ashakil: Chapter, 03: Gases Properties of Gases. 1st year

Chapter, 03: Gases

Properties of Gases.

(1) Very weak forces exist among gas molecules.

(2) Gas molecules are in constant motion in all possible directions.

(3) Gases have no definite shape and no definite volume.

(4) The gases exert pressure on the wall of container; the pressure of gases is due to

collisions of the gas molecules with each other and with the walls of container.

(5) The gases are compressible because there is larger distance among gas molecules.

(6) The gases flow from high concentrations to lower concentration region. This is

called diffusion of gases.

(7) The small sized gas molecules get out of container one by one through a narrow

hole, this is called effusion of gases.

Units of Pressure.

(1) Atmosphere(atm): is the force exerted by 76cm long Hg column on area of 1cm²at 0C^oat sea level i.e. 1atm = 76cm Hg or 1atm = 760mm Hg

(2) SI unit: is Nm^-2 or Pascal (1atm = 101325 Nm^-2)

(3) Torr: It is a smaller unit of pressure (1torr = 1mm Hg)

Gas Laws: Describe the relationship the number of moles(n),volume (V),Temperature(T) and Pressure of gases.

(1)Boyle’s Law: Volume of given mass of a gas is inversely proportional to pressure at constant temperature.

V α 1/P

V = K/P

PV = K

P₁V₁=P₂V₂= P₃V₃=K=Constant

This equation’s means that by changing pressure, the volume also changes but in such a way that product of pressure, volume remains constant.

Experimental verification of Boyle’s Law.

P₁V₁ = P₂V₂ = P₃V₃

2atmx1dm³ = 2atm dm³ , 4atmx1/2dm³ = 2atm dm³ , 6atmx 1/3dm³ = 2atm dm³

Example (1):-A gas having a volume of 10dm³ is enclosed in a vessel at and 2.5atm

pressure. The gas is allowed to expand until the new pressure is 2atm.

What will be the new volume, if temp is maintained at 273K.

Given that

Solution: As

so, According to Boyl,s law

Graphical explaination of Boyle’s law:-

(1)V vs P: at constant temperature.

The curve obtained by plotting pressure against volume at constant temperature is called “isotherm”which shows that if pressure increases, volume decreases and if pressure decreases volume increases.

At higher constant temperatures, the isotherms go away from both axis, because at higher temperatures, the gases occupy larger volumes.

(2) P vs 1/V : at constant temperature.

V^-1

This graph shows the direct relation between pressure and inverse of volume (V^-1) such that by increasing pressure, the inverse of volume also increases and vice versa.

The straight lines get close to pressure axis for higher temperature, because at higher temperature the gases occupy larger volumes whose inverses are low.

At very low pressure, the gases occupy such a large volume, that its inverse is very low and can be neglected, therefore, the straight lines, touch origin of both axis.

(3) PV vs P: at constant temperature.

This graph mean, that by changing pressure, the volume also get changed but in such a way that products of pressure volume remains constant.

Assignments:

Exercise Q.No.(4),a,b,c,d and Q.No.(15),d.

Exercise Q.No.(4)(a):What is Boyl,s law. Give its experimental verification.

Answer:---------------- See Boyl,s law and its verification.

Exercise Q.No.(4)(b):What are isotherms. What happens to positions of isotherms when they

are plotted at higher temperature for a particular gas.

Answer:----------------- See graphical representation of Boyle’s law

Exercise Q.No.(4)(c):Why do we get a straight line when pressure is plotted against inverse

of volume.This straight line change its position by varying temperature justify it.

Answer:-----------------See graph of P vs V^-1 at constant temperature.

Exercise Q.No.(4)(d):How will you explain that the value of constant K in the equation

depends on (1) Temperature of gas (2) Quantity of gas

Answer:------------------The equation

depends on

(1)Temperature of gas:By changing temperature volume of gas also changes to keep the pressure constant,therefore,product of pressure volume changes and as a result value of K changes.

(2). Quantity of gas:---By changing no. of moles of gas(quantity of gas) the pressure volume

product get changed at constant temperature,due to which the value of

K changes.

Exercise Q.No.(15)(i) The plot of PV vs P is a straight line at constant temperature and with

a fixed no. of moles of an ideal gas. Give reason.

Answer: ---------See graphical representation of Boyle’s law PV vs P at constant temperature.

Charles’s law: The volume of given mass of a gas is directly proportional to absolute temperature, if pressure is constant.

. If pressure is constant.

