CLS Aipmt 15 16 XI Che Study Package 2 SET 1 Chapter 5

Chapter

5

States of Matter Solutions SECTION - A School/Board Exam. Type Questions Very Short Answer Type Questions 1.

Define vapour pressure.

Sol. It is the vapour pressure exerted by the vapours in equilibrium with its liquid at a given temperature. 2.

What is the S.I. unit of surface tension?

Sol. Newton per metre (Nm–1) 3.

How is the volume and pressure of a gas related to each other at constant temperature?

Sol. At constant temperature, the pressure of a fixed amount of gas varies inversely with the volume of the gas. 4.

Convert 23°C into Kelvin temperature.

Sol. 1°C temperature = 273.15 K 23°C temperature = 23 + 273.15 K = 296.15 K 5.

Define isotherms.

Sol. Isotherms are the curves obtained when we plot a graph between pressure and volume at a constant temperature. 6.

What type of intermolecular forces are also known as London forces?

Sol. Dispersion forces that link the two non-polar molecules together are also known as London forces. 7.

How is the density and molar mass of a gaseous substance related?

Sol. Molar mass = M Density = d M= 8.

dRT P

What sort of deviation is shown by dihydrogen in the PV versus P plot?

Sol. Positive deviation. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

2 9.

States of Matter

Solution of Assignment (Set-1)

What is the standard boiling point of water?

Sol. Standard boiling point of water is 99.6°C. 10. Name two factors on which the viscosity of a liquid depends. Sol. (i) Intermolecular attractive forces (ii) Temperature Short Answer Type Questions 11. What are real gases? Under what conditions they tend to follow the ideal gas equation? Sol. Real gases are those gases which deviate from ideal gas behaviour and deviate from the gas laws at higher pressure and lower temperature. So, they follow ideal gas equation only at (i) High temperature (ii) Low pressure 12. What does the Z value greater than unity for H2 gas and Z value less than unity for CH4 gas indicate? Sol. Z value greater than unity (Z > 1) for H2 gas indicates that there are forces of repulsion which operate between the hydrogen gas at high pressure. Z value less than unity (Z < 1) for CH4 gas indicate high attractive forces between its molecules. 13. Define critical temperature and critical pressure of a gas. What is the value of critical temperature of carbon dioxide gas? Sol. Critical Temperature (TC) : Critical temperature of a gas may be defined as the temperature above which the gas cannot be liquefied however high pressure may be applied on the gas. TC of carbon dioxide gas = 30.98°C. Critical Pressure (PC) : At critical temperature a certain pressure is needed to liquefy the gas. So, this pressure at critical temperature is called the critical pressure. 14. Discuss hydrogen bonding. Why is HCl a gas while HF is liquid at room temperature? Sol. Hydrogen Bonding : Hydrogen bonding is a type of dipole-dipole interaction. Hydrogen bonding is a force of attraction between the hydrogen atom attached to the highly electronegative atom and the electronegative atom of the other polar molecules. e.g. H2O. HF is liquid and HCl is gas at room temperature because HF molecules are interacted by the hydrogen bonds whereas hydrogen bonding is absent in HCl molecules. Hydrogen bonding being very strong intermolecular interaction HF exists as liquid at room temperature. 15. Hydrogen gas is enclosed in a cylinder of volume 500 cm3 at a pressure of 760 mm Hg and temperature 30°C. Calculate the number of moles of hydrogen gas in the cylinder. Sol. V = 500 cm3 = 500 × 10–3 litre P = 760 mm Hg = 1 atm

(1 atm = 760 mm)

T = 30°C = 303 K R = 0.0820 atm L K–1 mol–1 Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

States of Matter

3

Ideal gas equation PV = nRT 1 × 500 × 10–3 = n × 0.0820 × 303 n = 0.021 mol

16. Which state of the substance has highest (i) Intermolecular forces of attraction? (ii) Thermal energy? (iii) Density? Sol. (i) Solid state (ii) Gaseous state (iii) Solid state 17. Why pressure cookers are employed at high altitudes to cook food? Sol. At high altitude, the atmospheric pressure is very low. We know that the boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. So, due to lower pressure at higher altitude the liquid starts boiling much before the food actually gets prepared. So, pressure cookers work to increase the pressure, so that the food gets cooked properly. 18. What is combined gas law? Sol. The Boyle’s and Charles’ law can be combined to give a relationship between the three variables P, V and T. The initial temperature, pressure and volume of a gas are T1, P1 and V1. With the change in either of the variables, all the three variables change to T2, P2 and V2. Then we can write P1V1 PV  nR and 2 2  nR T1 T2

