# Introduction To Fundamental Concepts Of Chemistry

Chapter:2 The three States Of Matter:- Gases, Liquids And Solids

**Question/Answers**

(1)In terms of the atoms, ions and molecules that comprise substances, why are some materials solids, some liquids and still others gases at 25^{o}C?

(A) When matter acquires gaseous form?

The state of matter depends upon the pattern in which particles (atoms, ions and molecules) are arranged in that substance. If the forces of attraction are weak, then particles are far apart from each other and free to move randomly, hence the matter acquires gaseous form.

When matter acquires liquid form?

If particles have strong forces of attraction between them and they are close and compact then the state of matter will be liquid.

When matter acquires solid form?

If these atoms, ions or molecules are arranged in fixed (with strongest possible attraction) pattern in three dimensions. The substance will be in a solid form. Thus it is on account of different forces between matter particles, that these are different states of matter i.e. solid, liquid and gas

(2)The process of diffusion occurs most rapidly in gas, less rapidly in liquids, and very slowly in solids. Why?

(A)Diffusion is rapid in gasses:

Diffusion means the spreading of molecules through their movement from one place to another. Gases have large spaces between molecules. In the empty space molecules can freely move. Because of this free movement of molecules the diffusion of gases is rapid.

Diffusion is slow in liquids:

Liquids have small spaces, besides this, their molecules move less freely. The entry and movement of molecules of one liquid’s molecules into spaces of other liquid molecule, is slow. Hence the diffusion process in liquids is slow.

Diffusion is slow in slow: Solids have very small spaces between their molecules due to which free movement of molecules cannot take place; hence diffusion in solids is extremely slow.

(3)Why does water spilled on a floor evaporate more rapidly than the same amount of water in a glass?

(A)Evaporation process depends upon the surface area of the exposed liquid. Greater the exposed surface area of a liquid the greater will be the evaporation. On amount of this, the water spilled on the floor, evaporates more rapidly than the same amount of water in a container having a small surface area i.e. a glass.

(4)Would a pressure cooker be of any value on the Mount Everest?

(A) Yes, it would be, since the build up of pressure on the inside of the cooker would be isolated from the lower air pressure on the outside.

(5)No liquid ionic compounds are known, but many of the known covalent compounds are liquids and some are gases. Account for these differences.

(A)Ionic compounds are solids at room temperature because the ions are arranged in three dimensional patterns with strong attractive forces between oppositely charged ions. Very large would be required to break these electrostatic forces. Hence they are solids at room temperature.

In the case of covalent bonds, the particles are attracted by weak Vandeer Waal’s forces.Due to these weak forces of attraction, the particles are loosely held with each other and this result in the formation of liquid state or ofgaseoud state. Therefore many covalent compounds are liquid and some are gases.

(6)What is kinetic theory and how does it account for the following properties of gases, liquids and solids?

(i) Diffusibility (ii)Compression (iii)Expansion

(A) (i) Diffusibility:- We know from the kinetic theory, that due to the little space between solid molecules, the movement of particles restricted. However due to this limited vibrational movement very small diffusion takes place.

(ii)Compression:-In solid, the particles are in close contact with each other, they have very small unfilled space between them. When high pressure is applied deformation or shattering of the mass takes place.

(iii)Expansion:-When a liquid is heated, the kinetic energy of its molecules increases. As a result their motion increases and the attraction between the molecules become weaker. They go further apart and hence the liquid expands.

(7)1.40 dm^{3} volume of a gas measured at a temperature 27^{o}C and a pressure of 900 torr was found to have mass 2.273g. Calculate the molecular mass of the gas.[Hint: Calculate “n” by the formula PV=nRT and then molecular mass] (33.72 a.m.u Ans.).

(A) Solution:-

Given:-Apply the Gas Equation

PV=nRT, where

P=900/760=1.184 atom

V=1.40dm^{3}

R=0.0821 dm^{3} atm/kelvinxmole

T=27+273=300^{o}A

N=1 mole

Substitute the value in the above equation

PV=nRT

N=PV/RT=(1.184x1.4)/(0.0821x300)=0.067 mole

N=0.067 mole

Mol. Wt.= Wt. in g/mole=2.273/0.067=34 a.m.u.