Charles’s law can also be defined as Volume of given mass of a gas

increases or

decreases by factoer

for each

increase or decrease of temperature, if pressure is constant.

Experimental verification of Charles’s law:

If heat is supplied to a gas in a cylinder then volume increases but

Constant

Graphical representation of Charles’s law(V vs T at constant Pressure)

-273C^o T(C^o)

This graph mean the volume of a gas is directly proportional to temperature, if pressure is constant i-e. If temperature decreases volume also decreases and vice versa.

The dotted part of straight line shows the change of gas into liquid or solid just before

The straight line touch temperature axis at

, it means that volume of a gas is zero at this temperature theoretically, called “Absolute zero temperature”

The slope of line depend, on the no. of moles of gases such that greater the no. of moles high will be the volume of gas, the straight line will be more close to the volume axis,therefore,greater will be the slope of the straight line.

Example (2):-of H₂ is cooled from to by maintaining the

pressure constant calculate the new volume of gas at low temperature.

Solution:

By applying the Charles’s law:

Absolute Zero Temperature: The volume of ideal gases becomes zero theoretically just before

or 0K. Which is known as, absolute Zero temperature.

Derivation of absolute zero temperature:

Let volume of a gas at

= V_o

If temperature is decrease to

C^o,

according to Charles’s law, V_-1C^o = V_o - V_o

₂₇₃

If temperature is decreased to

( 1 - 1)

Zero.

Scales of Temperature:

(1) Centigrade C^o(2).Fahrenheit

(3). Kelvin

Interconversion Formulae:

Assignments

Q.No.1(i) Q.No.5 a, b, c Q.6.(a)

Exercise Q.No.(1)(i): If pressure is constant, at which temperature the volume of a gas will

become twice of what it is

(a)

(b)

(c)

(d)

For help: O C^o = 273 K, if Volume of gas will become twice at constant pressure, then

according to Charles’s law temperature will also be doubled i-e 546 K.

Exercise Q.No.(5)(a):What is the Charles’s law, which scale of temperature is used to verify

that

Answer: See Charles’s law.

Exercise Q.No.(5)(b): A sample of CO₂ gas occupy 150 ml at

. It is then cooled at

constant pressure, until it occupies 100ml. What is the new temperature.

Solution: Pressure = constant,

Exercise Q.No.(5)(c): Do you think that volume of any quantity of a gas become zero at

. Is it not against the law of conservation of mass. How do you

deduce the idea of absolute zero from this information.

Answer: I think that volume of any quantity of a gas become zero theoretically but practically

the gases get changed into liquid or solid just before -273C^o, therefore, it is not

against the law of conservation of mass.

So, i concluded that absolute zero

is the lowest temperature, just before

which gases changes into liquid or solid.

Exercise Q.No.(6)(a):Throw some light on the factor

in the Charles’s law.

Answer: Charles’s law states, That volume of given mass of a gas at

, increases or decreases by factor

of its original volume for each increase / decrease of

, if pressure is constant.

Applying the above statement and decreasing the temp – of a gas upto

, the volume

of gas become zero Theoretically.

Avogadro’s law: Volume of an ideal gas is directly proportional to no. of moles of gas such

that equal volumes of different gases contains equal no. of moles at STP.

i-e

at STP.

This equation means that if

then

e.g at STP, 22.4 dm³

mole

molecules

22.4 dm³

mole

molecules

Although masses and sizes of

are different but volume of gases do not depends on sizes and molar masses because there are larger distances among gas molecules.

General Gas equation:

,If temp – constant (Boyle’s law.)

,If pressure constant (Charles’s law.)

, at STP (Avogadro’s law.)

, This is the first form of general gas equation. where R is general gas constant.

Other forms of general gas equation.

PV = nRT n = m/M

PV = m RT

PV = dV RT

MP = dRT This is 2^nd form of general gas equation.

we know that

PV = nRT, if

PV = RT

PV = R

P₁V₁ = P₂V₂= R (This is third form of general gas equation.)

T₁ T₂

Applications of general gas equation:

General gas equation is used to calculate

(i) Unit and Value of R = PV/nT

(ii) Mass of gas

but n = m/M

(iii)Density of gas,

but m/M

but m = dv

Unit and Value of R:

For 1 mol gas at STP.

or. .0821 x 760x1000mmHg cm³mol^-1K^-1

SI unit and value of R:

We know that SI unit for volume is m³ and for pressure is

1 atm = 101325Nm^-2 and 1 m³ = 1000dm³

for 1mol at STP

As 1Nm = 1 J

So,

As 1cal = 4.184 J.