Combining the equations we get P1V1 P2 V2 = T1 T2

The above relation is called the combined gas law. 19. Define surface energy. What are its dimensions? Sol. Surface energy of the liquid is the energy required to increase the surface area of the liquid by one unit. Dimensions = Jm–2 20. Define (i) Vaporisation, (ii) Condensation. Sol. (i) Vaporisation : Vaporisation is the process by which liquid molecules go into the vapour phase. Liquid State   Vapours

(ii) Condensation : It is the process of conversion of vapours into the liquid state.

Vapours   Liquid State Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

4

States of Matter

Solution of Assignment (Set-1)

21. F2 and Cl2 are gases, Br2 is a liquid and I2 is a solid at room temperature. Why? Sol. All of these molecules are completely non-polar and according to theory, not attracted to each other, so it is predicted that they all should be gases at room temperature. But is not so, F2 and Cl2 are gases, Br2 is a liquid and I2 is a solid. The reason is that more are the number of electrons, more is the polarizability, diffuse electron cloud can be easily distorted. As I2 has maximum electrons, so maximum extent of induced dipole intermolecular forces among the halogens. These interactions decrease as electrons decrease (Size decreases) moving from Br2 to Cl2 to F2. 22. The density of CO2 is 0.326 gL–1 at 27°C and 260 mm pressure. If the pressure remains constant, then what is the density of the gas at 40°C? Sol. d1 = 0.326 gL–1

d2 = ?

T1 = 27°C

T2 = 40°C

= (27 + 273) K

= (40 + 273) K

= 300 K

= 313 K

d1T1 = d2T2 (R, P and M constant) d1T1 d2 = T 2

=

0.326  300 313

d2 = 0.312 gL–1 23. Calculate the total pressure in mixture of 8 g of oxygen and 4 g of hydrogen confined in a total volume of one litre at 27°C. Sol. Moles of oxygen =

Moles of hydrogen =

8 = 0.25 mol 32 4 = 2 mol 2

Total number of moles = 0.25 + 2 = 2.25 mol Using Ideal gas equation PV = nRT P × 1 = 2.25 × 0.0821 × 273 P = 50.42 atm

24. (i) Will a real gas occupy the same volume at high pressure as that of ideal gas under similar conditions? (ii) Which van der Waals constant signifies the intermolecular forces between the molecules? Sol. (i) No, a real gas at high pressure will occupy more volume than an ideal gas under similar conditions. This is because the volume of the molecules themselves will have a significant value in case of a real gas at high pressure while the volume of the molecules of an ideal gas will be zero. (ii) van der Waals constant ‘a’ signifies the intermolecular forces between the molecules. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

States of Matter

5

25. Four litres of CO2 are kept at 27°C. What will be the volume if the temperature is lowered to 150 K at the same pressure? Sol. V1 = 4 L

V2 = ?

T1 = 27°C

T2 = 150 K

= (27 + 273) K = 300 K

V1 V2 Applying Charles law as pressure is constant T = T 1 2 

V 4 = 2 300 150



V2 =



150  4 =2L 300

V2 = 2 L

26. Working of syringe is an application of Boyle’s law. Explain. Sol. A syringe is a device used in a hospital to draw blood samples or give injections. When the plunger of the syringe is pulled back the volume of the syringe container increases, decreasing the pressure inside, since the same amount of gas is now spread over a greater volume, this is what is stated by Boyle’s law. To balance this effect of low pressure, air or blood is sucked in through the needle, thus balancing the pressure inside and outside the container. 27. One litre of a gas weighs 3 g at 300 K at 1 atm pressure. If the pressure is made 0.85 atm, at which temperature will one litre of the same gas weigh one gram? Sol. P1V1 = n1RT1 P2V2 = n2RT2 

n1RT1 P1V1 = n P2 V2 2RT2

w1 n1 = M 1

w2 n2 = M 2

As the gas is same, so M1 = M2 

P1V1 w RT = 1 1 P2 V2 w 2RT2



1 1 3  300  0.85  1 1  T2

 T2 = 3 × 300 × 0.85 

T2 = 765 K

Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

6

States of Matter

Solution of Assignment (Set-1)

28. Pressure remaining constant, at what temperature the density of the gas becomes 0.625 d, when the density at 27°C is d?