Mol. Wt. is 34

(8)A12.5 dm^{3} vessel contains 4 g CH_{4}, 1.8g N_{2} and 10 g X_{e}. What is the pressure in the vessel at 0^{o}C? (0.698 atm. Ans.).[Hint: Find out the total moles of the gases and then apply PV=nRT].

(A)Given:-

Volume of the container = 12.5 dm^{3}

(i)Amount of CH_{4}= 4 g

Therefore No. of moles of CH_{4}=4/16=0.250 mole

(ii)Amount of N_{2}=1.8g

Therefore No. of moles of N_{2}=1.8/28=0.064 mole

(iii)Amount of X_{e}=10g

Therefore No. of moles of N_{2}=10/131.3=0.076

Total of moles=0.250+0.064+0.076=0.390 moles

Apply the gas equation PV=nRT

P=nRT/V=(0.390x0.0821x273)/12.5

P=8.7305/12.5=0.698 atm.

P=0.698

(9)(a)Stat and explain the Graham’s law of Diffusion.

(b)Compare the rates of diffusion of the following pairs of gases:

(i)H_{2} and D_{2} (ii)CH_{4} and He (iii)SF_{6} and SO_{2} [(i)1.41/1, (ii)2/1, (iii)1.5/1]

(A)(a)Graham’s Law Of Diffusion Of Gasses:

Definition Of Diffusion:-“The spontaneous mixing of molecules of different gases by random motion and collisions until the mixture becomes homogeneous is called diffusion of gases”.

Graham established a relationship between the rates of diffusion gases and their densities. This is called Graham’s Law os Diffusion.

Statement Of Law:- The law states that the rates of diffusion of gases are inversely proportional to the square root of the density of the gas.

Mathematical Expression Of Graham’s Law:-

rα1/

Where r= rate of diffusion

d= density of gas

We consider two gases with densities d_{1} and d_{2}, then

r_{1}α1/

Or r_{1}=K_{1}x1/ -------------------------------(i)

For 2^{nd} gas,

r_{2}α1/

Or r_{2}=K_{2}x1/ -------------------------------(ii)

From the eq.(i) and (ii) we get:-

i.e. r_{1}/r_{2}=

Because the density of a gas is directly proportional to its molecular mass, the above eq. becomes:

i.e. r_{1}/r_{2}=

Where M_{1} and M_{2} are the molecular masses of two gases.

(A)(b)(i) H_{2} and D_{2}

According to Graham’s Law of Diffusion:

r_{1}/r_{2}=

Where,

r_{1}= rate of diffusion of H_{2}

r_{2}= rate of diffusion of D_{2}

M_{1}= Mol. Wt. of H_{2}= 2

M_{2}= Mol. Wt. of D_{2}= 4

Thus,

r_{1}/r_{2}= =

The rate of diffusion of H_{2} and D_{2} is in the ration 1:4:1 Ans.

(ii)CH_{4} and He:- A/c to Graham’s Law of Diffusion:

r_{1}/r_{2}=

Where,

r_{1}= rate of diffusion of He

r_{2}= rate of diffusion of CH_{4}

M_{1}= Mol. Wt. of He= 4

M_{2}= Mol. Wt. of CH_{4}= 16

Therefore r_{1}/r_{2}= =

The ratio in diffusion is 2:1 Ans.

(c)SF_{6} and SO_{2}:- According to Graham’s Law of Diffusion:

r_{1}/r_{2}=

Where,

r_{1}= rate of diffusion of SO_{2}

r_{2}= rate of diffusion of SF_{6}

M_{1}= Mol. Wt. of SO_{2}=6 4

M_{2}= Mol. Wt. of SF_{6}= 146

Therefore r_{1}/r_{2}= =

The ratio in diffusion is 1.5:1

(10)State the following gas laws and explain in terms of Kinetic Theory.

(a)Boyle’s law (b)Charles’s law (c)Dalton’s law of partial pressure.

(A)(a)Statement of Boyle’s Law:-According to it, at constant temperature, the volume of given mass of gas is inversely proportional to pressure.

Explanation of Boyle’s Law in terms of kinetic theory:-If the temperature remains constant, the volume of a gas is decreased, then the average velocity of gas molecules will also remain constant. Because of decrease in volume, the molecules come close to each other. Hence they collide more frequently with the wall of the vessel producing higher pressure.

