STATES OF MATTER.doc

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STATES OF MATTER

Water can exist as ice, which is a solid; it can exist as liquid; or it can exist inthe gaseous state as water vapour or steam

Chemical properties of a substance do not change with the change of itsphysical state;

nature of intermolecular forces, molecular interactions and effect of thermalenergy on the motion of particles because a balance between these determinesthe state of a substance.

Threestatesis depictedas follows:

INTERMOLECULARFORCES

Intermolecular forces are the forces of attraction and repulsion between interacting particles (atoms

and molecules

Dispersion Forces or London Forces---

Atoms and non polar molecules are electricallysymmetrical and have no dipole

momentbecause their electronic charge cloud issymmetrically distributed. But a

dipole maydevelop momentarily in such atoms andmolecules.The intermolecularforces of attractioninbetween two temporarydipoles non polar substances isnown as London forceor dispersion force.
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Theseforces are always attractie and interactionener!y is inersely

proportional to the si"thpower of the distance between two interactin!

particles!i.e.,#/r$wherer is the distancebetween two particles". #hese forces wor

at short distances !$%&& pm" and their magnitude depends on the polarisability of the

particle.

Dipole% Dipole Forces

'ipole(dipole interactions occur between molecules that have permanent net

dipoles

between )Cl* molecules, +Cl molecules, C-Cl molecules and -Cl molecules.

#he dipole ( dipole interactions are stronger than the ondon forces.

#he dipole(dipole interaction energy between stationary molecules is

proportional to /0r and for rotating molecules is proportional to /0r1, where r is

the distance between the polar molecules.

Dipole % &nduced dipole

A dipole(induced dipole attraction is a wea attraction that results when a polar

molecule induces a dipole in an atom or in a non polar molecule by disturbing

the arrangement of electrons in the non(polar species.

#he interaction energy is proportional to /0r1 where r is the distance between

two molecules.
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'ydro!en (ondin! 2n a molecule, whenever the hydrogen atom is directly lined to a highly

electronegative atom !lie 3, 4 or 5", the shared pair of electron is pulled moretowards this atom and it acquires a slight negative charge !d6" and thehydrogen atom acquires a slight positive !d(" charge.

#here occurs a wea bond formation between the negative end of one moleculeand the positive end of the other. #his bond is nown as a hydrogen bond,

which is represented by dotted lines. 2t involves dipole(dipole interactions.

THE GASLAWS


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E"periments of (oyle) in a *uantitatie manner proe that !ases are hi!hly

compressible because when a !ien mass of a !as is compressed) the same

number of molecules occupy a smaller space+ This means that !ases become

denser at hi!h pressure+

A relationship can be obtained between density and pressure of a !as by usin!

(oyle,s law

By definition, density 7d8 is related to the mass 7m8 and the volume 7V8 by the relation

d9m0v. 2f we put value of V in this equation from Boyle8s law equation, we obtain the

relationship.

#his shows that at a constant temperature, pressure is directly proportional to the density of

a fixed mass ofthe gas.

Problem5.1

A balloon is filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1

bar pressure the gas occupies 2.27 L volume, upto what volume can the balloon be expanded ?

Solution

According to Boyles Law p1V1=p2V2

if p1is 1 bar, V1will be 2.27L

if p2= 0.2 bar, then V2=p1V1/P2

:V2=1 bar2.27L/0.2bar=11.35L

Since balloon bursts at 0.2 bar pressure,the volume of balloon should be less than 11.35 L.


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Each line of the volume vs temperature graph is called isobar.


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All gases obey Charles8 law at very low pressures and high temperatures.

Problem5.2

On a ship sailing in pacific ocean where temperature is 23.4 C , a balloon is filled with 2 L air. What

will be the volume of the balloon when the ship reaches Indian ocean, where temperature is 26.1C ?

Solution

V1= 2L

T1= (23.4 + 273)K = 296.4 K

T2= 26.1 + 273 = 299.1 K

From Charles law

V1/T1=V2/T2:V2=(V1T2)/T1:V2=2L 299.1K / 296.4K

=2L 1.009

=2.018L

.ay Lussac/s Law 01ressure-Temperature Relationship2,


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R =(105Pa) (22.71 103m3)/(1mol) (273.15K)

=8.314 Pa m3k-1mol-1

=8.314 X10-2bar L K-1mol-1

=8.314 JK-1mol-1


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At STP conditions used earlier (00C and 1 atm pressure), value of R is 8.20578 X10-2L atm K-1mol-1.

Ideal gas equation is a relation between four variables and it describes the state of any gas, therefore,

it is also called equation

of state.

Let us now go back to the ideal gas equation. This is the relationship for the simultaneous variation of

the variables. If temperature, volume and pressure of a fixed amount of gas vary from T 1, V1and p1to

T2, V2and p2then we can write

p1V1/T1and p2V2/T2:p1V1/T1= p2V2/T2(5.19)

Problem 5.3

At 25C and 760 mm of Hg pressure a gas occupies 600 mL volume. What ill !e its pressure at a height

here temperature is "0C and volume of the gas is 6#0 mL.