Sol. d =

PM or RT

d1 = d

d1 T2 = d2 T1

T1 = 27°C

d2 = 0.625 d

= (27 + 273) K = 300 K T2 = ?

T d = 2 0.625 d 300 T2 = 480 K

Long Answer Type Questions 29. Derive the ideal gas equation. What is equation of state? Sol. Ideal gas equation is an equation which is followed by the ideal gases. A gas that would obey Boyle's and Charles’ Law under all the conditions of temperature and pressure is called an ideal gas. As discussed the behaviour of gases is described by certain laws as Avogadro's law, Boyle's law, Charles' law. According to Avogadro's law ; V  n (P and T constant) According to Boyle's law ; V 

1 (T and n constant) P

According to Charles law ; V  T (P and n constant) Combining the three laws ; we get V



nT P

V =R

nT P

where; 'R' is the proportionality constant On rearranging the above equation we get, PV = nRT This is the ideal gas equation as it is obeyed by the hypothetical gases called ideal gases under all conditions of temperature and pressure. However there is no gas that is perfectly ideal. But the gases may show nearly ideal behaviour under the conditions of low pressure and high temperature and are called real gases. Equation of state : As the ideal gas equation expresses the quantitative relationship between the four variables that describe the state of a gas therefore it is called equation of state for gases. When 1 mole of a gas is considered, then PV = RT ∵ n  1mol Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

States of Matter

7

30. What is the effect of temperature on the compressibility of the gas above and below Boyle’s temperature? Sol. Pressure alone does not decide the behaviour of gases. Temperature is also an important factor which plays role. Boyle Temperature or Boyle’s Point : It is the temperature at which real gases obey ideal gas laws over an appreciable range of pressure. Boyle temperature depends on the nature of the gas. Effect of Temperature on the Compressibility Factor (a) Above the Boyle temperature : Real gas show positive deviation (Z > 1) from ideality. This is because with increase in temperature, the molecules move far from each other So, volume increases thereby the forces of attraction between the molecules become feeble. (b) Below the Boyle's temperature : Below the Boyle's temperature 'Z' value first decreases and reaches a minimum with increase in pressure because of forces of attraction which start operating between the molecules. Later, on further increase in pressure force of repulsion operates, So, now the value of Z increases continuously. 31. (a) Define surface tension. (b) Explain why water has concave and mercury has convex surface in glass tubes, pipettes, burettes etc. (c) Shaving blade floats on water surface. Why? Sol. (a) Surface Tension : Surface tension may be defined as the force acting per unit length perpendicular to the line drawn on the surface of liquid. (b) It is due to surface tension. Surface of water is concave because the adhesive forces between glass and water molecules are greater than intermolecular forces of attraction between molecules. Whereas, the surface of mercury is convex because, glass-Hg adhesive forces are less than intramolecular forces of attraction between mercury molecules. (c) A shaving blade, if placed gently floats on water surface due to surface tension of water. 32. (a) State Boyle’s law. (b) A sample of gas occupies 200 litres at 0.5 atm pressure at 2°C. If the volume of the gas is to be reduced to 20 litres at the same temperature, what additional pressure must be applied? Sol. (a) Boyle’s Law : Boyle’s law states that “at constant temperature, the pressure of a fixed amount of gas varies inversely with the volume of the gas”. (b) P1 = 0.5 atm

P2 = ?

V1 = 200 litres

V2 = 20 litres

T1 = 2°C

T2 = 275 K

= (2 + 273) = 275 K Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

8

States of Matter

Solution of Assignment (Set-1)

As temperature is constant, so applying Boyle’s law P1V1 = P2V2 

0.5 × 200 = P2 × 20



P2 =

0.5  200 = 5 atm 20

Additional pressure that should be applied P2 – P1 = 5 – 0.5 atm P2 – T1 = 4.5 atm

33. Explain in detail the concept of viscosity. Explain the factors on which it depends. Sol. Viscosity is one of the characteristic properties of liquids. All liquids tend of flow. But some liquids flow faster than the other. This is because of the difference of intermolecular forces of attraction. Example, ink flows faster than honey.