Mathematical Representation of Boyle’s Law:-

Mathematically, Boyle’s Law is represented as

Vα1/P [T(temperature) is being constant ]

V=K1/P (Where K is constant)

Or PV=K (1^{st} form of Boyle’s law)

PV=K

i.e. “The product of pressure and volume of a given gas at constant temperature is always constant.”

If P_{1} and V_{1} are the initial pressure and volume and P_{2}V_{2} are changed pressure and volume then according to Boyle’s law:

P_{1}V_{1}=K

P_{2}V_{2}=K (2^{nd} form of Boyle’s law)

P_{1}V_{1}= P_{2}V_{2}

This mathematical expression generally used in solving numerical problems.

(b)Charle’s law:-At constant pressure, the volume of a given mass of a gas varies directly with absolute temperature.

Explanation of Charle’s law In Terms Of Kinetic Theory:-The average molecular velocity decrease with the decrease of temperature.The decrease velocity causes the gas to shrink, hence the volume occupied will be small (the reverse will be the case with increase of temperature) this is the Charle’s Law.

Mathematical Representation Of Charle’s Law:-

The Law represented as

VαT (P is being constant)

Or V=KxT (Where K is a constant)

Or V/T=K

i.e. at constant pressure, the ratio of volume of a gas to its absolute temperature is always constant.

If V_{1} and V_{2} are the initial and final volume and T_{1} and T_{2} are the initial and final absolute temperature of a gas respectively then according to Charle’s law.

V_{1}/T_{1}=K

Similarly V_{2}/T_{2}=K

Or V_{1}/T_{1}= V_{2}/T_{2 } (Second form of Charle’s law)

V_{1}/T_{1}= V_{2}/T_{2}

This equation is used to solve the numerical problems, based on Charle’s Law.

(c)Dalton’s Law Of Partial Pressure:-All gases are miscible in all proportions. John Dalton 1801 studied mixtures of gases evolved a law, called Dalton’s Law of partial pressure.

Statement of Dalton’s Law Of Partial Pressure: The sum of the partial pressure of all the different gases in a mixture is equal to the total pressure of the mixture.

Mathematical form of Dalton’s Law Of Partial Pressure:

The law can be written as,

P=P_{1}+P_{2}+P_{3}--------------

Where P=Total pressure exerted by all the gases

P_{1}=Total pressure exerted by all the gas “1”

P_{2}=Total pressure exerted by all the gas “2”

P_{3}=Total pressure exerted by all the gas “3”

The general gas equation is PV =nRT (a)

Therefore,

P_{1}V=n_{1}RT (b) (For gas-1)

P_{2}V=n_{2}RT (c) (For gas-2)

P_{3}V=n_{3}RT (d) (For gas-3)

Adding (b),(c) and we get,

V(P_{1}+P_{2}+P_{3})=(n_{1}+n_{2}+n_{3})RT

As n=n_{1}+n_{2}+n_{3} (n=Total number of moles)

Substitute the value of n_{1}+n_{2}+n_{3} in the above equation (ii), we get

V(P_{1}+P_{2}+P_{3})= nRT

Dividing equation (b) by (a) we get,

P_{1}V/PV=n_{1}RT/nRT

Or P_{1}/P=n_{1}/n

Or P=P_{1}x(n_{1}/n)

(Where n=mole of gas 1, P_{1}=Partial pressure of gas 1 )

Or P_{gas}=P_{total}x(n_{gas}/n_{total})

Partial pressure of gas= Total pressure Number of moles of gas/Total no. of moles

By the above formula, the partial pressure of a gas can be calculated.

Explanation of Dalton’s Law Of Partial Pressure:-In an inert mixture of gases, the individual gas exerts its own pressure due to the collision of its own molecules with the walls of the container. Thus the total pressure produced on the walls of the container will be the sum of the pressure of all the individual gases present in the mixture.

(11)Define the following: (i)Melting Point (ii)Boiling point (iii)Diffusion (iv)Latent heat of fusion

(A)(i)Melting:- When solids are heated they change to liquid state. This is called as melting.

