2.

Question 36. Name two phenomena that can be explained on the basis of surface tension.
Answer:

The two phenomena that can be explained on the basis of surface tension are-
(i) The spherical shape of liquid drops
(ii) Capillary action i.e. rise and dip of a liquid in the column of a capillary.

Question 37. Viscosity of a liquid arises due to strong intermolecular forces existing between the molecules. Stronger the intermolecular forces, greater is the viscosity. Name the intermolecular forces existing in the following liquids and arrange them in the increasing order of their viscosities. Also give reason for the assigned order in one line. Water, hexane $(CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{3})$, glycerine $(CH_{2}OH \; CH(OH) CH_{2}OH)$.
Answer:

(a) Hexane has Vander Waal force of attraction present as the intermolecular force.
(b) Water has hydrogen bonding present in the form of intermolecular forces
(c) Glycerin has hydrogen bonding present as the intermolecular force
Hexane < water < glycerine
The reason behind this order is that in hexane, weak bonding forces exist; therefore, it has the least viscosity, whereas, in water and glycerine dipole-dipole interaction and extensive H-bonding are present. Glycerine has the strongest intermolecular forces. Therefore, the order of viscosity is-

Question 38. Explain the effect of increasing the temperature of a liquid, on intermolecular forces operating between its particles, what will happen to the viscosity of a liquid if its temperature is increased?
Answer:

When there is an increase in the temperature, the intermolecular force which operates between the particle reduces, subsequently the strength of the bond increases along with the kinetic energy. Thus, when there is an increase in the temperature, there is a reduction in the viscosity. This is a result of the fact that viscosity decreases when there is reduction in the intermolecular forces.

Question 39. The variation of pressure with the volume of the gas at different temperatures can be graphically represented as shown in Fig. 5.3. Based on this graph answer the following questions.

Answer:

(i) According to Boyle's law, The pressure of gas $\alpha$ 1/volume of gas, at a constant temperature.
Therefore, if volume decreases at a constant temperature, then the pressure increases and vice versa. For example, at 200K if pressure increases from p₁ to p₂, volume decreases from v­₁ to v₂(v₂ < v₁)
(ii) According to Charles's law, The volume of gas α temperature, at constant pressure.Therefore, if temperature decreases from 200K to 400K, at a constant temperature, the volume of gas increases.

Question 40. Pressure versus volume graph for a real gas and an ideal gas is shown in Fig. 5.4. Answer the following questions based on this graph.

(i) Interpret the behaviour of real gas with respect to an ideal gas at low pressure.
(ii) Interpret the behaviour of real gas with respect to an ideal gas at high pressure.
(iii)Mark the pressure and volume by drawing a line at the point where real gas behaves as an ideal gas.
Answer:

(i) Real gases show a very small deviation from ideal behaviour at low pressure because both the curves coincide with each other.
(ii) Real gases show large deviation at high pressure as the curves fall apart.
(iii)The point at which the curves intersect each other shows that the real gas behaves like an ideal gas.

Question 42. Match the following gas laws with the equation representing them.

Answer:

(i) →(e); (ii) →(d); (iii) →(b); (iv) →(a)
Explanation:
(i) pressure of gas α 1/volume of gas (T constant) ….. (Boyle's law)
(ii) The volume of gas α temperature of the gas, at constant pressure …… (Charles's law)
(iii) At constant V and T, the total pressure of the gaseous mixture of two or more non-reacting gases is equal to the algebraic sum of their partial pressures ……. (Dalton's law)
(iv) Under same temperature and pressure conditions, equal volumes of all gases contain equal no. of moles. V α n. ……. (Avogadro's law)

Question 43. Match the following graphs of ideal gas with their co-ordinates :

Graphical representation x and y co-ordinates

Answer:

(i) →(b); (ii) →(c); (iii) →(a)
Explanation:
(i) pressure of gas $\alpha$ 1/volume of gas (T constant)
(ii) The pressure of gas $\alpha$ 1/V
(iii) Product of pressure and volume is constant. PV = constant

Question 45. In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): At constant temperature, pV vs V plot for real gases is not a straight line.
Reason (R) : At high pressure all gases have Z > 1 but at intermediate pressure most gases have Z < 1.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the option (i). Both A and R are true and R is the correct explanation of A.

For a real gas pV vs V plot is not a straight line due to intermolecular forces and finite molecular volume.

At high pressures, repulsive forces dominate → gas occupies more volume than ideal → Z>1

At intermediate pressures, attractive forces dominate → gas compresses more → Z<1.