Ink

Beaker A

Honey

Beaker B

Fig. : An ink drop diffuses faster in water than does a drop of honey because honey is more viscous than ink Let us understand the concept of viscosity. Viscosity is actually the measure of resistance to the flow of the liquid. The flow of liquid molecules can be analysed in terms of molecular laminar layers. Laminar Flow : The liquid is considered to be consisting of molecular layers arranged one over the other. When the liquid flows over a glass surface then the layer of molecules immediately in contact with the glass surface are stationary with zero velocity. But layer immediately above it is not stationary but flows with some velocity. Further the next layer above it flows still more faster and this continues and the top most layer of molecules flow with maximum velocity. So, this type of flow in which there is gradual gradation in the velocities on passing from one layer to the another is called laminar flow.

Q(V + dv) P(V) R(V – dv)

Fig. : Gradation of Velocity in the laminar flow If any molecular layer (say P) we take, then the layer above it (say Q) accelerates its flow because the molecules from the above layer (Q) with higher velocity speeds the lower layer (P) molecules also. But the Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

States of Matter

9

layer (R) below retards its flow because the molecules in the lower layer (R) are with lower velocity. This whole results in the friction between the two layers which give rise to viscosity. So, viscosity is defined as the internal resistance of flow in liquids which arises due to the internal friction between the layers of liquid as they slip past one another while liquid flows. Viscosity Depends upon the following two factors : (a) Intermolecular attractive forces : Greater the intermolecular forces of attraction greater is the viscosity. Hydrogen bonding and van der Waals forces are strong enough to cause high viscosity. More viscous liquids flow slowly. (b) Temperature : With the rise in temperature, the kinetic energy of the molecules increases. Molecules with greater kinetic energy can overcome the intermolecular forces to slip past one another between the layers. So, we can say that the viscosity of the liquids decreases with rise in temperature. 34. (a) Explain Dalton’s law of partial pressure. (b) How is the pressure of a dry gas calculated over water? Sol. (a) According to ideal gas equation, if n1 is the number of moles of one constituent gas of the gaseous mixture then P1 is the pressure exerted by the gas at temperature (T) enclosed in the volume (V) P1 =

n1RT V

...(i)

Similarly for the other two constituting gases of the gaseous mixture P2 =

n2RT V

P3 =

n3RT V

According to Dalton's law of partial pressures PTotal = P1 + P2 + P3 + ... =

n1RT n2RT n3RT + + + ... V V V

Ptotal =  n1 + n2 + n3 

RT V

...(ii)

That means the total partial pressure of the mixture is determined by the total number of moles present. Dividing equation (i) by (ii) we get total number of moles present. ⎛ ⎞ RTV P1 n1 = ⎜ Ptotal ⎝ n1 + n2 + n3 ⎟⎠ RTV

n1 = n +n +n 1 2 3 =

n1 n

Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

10

States of Matter

Solution of Assignment (Set-1)

where, n = n1 + n2 + n3

Now,



n1 moles of 1st gas



n  Total number of moles

= Mole fraction of first gas (x1)

Mole Fraction : It is the ratio of the number of moles of an individual gas to the total number of moles of all gases in a molecule. P1 = x1 Ptotal

Thus P1 = x1 Ptotal

Similarly P2 = x 2 Ptotal Therefore the generalised equation becomes Pi = x i Ptotal

Where Pi = partial pressure of ith gas xi = mole fraction of ith gas Thus, the partial pressure of a gas in the mixture of gases is the product of its mole fraction and the total pressure of the mixture. Therefore the above equation is used to find out pressure exerted by an individual gas in the mixture of gases. (b) Calculation of the pressure of dry gas collected over water : When the gas is collected over water it is moist because of the water vapours. Saturated water vapour exert its own partial pressure called aqueous tension. So, in order to calculate the partial pressure of dry gas, aqueous tension is subtracted from the pressure of moist gas (Pmoist gas or PTotal) Pdry gas = Ptotal – Aqueous tension

35. Explain the four parameters which are used to describe the characteristics of gases along with various interconversions of units. Sol. There are certain parameters or measurable properties which are used to describe the characteristics of gases : (a) Volume (V) : The volume of the container is the volume of the gas sample as gases occupy the entire space available to them S.I. unit = m3 C.G.S. unit = cm3 Commonly used unit = L 1 L = 1000 mL 1 mL = 10–3 L 1 m3 = 103 dm3 = 103 L = 106 cm3 = 106 mL Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