(ii)Boiling Point:- The boiling point of a substance is the temperature at which the vapour pressure of the liquid equals the pressure surrounding the liquid and the liquid changes into a vapour.

(iii)Diffusion:-“The spontaneous mixing of molecules of different gases by random motion and collisions until the mixture becomes homogeneous is called diffusion of gases”.

(iv)Latent heat of fusion:-It is defined as the heat energy required to change 1g of a solid into liquid at its melting point.

(12)Four containers of equal volume are filled as follows:

(i)2.0g H_{2} at 0^{o}C (ii)1.0g H_{2} at 273^{o}C (iii)24g O_{2} at 0^{o}C (iv)16g CH_{4} at 273^{o}C

Which container (a)is at the greatest pressure (b)is at the lowest pressure.[a=(iv) b= (iii) Ans.]

Solution: Apply the PV=nRT formula

Suppose the volume of container is 1dm^{3}.

For (i) 2 g of H_{2 }at 0^{o}C:

No. of mole of H_{2} = 2/2=1 mole

T=273^{o}A

R=0.0821

P=?

P=nRT/V By putting the values

=(1x0.0821x273)/1=22.4 atm.-------------------------(i)

For (ii) 1.0g H_{2} at 273^{o}C:

No. of mole of H_{2} = 1/2=0.5 mole

T=273+273=546^{o}A

R=0.0821

P=?

V=1 dm^{3}

P=nRT/V By putting the values

=(0.5x0.0821x546)/1=22.41 atm.-------------------------(ii)

For (iii)24g O_{2} at 0^{o}C

No. of mole of N_{2} = 42/28=1.5 mole

T=273+273=546^{o}A

R=0.0821

P=?

V=1 dm^{3}

P=nRT/V By putting the values

=(1.5x0.0821x546)/1=67.24 atm.-------------------------(iii)

For (iv)16g CH_{4} at 273^{o}C

No. of mole of CH_{4} = 32/16=2 moles

T=273+273=546^{o}A

R=0.0821

P=?

V=1 dm^{3}

P=nRT/V By putting the values

=(2x0.0821x546)/1=89.65 atm.-------------------------(iv)

Result:-Thus (a) greatest pressure is in container (iv) 89.65 atm.

(b) lowest pressure is in container (iii) 16.8 atm.

(13)Explain the relationship between intermolecular attractions and the kinetic energy of molecules in determining the physical state of a substance.

(A)

(14)Describe gases, liquids and solids on a molecular basis and explain their behaviour and change of state.

(A)

(15)Co-relate the energy changes accompanying changes of state.

(A)When solids are heated, their particles acquire greater kinetic energy and vibrate more violently. When the particles of solids got sufficient amount of energy, their vibration overcome the attractive forces and the particles become mobile. The solid changes to liquid.

When a liquid is heated, the kinetic energy of its molecule increases. At sufficient heating, the kinetic energy of some of the molecules becomes so high that they leave the surface of the liquid in the form of vapours i.e. in a gaseous state.

When the temperature of a gas is lowered, the kinetic energy of its molecule becomes less. As a result of it, the intermolecular attractions become dominant. At sufficient low temperature, the kinetic energy of the molecules become so low that attractive forces become capable of holding molecules firmly, i.e. the gas changes to liquid.

The same is true when the temperature of liquid is lowered further, the molecules hold together very firmly and the liquid changers to solid.

(16)Explain in terms of kinetic molecular model:

(a)Why liquids and solids cannot be compressed as gases can?

(b)Why solids do not flow as liquids and gases do?

(c)Why food is cooked more quickly in a pressure cooker than in a covered pot?

(17)Explain (i)Viscosity (ii)Surface tension (iii)Vapour pressure.

(A)(i)Viscosity:- The liquids have the ability to flow because molecules of liquids can slide over each other and the resistance of a liquid to flow is called Viscosity.

A liquid in a tube is considered as made up of a series of molecular layers. The layer of the liquid in contract with the walls of the tube remains stationary. The layer in the centre of the tube has highest velocity. Each layer exerts a drag on the next layer and cause resistance to flow.

(ii) Surface tension:-It is defined as: “The force, in dynes, acting upon a line of 1 cm length on the surface of the liquid.”

It is denoted by the Greek letter ϒ (Gamma).