So,

Note:

But 1 J = 10⁷ ergs

So,

Example(3):-A sample of nitrogen gas is enclosed in a vessel of volume 380 cm³ at and pressure of . This gas is transferred to a flask and cooled to . Calculate the pressure of gas at .

Solution:-

Give that

Applying the formula

Example(4):Calculate the density of at . What will happen to the density

a) Temperature is increased to .

b) Pressure is increased to at .

Solution:

Give data

Molar mass of

Applying the formula

a) Density at

i.e.

b) Density at

and

Example(5):Calculate the mass of 1dm³ of NH₃ gas at and pressure considering that NH₃ is behave ideally.

Solution:-

Give data Applying the formula

grams.

Molar mass of

Mass (m) of

Assignments:

ExerciseQ.1(iii)(iv) (vi).ExerciseQ. 7(a, b, c, d) ExerciseQ. 8( a, b, c)Exercise Q. 9( a, b,c)

Exercise Q .16,17,18,19,20.

Exercise Q.No.(1)(iii):Which of the following have same no.of molecules at STP?

(a) 280cm³ of CO₂ and 280 cm³ of N₂O (b) 11.2dm³ ofO₂ and 32g of O₂

(c) 44g of CO₂ and 11.2 dm³ of CO (d) 28g of N₂ and 5.6 dm³ of Oxygen

For help:Equal volumes of different gases have equal no.of moles at STP according to

Avagadros law

Exercise Q.No.(1)(iv): If absolute temperature is doubled and pressure is reduce to one

half, the volume of gas will be

(a) Unchanged (b) Increases four times (c) Reduce

(d) Be doubled

For help: let T₁ = 10 K, T₂ = 20 K and P₁ = 2atm , P₂ = 1atm

Now if, V₁ = 1dm³, V₂ = ?

According to G.G.Equation P₁V₁ = P₂V₂ OR V₂ = P₁V₁_x T₂

T₁T₂P₂T₁

Exercise Q.No.(1)(vi): Molar volume of CO₂ is maximum at

(a) STP (b)

atm (c)

atm (d)

atm

For help: Molar volume V = RT

Exercise Q.No.(7)(a):What is general gas equation? Derive it in various forms.

Answer: See derivation of general gas equation.

Exercise Q.No.(7)(b):Can we determine the molecular mass of an unknown gas, if we

know the pressure, temperature, volume and mass of that gas.

Asnwer :Yes, by applying

Exercise Q.No.(7)(c):How do you justify from general gas equation that increase in

temperature or decrease in pressure decrease the density of gas.

Answer: MP = dRT shows that

and

Exercise Q.No.(7)(d):Why do we feel comfortable in expressing the density of gases in

the units of g dm^-3 rather than gem^-3, a unit which is used to

express the densities of liquid or solids.

Answer:Because the gases occupy the volume available,therefore,they have very low

densities. So this is the reason that the densities of gases are expressed in gram/dm³.

Exercise Q.No.(8)Derive units for General gas constant (R)

a) When pressure is in atm and volume in dm³

b) When pressure is in Nm-² and Volume in dm³ .

c) When energy is in ergs.

Answer: See unit and value of R.

Exercise Q.No.(9)(a): What is Avogadro’s law?

Answer:See Avogadro’s law.

Exercise Q.No.(9)(b):Do you think that 1mol of H₂ and 1mol of NH₃at O, 1atm will

have Avogadro’s number of particles?

Answer:Yes.

Exercise Q.No.(9)(a):Justify that 1cm³ H₂ and 1cm³ CH₄ at STP will have same No. of molecules, when 1 molecule of CH₄ is 8 times heavier than that of H₂.

Answer: According to Avogadro’s law

at STP

V = Kn

V = K

constant

This equation means that If

, then

, therefore,it is justified that If volumes of different gases are equal, then no.of moles will be equal and they will have equal no.of particles.

Exercise Q.No.(16):Helium gas in a 100 cm³ container at a pressure of 500 torr is transferred to a container with a volume of 250 cm³.What will be the new pressure

(a) if no change in temperature occurs.

(b) if its temperature changes from 20C^o to 15C^o?

Solution:

V₁ = 100cm³ , V₂ = 250cm³, P₁ = 500torrs, P₂ = ?

(a) if no change in temperature occurs.

Applying the Boyles law P₁V₁ = P₂V₂ OR P₂ = P₁V₁= 500torrs x 100cm³= 200torrs.