Solution

p"$ 760 mm Hg% &%su!'gt("$ 600 mL

)"$ 25 * 27+ $ 2,-

&2$ 6#0 mL and )2 $ "0 * 27+ $ 2-+

According to Com!ined gas la

p"&"/)"$ p2&2/)2

:p2$p"&")2/ &"/)2

:p2$760 mm Hg1600 ml12-+ 1/6#0 ml12,- 1$676.6 mm Hg


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ases are generally collected over water and therefore are moist. +ressure of dry gascan be calculated by subtracting vapour pressure of water from the total pressure ofthe moist gas which contains water vapours also. +ressure exerted by saturated watervapour is called aqueous tension


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p'ry gas9 p#otal< Aqueous tension !%.*="

Partial pressure in terms of mole fraction

1artial pressure in terms of mole fraction )uppose at the temperature #, three gases,

enclosed in the volume V, exert partialpressurep/,p*andprespectively. then,

here n/n*and nare number of moles of these gases. #hus, expression for total pressure will

be

p#otal9p/6p*6p

4n dividingp/byp totalwe get

here n 9 n/6n*6n

x/is called mole fraction of first gas.

#hus,p/9x/p total

)imilarly for other two gases we can write p*9x*ptotalandp9xptotal

#hus a general equation can be written as pi9xiptotal

wherepi and xi are partial pressure and mole fraction of i thgas respectively.

Problem 5.4

A neon3dio4gen mi4ture contains 70.6 g dio4gen and "67.5 g neon. f pressure of the mi4ture of gases in

the clinder is 25 !ar. What is the partial pressure of dio4gen and neon in the mi4ture

8um!er of moles of dio4gen $70.6g/+2gmol9"$ 2.2" mol

8um!er of moles of neon $"67.5g/20gmol9"$ -.+75 mol

:ole fraction of dio4gen $ 2.2"/2.2" * -.+751 $ 2.2"/"0.5-5 $ 0.2"

:ole fraction of neon $ -.+75/2.2"* -.+751 $ 0.7,

Alternativel%

mole fraction of neon $ " ; 0.2" $ 0.7,


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EXERCISES

5.1 What will be the minimum pressure required to compress 500 dm3of air at 1 bar to 200 dm3at

30C?

5.2 A vessel of 120 mL capacity contains a certain amount of gas at 35 C and 1.2 bar pressure. The

gas is transferred to another vessel of volume 180 mL at 35 C. What would be its pressure?

5.3 Using the equation of state pV=nRT; show that at a given temperature density of a gas is

proportional to gas pressure p.

5.4 At 0C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What

is the molecular mass of the oxide?

5.5 Pressure of 1 g of an ideal gas A at 27 C is found to be 2 bar. When 2 g of another ideal gas B is

introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship


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between their molecular masses 5.7 What will be the pressure exerted by a mixture of 3.2 g of

methane and 4.4 g of carbon dioxide contained in a 9 dm3flask at 27 C ?

5.8 What will be the pressure of the gaseous mixture when 0.5 L of H2at 0.8 bar and 2.0 L of dioxygen

at 0.7 bar are introduced in a 1L vessel at 27C?

5.9 Density of a gas is found to be 5.46 g/dm3at 27 C at 2 bar pressure. What will be its density at

STP?

5.10 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 C and 0.1 bar pressure. What is the

molar mass of phosphorus? 5.7 What will be the pressure exerted by a mixture of 3.2 g of methane

and 4.4 g of carbon dioxide contained in a 9 dm3flask at 27 C ?

5.8 What will be the pressure of the gaseous mixture when 0.5 L of H2at 0.8 bar and 2.0 L of dioxygen

at 0.7 bar are introduced in a 1L vessel at 27C?

5.9 Density of a gas is found to be 5.46 g/dm3at 27 C at 2 bar pressure. What will be its density atSTP?

5.10 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 C and 0.1 bar pressure. What is the

molar mass of phosphorus?

5.11 A student forgot to add the reaction mixture to the round bottomed flask at 27 C but instead

he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a

pyrometer he found the temperature of the flask was 477 C. What fraction of air would have been

expelled out?

5.12 Calculate the temperature of 4.0 mol of a gas occupying 5 dm3at 3.32 bar. (R = 0.083 bar dm3K-1

mol-1).

5.13 Calculate the total number of electrons present in 1.4 g of dinitrogen gas.

5.14 How much time would it take to distribute one Avogadro number of wheat grains, if 1010 grains

are distributed each second ?

5.15 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a

vessel of 1 dm3 at 27C. R = 0.083 bar dm3K-1mol-1.

5.16 Pay load is defined as the difference between the mass of displaced air and the mass of the

balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at

1.66 bar at 27C. (Density of air = 1.2 kg m -3and

R = 0.083 bar dm3K-1mol-1).

5.17 Calculate the volume occupied by 8.8 g of CO2 at 31.1C and 1 bar pressure. R = 0.083 bar L K -1

mol-1

.

5.18 2.9 g of a gas at 95 C occupied the same volume as 0.184 g of dihydrogen at 17 C, at the

same pressure. What is the molar mass of the gas?


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5.19 A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of

dihydrogen. Calculate the partial pressure of dihydrogen.

5.20 What would be the SI unit for the quantity pV2T2/n ?

5.21 In terms of Charles law explain why -273 C is the lowest possible temperature.

5.22 Critical temperature for carbon dioxide and methane are 31.1 C and -81.9 C respectively.

Which of these has stronger intermolecular forces and why?

5.23 Explain the physical significance of v 5.1 2.5 bar

5.2 0.8 bar

5.4 70 g/mol

5.5 MB= 4MA

5.6 202.5 mL

5.7 8.314 x 104Pa

5.8 1.8 bar

5.9 3g/dm3

5.10 1247.7 g

5.11 3/5

5.12 50 K

5.13 4.2154 x 1023electrons

5.14 1.90956 x 106year

5.15 56.025 bar

5.16 3811.1 kg

5.17 5.05 L

5.18 40 g mol1

5.19 0.8 bar


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an der Waals parameters.

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