States of Matter

11

(b) Pressure (P) : Pressure of the gas is the force exerted by the gas per unit area on the walls of the container in all directions. S.I. unit = Pascal (Pa) 1 Pa = 1 Nm–2 = 1.013 × 105 Pa Conversions 1 bar = 105 Pa = 0.987 atm 1 atm = 760 mm Hg = 760 torr = 1.013 × 105 Pa. (c) Temperature : It is the measure of hotness of the system and energy of the energy. S.I. unit = kelvin (K) K = °C + 273 °C  centigrade degree (°C) or celsius degree (d) Mass : The mass of a gas can be determined by weighing the container in which the gas is enclosed and again weighing the container after removing the gas. The mass of the gas is related to the number of moles of the gas i.e. Moles of gas (n) =

Mass in grams m = Molar mass M

36. (a) What is boiling? (b) Whether boiling occurs in an open or closed vessel. (c) Explain the difference between normal boiling point and standard boiling point. Sol. (a) Boiling : Boiling is a bulk phenomenon. At the boiling point, the vaporisation occurs throughout the bulk of the liquid and the vapours expand freely in the surroundings. (b) Boiling occurs in an open vessel it cannot occur in a closed vessel. (c) Normal Boiling Point : As the atmospheric pressure varies with altitude and other conditions, the boiling points are reported at 1 atm. So, the normal boiling point of a liquid is the temperature at which the vapour pressure of the liquid is 1 atm. e.g. Normal boiling point of water = 100°C Standard Boiling Point : It is the temperature at which the vapour pressure of the liquid is 1 bar. e.g. Standard boiling point of water = 99.6°C Standard boiling point of the liquid is slightly lower than the normal boiling point of the liquid. 37. (a) What is the effect of hydrogen bonding on the viscosity of a liquid? (b) Liquids have a definite volume, but no definite shape. Explain Sol. (a) Hydrogen bonding is a type of intermolecular force of attraction. So, it links the various units of liquid molecules, so the effective size of the moving unit in the liquid increases. Due to an increase in the size and mass of the molecule, their is greater internal resistance to the flow of the liquid. Therefore, viscosity of liquid increases. (b) The intermolecular forces in liquids are strong enough to hold the molecules together, but not strong enough to fix them into definite positions as in solids. Therefore, liquids have a definite volume, but no definite shape. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

12

States of Matter

Solution of Assignment (Set-1)

38. (a) What is the volume of the 6 g of chlorine at 27°C and 101 kPa pressure? (b) What is the volume of the chlorine in dm3 and cm3? Sol. (a) T = 27°C = (27 + 273) K = 300 K P = 101 kPa = 101 × 1000 = 101000 Pa Molar mass of chlorine = 71 g Given mass of chlorine = 6 g

Moles of chlorine (n) =

6 = 0.084 mol 71

R = 8.314 Pa m3 K–1 mol–1 Applying ideal gas equation PV = nRT 

101000 × V = 0.084 × 8.314 × 300



V=



V = 2.074 × 10– 3 m3



V = 0.002074 m3

0.084  8.314  300 101000

(b) 1 m3 = 1000 dm3 = 106 cm3 So, the volume in dm3 = 2.074 dm3 Volume in cm3 = 2074 cm3 39. 0.64 g of an oxide of sulphur occupies 0.224 L at 2 atm and 273°C. Identify the compound. Also find out the mass of one molecule of the gas. Sol. P = 2 atm V = 0.224 L Given, mass (m) = 0.64 g T = 273°C = (273 + 273) K = 546 K R

= 0.082 L atm mol–1 K–1

PV = nRT PV =

M=

m RT M mRT PV

Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

M =

States of Matter

13

0.64  0.082  546 2  0.224

= 63.96  64 The oxide is SO2 as its molar mass = 32 + 2 × 16 = 64 g/mol Mass of 6.022 × 1023 molecules of oxide = 64 g