V₂ 250cm³

(b) if its temperature changes from 20C^o to 15C^o?

T₁ = 20C^o + 273 = 293 K and T₂ = 15C^o + 273 = 288 K

Applying P₁V₁ = P₂V₂ OR P₂ = P₁V₁_xT₂=500torrs x 100cm³x288K = 196.56torrs

T₁ T₂T₁V₂293K X 250cm³

Exercise Q.No.(17)(a): What are the densities in Kg/m³ of the following gasses at STP.

(P=101325Nm^-2,T= 273 K,molecular masses are in Kg/mol^-1)

(i)Methane (ii) Oxygen(iii) Hydrogen

Solution: P=101325Nm^-2,T= 273 K,

Molar masses of CH₄, O₂,H₂ are 0.016kg/mol,0.032kg/mol,and 0.002kg/mol respectively.

Required: d_CH in kg/m³ = ? d_O in kg/m³ = ? d_H in kg/m³ = ?

We know that MP = dRT Or d = MP/RT

d_CH in kg/m³ = 0.016kgmol^-1 x 101325Nm^-2 = 0.714kgm^-3

8.134NmK^-1mol^-1 x 273K

d_O in kg/m³ = 0.032kgmol^-1 x 101325Nm^-2 = 1.428kgm^-3

8.134NmK^-1mol^-1 x 273K

d_H in kg/m³ = 0.002kgmol^-1 x 101325Nm^-2 = 0.089kgm^-3

8.134NmK^-1mol^-1 x 273K

Exercise Q.No.(17)(b): Compare the values of densities in proportion to their molar masses.

d_CH : d_{O :} d_H

0.714 : 1.428 : 0.089

0.089 0.089 0.089

8 : 16 : 1

But ratio of molar masses

CH₄ : O₂ : H₂

16 : 32 : 2

2 2 2

8 : 16 : 1

Exercise Q.No.(17)(c):How do you justify that increase of volume upto 100dm³ at 27 C^o

Of 2 moles of NH₃ will allow the gas behave ideally as compared to STP conditions.

Answer: Increase in volume upto 100dm³ and Temperature upto 27C^o as compare to STP condition for 2 moles NH₃ results molecules to go away from each other, no attractive forces exist among them,therefore,NH₃ behaves ideally.

Exercise Q.No.(18): A sample of Krypton-with a volume of 6.25dm³,a pressure of 765 torr and a temperature of 20C^o is expanded to a volume of 9.55 dm³ and a pressure of 375 torr.What will be its final temperature in C^o?

Solution: V₁ = 6.25dm³ V₂ = 9.55dm³

P₁ = 765torrs P₂ = 375torrs

T₁ =20C^o + 273 = 293K T₂ in C^o = ?

P₁V₁ = P₂V₂ Or T₂=P₂V₂ x T₁= 375torr x 9.55dm³ x 293K=219.46K-273=-53.6C^o

T₁ T₂P₁V₁ 765torr x 6.25dm³

Exercise Q.No.(19):Working at a vacuum line, a chemist isolated a gas in a weighing bulb with a volume of 255cm³,at a temperature of 25C^o and under a pressure in the bulb of 10. torr.The gas weighed 12.1mg. What is the molecular mass of this gas?

Solution: V = 255cm³ = 0.255dm³ T = 25C^o + 273 = 298K.

P = 10.0 torrs = 0.0131atm Mass of gas = 12.1mg = 0.0121g

760

Required:M = ?

PV = m RT

M = mRT = 0.0121g x 0.0821atm dm³ K^-1 mol^-1 x 298K = 86.62gmol^-1

PV 0.0131atm x 0.255dm³

Exercise Q.No.(20): What pressure is exerted by a mixture of 2g of H₂ and 8g of N₂ at 273K in a 10dm³ vessel?

Solution: Mass of H₂ = 2.00g , Mass of N₂ = 8.00g T = 273K , V = 10dm³ P = ?

Moles of H₂ = 2/2 = 1 mol , Moles of N₂ = 8/28 = 0.2857mol

Total no. of moles = 1 + 0.2857mol = 1.2857moles

PV =nRT Or P = nRT = 1.2857mol x 0.0821atm dm³ K^-1 mol^-1 x 273K = 2.88atm

V 10dm³

Partial pressure: The pressure of each in a mixture of gases is called partial pressure of

that gas.

Calculation of partial pressure of a gas: Consider a mixture of gases A and B with

total volume v, at constant temperature (T).