Question 46. In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): The temperature at which vapour pressure of a liquid is equal to the external pressure is called boiling temperature.
Reason (R) : At high altitude atmospheric pressure is high.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the option (iii) A is true, but R is false.
Explanation: Boiling point of a liquid is the temperature at which the vapour pressure of the liquid is equal to the external pressure. As the altitude is high, the pressure is low.

Question 47. In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Gases do not liquefy above their critical temperature, even on applying high pressure.
Reason (R) : Above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the option (i)Both A and R are true, and R is the correct explanation of A.
Explanation: Gases do not liquefy even on applying high pressure if they are above the critical temperature as the Intermolecular forces of attraction cannot hold the molecules together, and the molecular speed is also high.

Question 48. In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): At critical temperature liquid passes into gaseous state imperceptibly and continuously.
Reason (R) : The density of liquid and gaseous phase is equal to critical temperature.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the optio (i) Both A and R are true, and R is the correct explanation of A.
Explanation: Density of liquid becomes equal to its vapour phase at a critical temperature which causes the liquid to change into gaseous state imperceptibly and continuously.

Question 49. In the following questions, a statement of Assertion (A) followed by a statement of the reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Liquids tend to have maximum number of molecules at their surface.
Reason (R) : Small liquid drops have spherical shape.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) A is false but R is true.
Answer:

The answer is the option (iv) A is false, but R is true.
Explanation: Liquid drops have a spherical shape as they try to reduce the no. of molecules and tend to reduce surface tension at their surface.

Question 51. The variation of vapour pressure of different liquids with temperature is shown in Fig. 5.6.

(i) Calculate graphically boiling points of liquids A and B.
(ii) If we take liquid C in a closed vessel and heat it continuously. At what temperature will it boil?
(iii) At high altitude, atmospheric pressure is low (say 60 mm Hg). At what temperature liquid D boils?
(iv) Pressure cooker is used for cooking food at hill station. Explain in terms of vapour pressure why is it so?
Answer:

(i) Approximately, BP of A is 315K, and that of B is 345K.
(ii) As pressure is kept on increasing in the closed vessel, liquid C will not boil.
(iii) 313K is the temperature corresponding to 60 mm Hg.
(iv) Since atmospheric pressure is low at high altitude, water boils at low temperature on hills.

Question 52. Why does the boundary between liquid phase and gaseous phase disappear on heating a liquid upto critical temperature in a closed vessel? In this situation what will be the state of the substance?
Answer:

As a liquid is heated in a closed vessel, its vapour pressure increases and the density of the vapour also increases, while the density of the liquid decreases. At a particular temperature known as the critical temperature, the densities of the liquid and vapour become equal. As a result, the clear boundary that separates the two phases disappears. At the critical temperature, the substance exists as a homogeneous single phase called a supercritical fluid. This supercritical state has unique properties like a gas, it can fill a container, and like a liquid, it can dissolve substances making it very useful in industrial applications.

Question 53. Why does sharp glass edge become smooth on heating it to its melting point in a flame? Explain which property of liquids is responsible for this phenomenon.
Answer:

When a sharp glass edge is heated to its melting point, the glass softens and behaves like a viscous liquid. In this state, the molecules of the glass have enough energy to move slightly and rearrange themselves. The property responsible for the smoothing of the sharp edges is surface tension. Surface tension tends to minimize the surface area of a liquid. As the glass melts, surface tension pulls the molecules into a more stable shape with the least surface area which is smooth and rounded. Thus, sharp edges become smooth when heated due to the action of surface tension in the softened glass.

Question 54. Explain the term ‘laminar flow’. Is the velocity of molecules the same in all the layers in laminar flow? Explain your answer.
Answer:

Laminar flow can also be defined as the flow in which there is a regular gradation of velocity in passing from one layer to the next.
When a liquid flows over a fixed surface, it comes in contact with it. The layer of molecules in the immediate contact of the surface is stationary. As the distance of layers increases from the fixed layer, the velocity of the upper layer also increases.

Question 55. Isotherms of carbon dioxide gas are shown in figure. Mark a path for changing gas into liquid such that only one phase (i.e. either a gas or liquid) exists at any time during the change. Explain how the temperature, volume and pressure should be changed to carry out the change.

Answer:

In the figure above, we can increase the temperature to move from A to F vertically and can reach G by compressing the gas at constant temperature along isotherm, thus, increasing the pressure. Then by lowering the temperature, we can move vertically downwards to D. We get liquid as soon as we cross the point H on isotherm. The substances will remain in one phase if we carry out this process at a critical temperature. This is called continuity of state between the gaseous and liquid states.