64 Mass of 1 molecule of oxide = 6.022 × 1023 g = 1.063 × 10–22 g 40. (a) Explain dipole-dipole forces along with example. (b) How the interaction energy varies with the distances for stationary as well as rotating polar molecules? Sol. (a) Dipole-Dipole Forces : (Associated with Polar structures) : Polar molecules have a partially positive side and a partially negative side or a dipole. Dipole-dipole forces operate between the molecules which though neutral but possess permanent dipole. The separation of partial charges depends upon the electronegativity of the bonded atoms in a molecule. The partial charges are indicated by the Greek letter 'delta()'. In these type of forces the partial positive end of the one molecule is attracted towards the negative end of the other molecule. Example : Dipole-dipole forces are present between the two HCl molecules. Chlorine being more electronegative pulls the shared pair of electrons towards itself. So, it has a partial negative charge (–) on it and hydrogen atom has a partial positive charge(+)

Polar Polar Molecule |||||||||||||| Molecule + + – –   H

Cl

Covalent Bond

H

Cl

Covalent Bond

Dipole-dipole forces (interactions) (b) The interaction energy is dependent upon the distance between the polar molecules. Stationary polar molecules : Dipole-dipole interaction energy is inversely proportional to the third power of the distance between stationary polar molecules (in solids)

Interaction Energy 

1 r3

Rotating polar molecules : Dipole-dipole interaction energy is inversely proportional to the sixth power of the distance between the rotating polar molecules (like water)

Interaction Energy 

1 r6

Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

14

States of Matter

Solution of Assignment (Set-1)

41. (a) State Gay Lussac’s Law. (b) Give its graphical representation. (c) What are isochores? Sol. (a) Gay Lussac's Law or Pressure–Temperature Law can be stated as ‘at constant volume, the pressure of a fixed amount of a gas is directly proportional to the temperature’. The law may be expressed mathematically as P  T (V, n are constant) or P = k3 T P = k 3  constant  T

P1 P =k = 2 T1 T2



P1 P2  T1 T2

When we know P1, T1 and T2 then we can easily calculate the value of P2. (b) Graphical Representation

Above is the temperature (kelvin) vs pressure graph at different fixed volume of the gas. (c) Isochores : The lines showing the pressure temperature behaviour plotted at fixed volumes are called isochores. 42. (a) Explain the pressure correction applied to ideal gas equation to make it fit for real gases. (b) Explain the isotherm of CO2 at 30.98°C temperature. Sol. (a) Pressure Correction (Correction due to intermolecular forces of attraction) In the derivation of ideal gas equation, it is assumed that there are no intermolecular forces of attraction. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

States of Matter

15

But as now we have identified this faulty assumption, so there is need to incorporate results obtained from the intermolecular force of attraction between the gas molecules.

Molecule in midst of the vessel

Molecules near the wall

Net inward force

No net force

Fig. Correction due to Molecule attraction For that let us take an example, molecule A when present in the midst of the vessel is attracted uniformly by the other molecules from all the directions. So, there is no net attractive force of the molecules but when this molecule approaches the wall of the vessel, then it only experiences attractive force from the bulk molecules behind it. So, now the molecule strikes the wall of the vessel with a lower velocity as it is dragged back by the other molecules. Therefore, the pressure exerted by the molecule now is lower than it would have exerted if there was no force of attraction. In other words, we can say that the pressure exerted by a real gas is less than that exerted by an ideal gas. Therefore, we need to add the pressure correction term to make the ideal gas equation fit for real gases also.

Pideal = Preal 





observed pressure

an2 V2

...(i)

⎛ pressure ⎞ ⎜ correction⎟ ⎜⎝ term ⎟⎠

Here 'a' is van der Waals constant. Its value is (i)

Dependent upon the intermolecular forces of attraction within the gas.

(ii)

Independent of temperature and pressure.

(b) Isotherm at temperature 30.98°C : At 30.98°C temperature gas changes into liquid while passing through a phase in the form of a point when liquid state and gaseous state of carbon dioxide co-exist. Above this temperature, the isotherm is continuous. That means carbon dioxide just exists in the gaseous state and does not liquefy even at very high pressure. So, this point or temperature (i.e. 30.98°C) is the critical temperature of carbon dioxide. In other words, we can say that at temperature 35°C or 50°C, carbon dioxide does not liquefy.

SECTION - B Model Test Paper Very Short Answer Type Questions : 1.

What is the S.I. unit of surface tension?

Sol. Newton per metre (Nm–1) 2.