Apply the general gas equation to mixture,

eq (1)

eq (2)

eq (3)

Divide equation (2) by eq (1)

and similarly

X_A.P_t and

Where X = mole fraction = moles of a gas

total moles of mixture

Example(6):There is a mixture of H₂, He and CH₄ occupying a vessel of volume 13dm³ at and pressure 1atm. the masses of H₂ and He are 8g and 12g respectively. Calculate the partial pressures in torrs of each gas in the mixture.

Solution:

Mass of

Required:

Total No. of moles of gas will be

moles.

Moles of

moles

Moles

moles,

So,

Moles of

moles.

Applying the formula,

Partial pressure of a gas

total pressure

So,

Dalton’s law of partial pressure:

Total pressure of a mixture of gases is equal to the sum of partial pressures of all gases in it,at constant temperature and pressure.e.g Consider a mixture of H₂,O₂ , CH₄ and apply PV =nRT

So, P_H = n_HRT , P_O = n_ORT , P_CH = n_CHRT

V V V

According to Dalton’s law of partial pressure:

constant

Applications of Dalton’s law of partial pressure:

(1)Collection of gases over water:

Some gases are collected over water which contains water vapours, therefore, according to Dalton’s law the total pressure P_t = P_gas + P_w.vap

Note: Aqueous tension means partial pressure of water vapours.

(2)Normal Breathing:

Partial pressure of oxygen in lungs is 116 torrs while partial pressure of oxygen in air is 159 torrs, this difference in partial pressures of oxygen is reason, for comfortable breathing.

(3)Breathing at high altitude:

At high altitudes the partial pressure of oxygen decreases from 159 torrs, that is reason for uncomfortable breathing at high altitudes.

So, pressurized air should be used for breathing at high altitudes.

(4)Breathing under the water:

3atm pressure increases over the body, as one go under the water, for each 100 feet depth, therefore, normal air cannot be used for breathing under the water.

The pressure of N₂ also increases in depth of sea which get diffused in blood. To avoid these two problems, the oxygen mixed with an inert gas like He is used for breathing under water.

Note: The partial pressure of O₂ should be adjusted according to depth of sea.

Assignment: Q. 1.viii. Q.10.a,b,c

Exercise Q.No.(1)(viii):

Equal masses of methane and oxygen are mixed in an empty container at . The fractional of total pressure exerted by methane is. (a) (b) (c) (d)

For Help:Consider equal masses of Methane = 32g and Oxygen = 32g

Moles of the Methane= 32g/16g mol^-1 =2 and Oxygen = 32g/32 g mol^-1 = 2

Total Moles = 1 + 2 = 3moles , Partial pressure of CH₄ = 1/3 x P_t

Exercise Q.No.(10)(a): Dalton’s law of partial is only obeyed by those gases which don’t

have attractive forces among their molecules. Explain it.

Answer: If in a mixture of gases, the attraction among molecules results decreases in total pressure, therefore, the total pressure of mixture is not equal to sum of the partial pressure of all gases. This is the reason that Dalton’s law of partial pressure is obeyed only by the gases which have no attractive forces.

Exercise Q.No.(10)(b):Derive an equation to find out the partial pressure of a gas knowing individual moles of component gases and total pressure of mixture.

Answer: See derivation of partial pressure formulae..

Exercise Q.No.(10)(c):

Explain that process of respiration obeys the Dalton’s law of partial pressure.

Answer: See Applications of Dalton’s law of partial pressure.

Diffusion:

The gases flow from high concentration to low concentration region, spontaneously. This is called diffusion of gases.

e.g. one can feel the smell of perfume through out the room is due to diffusion of perfume molecules.

Effusion:

Small sized gas molecules get escaped one by one through a narrow hole from a container. This is called effusion.

Graham’s law of diffusion and effusion:

Rate of diffusion or effusion of a gas is inversely proportional to square root of their density or molar mass at constant temp and pressure,

Note: The value of K is constant for all gases at constant temp and pressure.

Let apply the Graham’s law to two different gases 1 and 2

and

Compare the rates

and

Experimental verification of Graham’s law:

Take 100cm long glass tube and cover it with filter paper soaked in conc. HCl and NH₃ solutions. Both HCl and NH₃ start diffusion and found to meet at 60cm distance form NH₃ produces the white fumes of NH₄Cl as

100Cm

NH₄Cl

Now apply Grahm’s law

Example (7):250cm³ of sample of hydrogen effuses four times as rapidly as 250cm³ of unknown gas. Calculate the molar mass of unknown gas.