Units of Surface Tension:- The units used are ergs/cm or dynes/cm.

(iii)Vapour pressure:-The pressure exerted by vapours in equilibrium with its pure liquid at a given temperature is called vapour pressure.

Units of pressure: It is expressed in mm of Hg, atmosphere, torr or Newton.

(18)What are crystalline and amorphous solids? Explain different crystal systems.

(A)Crystalline Solids:- The solids in which structural units, (the atoms, ions or molecules are arranged in regular, repeating, three dimensional manner are called Crystalline solids.

Amorphous Solids:- The solids which are rigid and incompressible but do not have characteristic geometrical shapes are called amorphous solids. In such solids the ultimate particles do not have any definit pattern of arrangement. Examples are plastic, glass, rubber and graphite.

Atomic Crystals:- Atomic crystals are also called metallic crystals. The atoms are joined together by a special type of bond called metallic bond. A metal is pictured as an arrangement of positive ions called kernel (a metal atom less valence electron) immersed in a sea of mobile electrons.

Ionic Crystals: Ionic crystals, positive and negative ions are held together by electrostatic forces of attraction. The force of attraction between oppositely charged ions is very strong and hence a large amount of energy is required to separate the ions.

Covalent Crystals: In such crystals, atoms are held together by the continuous system of covalent bonds. Diamond and graphite are good examples of covalent crystals. Diamond is a conductor while graphite is a good conductor of electricity.

Molecular Crystals: Such crystals are composed of molecules which do not carry any charge. The binding forces with which the molecules are held together are:

(i)Vanfer Waal’s Forces

(ii)Dipole-Dipole Forces

(19)State and explain Avogadro’s law. How it helps in determining relative molecular mass?

(A)Statement of Avogadro’s Law:- Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.

Since equal number of molecules means equal number of moles and since the number of moles of any gas varies directly with the volume i.e.

Vαn (Where “n” is the number of moles)

Therefore we infer that one mole of all gases would occupy the same volume at a given temperature and pressure.

One mole of any gas contains the same number of molecules (Avogadro’s number=6.022x10^{23}) and Avogadro’s Law must occupy the same volume at any given temperature and pressure. It is found that one mole of a gas at S.T.P. occupies a volume of 22.4dm^{3}.

(20)What are solids? Describe Atomic, Ionic and Molecular solids.

(A) Crystalline Solids:- The solids in which structural units, (the atoms, ions or molecules are arranged in regular, repeating, three dimensional manner are called Crystalline solids.

Amorphous Solids:- The solids which are rigid and incompressible but do not have characteristic geometrical shapes are called amorphous solids. In such solids the ultimate particles do not have any definit pattern of arrangement. Examples are plastic, glass, rubber and graphite.

Atomic Solids:- Atomic crystals are also called metallic crystals. The atoms are joined together by a special type of bond called metallic bond. A metal is pictured as an arrangement of positive ions called kernel (a metal atom less valence electron) immersed in a sea of mobile electrons.

Ionic Solids: Ionic crystals, positive and negative ions are held together by electrostatic forces of attraction. The force of attraction between oppositely charged ions is very strong and hence a large amount of energy is required to separate the ions.

Covalent Solids: In such crystals, atoms are held together by the continuous system of covalent bonds. Diamond and graphite are good examples of covalent crystals. Diamond is a conductor while graphite is a good conductor of electricity.

Molecular Solids: Such crystals are composed of molecules which do not carry any charge. The binding forces with which the molecules are held together are:

(i)Vander Waal’s Forces

(ii)Dipole-Dipole Forces

(21)Explain the following:

(i)A falling drop of a liquid is spherical.

(ii)A drop of ink spreads on blotting paper.

(iii)Evaporation is a cooling process.

(v)Boiling point of a liquid remains constant although heat is continuously supplied to the liquid.

(vi)Honey is more viscous than water.

(vii)Mercury has its meniscus upward.

(A)(ii) Falling drop of a liquid is always spherical in shape due to surface tension. The inward forces on the surface molecules of the liquid droplet tend to cause the surface to volume ratio as small as possible. Since surface to volume ratio is minimum for the spherical shape that’s why falling drop of a liquid is spherical.