What is the value of Avogadro’s constant?

Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

16

States of Matter

Solution of Assignment (Set-1)

Sol. The value of Avogadro’s constant is 6.022 × 1023 3.

What is the effect of pressure on the boiling point of a liquid?

Sol. Boiling point of a liquid increases with increase in the pressure. 4.

Under which two conditions real gas tends to follow gas laws?

Sol. (i) High temperature (ii) Low Pressure 5.

What is the critical pressure of carbon dioxide gas?

Sol. The critical pressure (PC) of carbondioxide gas is 73 atm. 6.

How is the partial pressure of a gas in a mixture related to the total pressure of the gaseous mixture?

Sol. Partial pressure of a gas (Pi) = Mole fraction of the gas (xi) × total pressure (Ptotal) 7.

Which out of the three states of matter has highest thermal energy?

Sol. Gaseous state. Short Answer Type Questions : 8.

Define coefficient of viscosity and give its S.I. unit.

Sol. Coefficient of viscosity is the force when velocity gradient is unity and the area of contact is unit area. F = .A

dv dz

 = coefficient of viscosity S.I. unit = pascal second (Pa s) 9.

Define ideal gas. Give the relationship between the density and molar mass of a gaseous substance.

Sol. Ideal gas: A gas which obeys the gas laws and the gas equation strictly at all temperatures and pressures is said to be an ideal gas. Molar mass (M) =

dRT P

where d is the density of the gas. 10. What do you understand by ‘London Forces’? Sol. London forces are the type of intermolecular attractive forces. These are associated with the non-polar molecules. Sometimes the temporary dipoles are formed due to unsymmetrical electronic charge distributed on one molecule. Due to this molecule, the neighbouring molecule develops an induced dipole. So, the induced dipole induced dipole interactions are also known as London forces or dispersion forces. 11. Define compressibility factor. At what pressure conditions the gases are difficult to compress? Sol. Compressibility factor (Z) is defined as the ratio of product PV and nRT. Z=

PV nRT

When the pressure is very high then, Z > 1, so, the gases are difficult to compress under such conditions. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

States of Matter

17

12. What do you understand by the terms ‘isotherms’ and ‘isochores’? Sol. (i) Isotherms : Isotherms are the curves plotted by varying pressure against volume keeping temperature constant. (ii) Isochores : The lines showing the pressure temperature behaviour plotted at fixed volume are called isochores. 13. What is absolute zero temperature? Sol. Absolute zero temperature is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume. The absolute zero temperature is 0K or – 273°C. 14. What is equilibrium vapour pressure? Is boiling possible in a closed vessel? Sol. Equilibrium vapour pressure is the vapour pressure exerted by the vapours in equilibrium with its liquid at a given temperature. No, boiling is not possible in a closed vessel. 15. Calculate the volume of the vessel in which 1023 molecules of nitrogen gas are enclosed at a pressure of 0.829 atm at 28°C. Sol. Volume of vessel = ? Temperature = 28°C = 28 + 273 = 301 K Pressure = 0.829 atm R = 0.0821 atm L K–1 mol–1 6.022 × 1023 molecules = 1 mol

1019 molecules =

1019 mol = 0.166 mol 6.022  1023

Applying ideal gas equation PV = nRT 0.829 × V = 0.166 × 0.0821 × 301

V

0.166  0.0821  301 0.829

V = 4.96 L

Short Answer Type Questions : 16. Define the three critical constants. Sol. The three critical constants are critical temperature, critical pressure and critical volume.  Critical Temperature (TC) : Critical temperature of a gas may be defined as that temperature above which the gas cannot be liquefied however high the pressure may be applied. TC =

8a 27Rb

Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

18

States of Matter

Solution of Assignment (Set-1)

 Critical Pressure (PC) : At critical temperature, a certain pressure is needed to liquefy the gas. So, this pressure at critical temperature is called the critical pressure. PC =

a 27b2

 Critical Volume (VC) : The volume occupied by one mole of a gas at its critical temperature and critical pressure is known as the critical volume. VC = 3b 17. State Boyle’s law. Why mountainers have to carry oxygen cylinders with them to restore normal breathing? Sol. Boyle’s Law states that ‘at constant temperature, the pressure of a fixed amount of gas varies inversely with the volume of the gas.’ Mountaineers have to carry oxygen cylinders with them to restore normal breathing because at high altitude the density of air is less. That means less oxygen molecules occupy the same volume. Therefore, oxygen in air becomes insufficient for normal breathing. So, to avoid this situation oxygen cylinders are employed to restore normal breathing. 18. 600 cm3 of a gas at 16°C is heated to 30°C at the constant pressure. Calculate the volume of gas. (i) After expansion (ii) Expelled V1 V2 Sol. Applying Charles Law, T = T (P constant) 1 2

V1 = 600 cm3

V2 = ?