Solution: If

Required.

Apply Grahm’s law.

= 32g mol^-1

Assignments: Q 1(vii),Q.21.a,b,c

ExerciseQ.No.(1)(vii):The correct order of rate of diffusion of gases NH₃, SO₂, Cl₂ and CO₂

(a) NH₃>SO₂>Cl₂>CO₂ (b) NH₃> CO₂>SO₂>Cl₂

For Help: r α 1 / M

Exercise Q.No.(21)(a): The relative densities of two gas A and B are . Find out the volume of B which will diffuse in the same time in which 150dm³ of A will diffuse.

Solution:

Given data Applying the Grahm’s law

Exercise Q.No.(21)(b):H₂ diffuses through a porous plate at a rate of 500cm³ per minute at . What is the rate of diffusion of oxygen through the same porous plate at .

Solution:

500cm³ = 4 Or rO₂= 500 cm³ = 125 cm³

rO₂4

Exercise Q.No.(21)(c):The rate of effusion of an unknown gas A through a pinhole is found to be .279 times the rate of effusion of H₂ through the same pain hole. Calculate the molecular mass of unknown gas at STP.

Solution:

Apply Graham’s law

Kinetic Molecular theory of gases:

(1)All gases consist of molecules which are in constant motions in all possible directions.

(2) Gas molecules collide with each other and with walls of container. These collisions

involve no loss or gain of kinetic energy, called elastic collisions .

(3).The collisions of gas molecules develop the pressure of gases. This pressure is given

by following Kinetic Gas Equations as

where

No. of moles

mass of gas

mean square velocity.

pressure

volume

(4) Actual volume of gas molecules is negligible as compared to volume of container.

(5) No forces exists among gas molecules.

(6) Kinetic energy i.e.

is directly proportional to absolute temperature.

(7)The force of gravity has no effect on motions of gas molecules.

Types of velocities:

(1)Mean velocity: c^-

The sum of velocities of all molecules divided by total no.of molecules is called mean velocity

If there are

molecules with velocities

and

respectively, then mean velocity

will be

where

(2)Mean square velocity c^-2

It is the average of the square of all possible velocities

If there are

molecules with velocities

and

respectively, then mean square velocity

will be

(3)Root Mean square velocity c^-2

Square root of mean square velocity is known as root mean square velocity denoted as

Molar mass of gas

Absolute temp

General gas constant

Derivation of gas laws from kinetic molecular theory:

(1)Boyle’s law:

According to

Kinetic gas equation

But

By putty the value of K.E from equation (1) in equation (2) we get.

constant, then

constant.

(2)Charles’s law:

We know that

constant, then

(3)Avogadro’s law: (Equal volumes contains equal no. of moles)

consider equal volumes of two different gases at constant temp and pressure.

Kinetic gas equation for both gases will be

and

constant, so K.E per mole will also be equal for both gases i.e.

Divide equation (1) by (2) we get.

this is what, the Avogadro’s law say:

(4)Graham’s law of diffusion or effusion:(r α 1 )

We know that

, Then

PV = 1 Mc^-2 or

Take square root on both sides.

But

So,

constant

Kinetic Interpretation of temperature:

Kinetic gas equation for 1 mole General gas equation for 1 mole

equation – (2)

PV = 1 Mc^-2 Equation (1)

Compare equation – (1) and (2)

1 Mc^-2 = RT

or 3

2 . 1 Mc^-2

3 2

constant.

Assignments:

Qno.12(a)(b) Qno.13(b)

Exercise Q.No.(12)(a): What is kinetic molecular theory of gases. Give its postulates.

Answer: See KMT.

Exercise Q.No.(12)(b): How KMT explains the Boyles law, Charles’s law, Grahams law

and Avogadro’s law.

Answer: See derivation of gas laws from KMT.

Exercise Q.No.(13)(b):Do you think that some of postulates of KMT are faulty. Point

out these postulates.

Answer: Yes,Following two postulates are faulty at high pressure and low temperature.

(i) Actual volume of gases negligible as compared to volume of container.

(ii) No forces exist among gas molecules.

Liquefaction of gases:

Low temp: If temp is low, K.E of molecules is low, therefore molecules get close to each

other and intermolecular forces get dominated, therefore, gases areliqufied.

High pressure: At high pressure, molecules come close, intermolecular forces get

dominated, therefore, the gas changes into liquid state

Critical temperature:

For each gas there is a fixed temperature at which or below which the gas can be liquefied by applying pressure and above which gas cannot be liquefied. This fixed temperature is called critical temperature.