(ii) Blotting paper has a large number of pores (tiny holes) on its surface. Each pore acts like a capillary tube. When a drop of ink is placed on a blotting paper, it spreads on the paper due to capillary action.

(iii) According to kinetic theory, the temperature is a measure of average kinetic energy of the molecules of a liquid. During evaporation, the escape of high energy molecules from the surface of a liquid, lowers the average kinetic energy of the remaining molecules and therefore, the temperature of the liquids falls down, thus evaporation is a cooling process.

(iv) Boiling point of a liquid remains constant although heat is continuously supplied to the liquid because the heat that is applied is being used to change the liquid to a gas. Take H_{2}O for example. It takes 540 cal of heat to vaporize one gram of H2O (i.e. change it from a liquid to a gas). When all of the liquid has been vaporized, then you will begin to heat the gas that was produced and the temperature of the gas will start to increase.

(vi)It is due to the difference in viscosity. Honey is more viscous (thick) than water due to strong inter-molecular forces. On the other hand water is less viscous than honey. Therefore it is easier to pour water as compared to honey.

(vii) Formation of meniscus depends on cohesive and adhesive forces in a liquid.In water adhesive forces are larger than the cohesive forces, therefore water in a container stick to the wall of container and rises a little bit and form concave meniscus.In mercury cohesive forces are strong as compare to adhesive forces. Therefore mercury falls down from the sides attach to the wall and a concave or upward meniscus is obtained.

(22)40 dm^{3} of hydrogen gas was collected over water at 831 torr Hg pressure at 23^{o}C. What would be the volume of dry hydrogen gas at standard conditions? The vapour pressure of water at 23^{o}C is 21 torr of Hg.(Ans. :- 39.23 dm^{3}).

(A)Solution:

Volume of H_{2} gas V_{1} + 40 dm^{3}

Pressure of dry gas P_{1} =831-21=810 torr

T_{1} = 273+23+296^{o}K

V_{2} = ?

P_{2} = 760 torr

T_{2} = 273^{o}K

Apply the gas equation:

(P_{1}V_{1})/T_{1}= (P_{2}V_{2})/T_{2}

Or P_{1}V_{1}T_{1}= P_{2}V_{2} T_{2}

V_{2} = (P_{2}V_{1} T_{2})/(P_{2} T_{1})

By putting the values in the equation we get

V_{2} = (810x40x273)/(760x296)

= 39.32 dm^{3}

Volume of gas = = 39.32 dm^{3 }Ans.

(23)What is density of methane gas (CH_{4}) at 127^{o}C and 3.5 atmosphere? (Ans.:-density= 1.70g/dm^{3}).

(A)Solution:-

Molecular wt. of CH_{4} = 16

Pressure of gas = P = 3.50 atm.

Gas constant = R = 0.0821

Temperature = T = 273+127 =400^{o}K

Apply the gas equation, we have:

PV = nRT

PV = (m/mol. Wt.)xRT

d = m/v = (Pxmol. Wt.)/RT

Or d = (PxMol. Wt.)/RT {Therefore n=mass(g wt.)/mol. Wt.}

= (3.50x16)/(0.0821x400)

= 1.70 g/dm^{3}

Density = 1. 70 g/dm^{3 }Ans.

** **

**EXAMPLES**

Example No. 1)What is the pressure in torr of 3.5 atm?

Solution:Therefore 1 atm = 760 torr

Therefore 3.5 atm = (760/1)x3.5=2660 torr

Example No. 2)How many atmospheres correspond to 950 torr?

Solution:-Therefore 760 torr = 1atm

Therefore 950 torr = (1x950)/760=1.25 atm

Example No. 3)What volume does 400 cm^{3} sample of a gas at 700 torr occupy when the pressure is changed to 2 atm?

Solution:-Tabulate the data. (700 torr=700/760=0.92 atm)

P V T

Original 0.92 atm 400 cm^{3} K

Changed to 2.0 atm V K

According to the Boyle’s law,

P_{1}V_{1} = P_{2}V_{2}

V_{2} = (P_{1}V_{1}/ P_{2}) = (0.92x400)/2 = 184 cm^{3}

Example No. 4)A child’s ballon has a volume of 3.80 dm^{3} when the temperature is 35^{o}C. What is the volume, if the balloon is put into a refrigerator and cooled to 5^{o}C?Assume that pressure inside the balloon is equal to atmospheric pressure at all times.