T1 = 16 + 273

T2 = 30 + 273

= 289 K

= 303 K

Putting the values, V 600 = 2 289 303

V2 = 629.06 cm3 (i) Volume of expansion = 629.06 cm3 (ii) Volume of gas expelled = (629.06 – 600) = 29.06 cm3 19. Discuss viscosity in brief. Sol. Viscosity is one of the characteristic properties of liquids. All liquids tend to flow. But some liquids flow faster than the other. This is because of the difference of intermolecular forces of attraction. So, viscosity is actually the measure of resistance to the flow of the liquid. Viscosity depends upon the following two factors. (i) Intermolecular Attractive Forces : Greater the intermolecular forces of attraction, greater is the viscosity. More viscous liquids flow slowly. (ii) Temperature : With the rise in temperature the kinetic energy of the molecules increases. Molecules with greater kinetic energy can overcome the intermolecular forces to slip past one another between the layers. So, the viscosity of liquids, decreases with rise in temperature. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment (Set-1)

States of Matter

19

20. The total pressure of a gaseous mixture containing 6g of H2, 56g of N2, 88g of CO2 is 690 mm Hg. Calculate the partial pressure of CO2. Sol. Total pressure = 690 mm Moles of H2 =

6 = 3 mol 2

Moles of N2 =

56 = 2 mol 28

Moles of CO2 =

88 = 2 mol 44

nCO2 2 2 = = Mole fraction of CO2 = n + n + n 3  2  2 7 H2 N2 CO2

x CO2  0.285 Partial pressure of CO2 (pCO2) = xCO2 × Ptotal

PCO2 = 0.285 × 690 PCO2 = 196.65 mm Long Answer Type Questions : 21. Give the various postulates of kinetic theory of gases. Which two postulates are not applicable for real gases? Explain. Sol. Kinetic theory of gases provides the sound theoretical basis for the various gas laws. The kinetic theory of gases is based on the following postulates. (i) The actual volume of the gas molecules is negligible in comparison to the total volume of the container in which it is kept. (ii) There are no interaction between the gas molecules. (iii) The gas consists of extremely small particles which are in constant random motion. (iv) The particles of gas collide with each other and with the walls of the container. (v) The collisions are perfectly elastic. (vi) Different particles of the gas, have different speeds. (vii) The average kinetic energy of the gas molecules is directly proportional to the absolute temperature. Later from the experimental observations it was seen that two postulates are faulty made regarding the behaviour of gases. These are : (i) The forces of attraction between gas molecules are nil : This postulate is not valid at all pressures and temperature values. This is because at high pressure and low temperature, the molecules come close to each other. When the molecules are close to each other then the presence of intermolecular forces cannot be ruled out. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

20

States of Matter

Solution of Assignment (Set-1)

(ii) The volume occupied by the molecules themselves is negligibly small as compared to the total volume occupied by the gas. When the pressure is high and the temperature is lowered to a large extent, then the total volume of the gas decreases because of decrease in the empty space between the gas molecules whereas the volume of the individual gas molecules remains the same because the molecules themselves are incompressible. Hence under such conditions the volume occupied by the gas molecules will no longer be negligible in comparison to the total volume of the gas. 22. What is van der Waal gas equation and what is the significance of van der Waal gas constants a and b? Sol. van der Waal’s gas equation is ⎛ n2 a ⎞  P (V  nb)  nRT ⎜⎝ V 2 ⎟⎠

Where ‘a’ and ‘b’ are van der Waals gas constants. Constant ‘a’ gives the idea of intermolecular interactions of gas molecules, higher the value of a higher would be the intermolecular interactions and the gas would be easily liquefiable. Constant ‘b’ gives idea about the size of gas molecules, it is actually the excluded volume which is four times the volume of gas molecule.







Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456