Maximum temperature at which a gas can exists as liquid is called critical temperature.

Critical temp for a gas depends on.

(a)Intermolecular forces such that stronger the intermolecular forces, high will be critical temp and vice versa.

(b)Size of gas molecules such that larger the size of gas molecules high will be critical temperature and vice versa.

e.g. Substance Critical temp

405.6 K

647.6 K

304.3 K

154.4 K

126.1 K

Note: The gases having larger critical temperature values are liquefied easily than those having low critical temperature values. because the gases with larger critical temperature values posses stronger intermolecular forces.

Critical Pressure: The pressure requires to change a gas into liquid at its critical temperature

is called critical pressure.

Joule Thomson Effect:

A compressed gas causes cooling on expansion. This is known as Joule Thomson effect.

Strong forces exist among molecules of a compressed gas. If it is allowed to expand, then molecules provide energy form itself to overcome strong attractive forces, due to which temperature get fall.

Lind’s method of liquefaction of gases:

The basic principle involve is Joule Thomson effect.

Process:

Air is compressed upto 200atm by compressor.

The compressed air is passed through water

cooled pipes to remove heat of compression.

The compressed air is suddenly expanded

through a jet which causes fall of temperature.

By repeating this compression and expansion

of air, the air get liquefied.

Note: H₂ and He are non polar,small sized gases having very weak forces among their

molecules, therefore, they have such low critical temperature values which can not be

achieved through linds method.This is the reason that H₂ and He cannot be liquefied

through linds method of liquefaction.

Ideal and Non ideal Gases:

Ideal gases / perfect gases

(i) No attractive forces among their molecules.

(ii) Follow gas laws at all temperature and pressure.

(iii) All gases behaves ideally at high

Temperature and low pressure.

Non Ideal gases/real gases

(i) Attractive forces among their molecules.

(ii) (ii)Do not follow the gas laws at all temperature pressure..

(iii) All gases behaves non ideally at

Low temperature, high pressure.

Non ideal behaviors of gases:

It is experimental fact that for 1mole ideal gas the graph of compressibility factor

at all temperature pressure is always a straight line parallel to pressure axis as shown.

RT 1 ideal behavior

The deviations from ideal behavior depends on the temperature and pressure. e.g. the deviations from ideal behaviors has shown for H₂, He and CO₂ at O

and 100

. separately.

RT 1 ideal behavior PV 1 ideal behavior

P (0C^o) P(100C^o)

Assignments

Q.no.13(a) (c) ,15(i)(ii)(iv)(v)

Exercise Q.No.(13)(a):Gases show non ideal behavior at low temperature and high

pressure. Explain this with the help of a graph.

Answer: See non ideal behavior of gases.

Exercise Q.No.(13)(c): H₂ and He are ideal at room temperature but SO₂, Cl₂ are non

ideal. How will you explain this.

Answer:H₂, He are small sized non polar gases, have very very weak forces among their

molecules, therefore, they are ideal at room temperature.

But SO₂, Cl₂ are larged sized molecules have strong forces among their molecules,

therefore they are non ideal at room temperature

Exercise Q.No.(15):Explain the following facts.

Exercise Q.No.(15)(i):Plot of vs is a straight line at constant temperature and with

fixed No. of moles of an ideal gas.

Answer: This graph mean that by changing pressure volume also get change but product of

pressure and volume remains constant.

Exercise Q.No.(15)(ii): The straight line in (a) is parallel to pressure axis and goes away

from pressure axis at higher pressure for many gases.

Answer:At higher pressure the molecules of gases get closed to each other, intermolecular forces get dominated therefore, the gas behaves non ideally and the lines goes away from the line showing the ideal behavior.

Exercise Q.No.(15)(iv):Water vapours do not behave ideally at 273 K. Explain.

Answer:Due to polar nature, Strong attractive forces exists among water molecules at 273 K, therefore, it not behaves ideally at this temperature.

Exercise Q.No.(15)(v):SO₂ is comparatively non ideal at 273 K but behaves ideally at

327. Explain

Answer:SO₂ is a polar gas having attractive forces at 273 K, therefore, it behaves non ideally

but at high temperature like

, the molecules go away from each other and no

forces exists among them, therefore it behaves ideally at

Causes of deviations from ideal behaviors:

(i) Actual volume of gases is negligible as compared to container.

(ii) No attractive forces exist among gas molecules.