Solution:- Tabulate the data

Original

V_{1}=3.80 dm^{3}, T_{1}=35+273= 308 K, P=K

Changed to V_{2}=?, T_{2}=5+273=278K, P=K

According to Charle’s law

V_{1}/T_{1} = V_{2}/T_{2}

V_{2} = (V_{2}/T_{2})/T_{2}

= (3.8x278)/308 = 3.42 dm^{3}

Example No. 5)What will be the volume occupied by 14 g of nitrogen at 20^{o}C and 740 torr pressure?

Data: P = 740 torr = 740/760 = 0.974 atm.

V = ?

N = 14/28 = 0.5 mole

R = 0.0821 dm^{3} atm/K/mole

T = 20+273 =293 K

PV = nRT

V = nRT/P = (0.5x0.0821x293)/0.974 = 12.345 dm^{3}

Example No. 6)A certain mass of nitrogen gas at 20^{o}C and at 740 torr pressure occupies 12.345 dm^{3}. Calculate the volume that it will occupy at S.T.P.

Data: V_{1} = 12.345 dm^{3}

V_{2} = ?

P_{1} = 740 torr

P_{2} = 740 torr

T_{1} = 20+273 = 293 K

T_{2} = 273 K

We know that:_{}

(P_{2}/V_{2})/ T_{2} = (P_{1}/V_{1})/ T_{1}

V_{2 } = (P_{1}/V_{1})/ T_{1} x T_{2}/P_{2}

= (740x12.345)/293x(273/760) = 11.12 dm^{3}

Example No. 7)Calculate the volume that will be occupied by 0.8 moles of oxygen gas taken at 30^{o}C and at 800 torr of Hg pressure?

Data:-V=? Solution:-PV = nRT

n=0.8 mole V= (nRT)/P

R= 0.0821 dm^{3} atm/deg. Mole V= (0.8x0.821x303)/1.053

T=30+273=303 K =18.899 dm^{3}

P=(800/760)=1.053 atm.

Example No. 8)The ratio of rates of diffusion of two gases A and B is 1.5:1. If the relative molecular mass of gas is 16, find out the relative molecular mass of gas B?

Solution:- According to Graham’s law:

r_{A}/r_{B}=

1.5/1=

By taking square of both sides

(1.5/1)^{2} = ( )^{2}

Therefore M_{B} = (1.5)^{2}x16

= 2.25x16

= 36

Example No. 9)Compare the rates of Helium and sulphur dioxide.

Solution:- The molecular masses of Helium and sulphur dioxide are 4 and 64 respectively. Hence,

r_{He}/r_{SO2}= = =8/2 =4

Helium diffuses four times as fast as SO_{2}.

Example No. 10)In the preparation of oxygen in the laboratory, 500 cm^{3} of the moist gas was collected over water at 25^{o}C and 724 torr. What volume of dry oxygen at S.T.P. was produced?(Pressure of water vapours at 25^{o}C=24 torr).

Solution:- Presure of dry O_{2} is calculated by using Dalton’s law:

P_{dry} O_{2} = P_{moist gas} – P_{H2O}

= 724 torr- 24 torr

= 700 torr

Now calculate the volume of dry gas at S.T.P.

(P_{1}/V_{1})/ T_{1} = (P_{2}/V_{2})/ T_{2} (gas equation)

V_{dry gas} = P_{1}V_{1}T_{2}/P_{2}T_{1}

Example No. 11)A mixture of gases at 760 torr contains 2 moles nitrogen and 4 moles carbon dioxide. What is the partial pressure of each gas in torr?

Solution:- Total moles (n) = moles; total pressure = 760 torr

P_{gas} = P_{(total)}Xn_{(gas)}/ n_{(total)}

P_{N2} = (760x2)/6 = 253.33 torr

P_{CO2} = (760x4)/6 = 506.67 torr

Check: Total pressure = Sum of partial pressures = 760 torr

## Comments

**Abdul rehman** on October 25, 2017:

Please, sir can you upload progress tests for more chapters? Please......

**Arslan Qadir** on December 12, 2015:

Please, sir can you upload progress tests for more chapters? Please......

And thank you for this test..☺