The above two points in KMT are faulty, at high pressure and low temperature,therefore,

gases deviates from ideal behavior at high pressure and low temperature.

Vander wall’s equation for real gases:

Vander walls made the volume correction, pressure correction and apply it to general gas equation

getting another equation called Vander walls equation which is applicable to non ideal gases.

(1)Volume correction:

Vander wall pointed that the volume of a gas at low pressure and high temperature in a container can be neglected .

Neglected volume

But Volume of a gas at high pressure and low temperature cannot be neglected.

Non neglected volume (b)

This non neglected volume of gas is not available for free motions of gas molecules,therefore,

It is called excluded volume denoted by (b).

So, volume available for free movement of gas molecules is given as.

for 1mole

for n mole. Equation (1) .

(2)Pressure correction:

The net force on a molecule at anterior part of gas is zero, as the molecule is attracted equally from all sides as shown below.

B←A→B

But a molecule (A) near the wall is attracted by other molecules (B) laterally and inward only as Shown

A B

This inward pull decreases the pressure

by factor

Therefore , P_obs = P_I - P

P_I = P_obs + P Equation (2)

Where

shows the pressure decreased due to inward pull of molecules.

depends on conc.of A type and B type molecules such that.

But

P = a n² Where a is Vander Wall’s constant.

V²

By putting value of

in equation (2), we get.

P_I = P_obs + a n² equation (3)

V²

By applying the volume and pressure correction to general gas equation, we get.

(Vander Wall’s equation) for n mole of real gases.

Units of Vander Wall’s constants a, b.

We know that

SI unit will be

Excluded volume

b = V = dm³ mol^-1 or SI unit is m³ mol^-1

Note:

(i)The Vander Walls constant “a” shows the forces of attractions among gas molecules such that larger value of “a” for a gas shows stronger forces and vice versa.

(ii)The Vander Walls constant “b” shows the excluded volume which depends on size of gas molecules such that larger the size of gas molecules, high will be value of b, and vice versa.

(iii)Excluded volume (b) is not equal to actual volume of gas molecules but

(iv) Smaller the value of vander walls constant (a) and (b), more ideal will be gas.

Example (8):One mole of methane gas maintained at 300K. Its volume is 250 cm³.

Calculate the pressure exerted by the gas under the following conditions

(a)When the gas is ideal.

(b)When the gas is non ideal.

Solution:

(a)When gas is ideal,

P = ?

We know

(b) When gas is non ideal,

So, The net decrease

Assignment :

Q.14.a,b, Q.15.(iii)(vi) Q.24(a)(b)(c)

Exercise Q.No.(14)(a):Derive Vander Walls equation.

Answer: See Vander Walls equation.

Exercise Q.No.(14)(b):What is physical significance of Vander Walls constants a, b.

Give their Units.

Answer: See note on page 24

Exercise Q.No.(15)(iii): The pressure of NH₃ gas at given conditions (say 20atm pressure

and room temp) is less as calculated by Vander Waals equation

then that calculated by general gas equation.

Answer:NH₃ is polar have strong forces which decreases the pressure calculated by Vander Walls equation.

Exercise Q.No.(15)(vi):The Vander Walls constant “b” of a gas is four times the molar

volume of that gas. Prove that

Answer: Consider two molecules of a gas in contact with each other as shown

The space indicated by dotted sphere, is not available for pair of colliding molecules therefore, it is called excluded volume

for two colliding molecules.

The volume of dotted sphere is 4 π (2r)³

So, excluded volume for two molecules

Excluded volume for two molecules

Excluded volume per molecule

Excluded volume

Exercise Q.No.(24)(a):Two moles of NH₃ are enclosed in a 5dm³ flask at . Calculate

the pressure exerted by the gas assuming that

(i)It behaves like an ideal gas.

(ii)It behaves like a real gas.

a = 4.17atm dm⁶ mol^-2

Solutions:

Moles of NH₃ = 2moles.

Volume of NH₃ = 5dm³

Temperature = 27

Required (i)

(gas is ideal)

We know that

ii. If NH₃ behaves non ideally, then

Exercise Q.No.(24)(b):Also calculate the pressure lessened due to forces of attractions at

these conditions of volume and temperature.

Pressure lessened

9.852 – 9.813 = 0.039atm.

Exercise Q.No.(24)(c):Do you expect the same decrease in pressure of two moles of NH₃

having a volume of 40dm³ and at temperature of .

Answer: No, the decrease in pressure should be different for different volumes, if calculated.

Plasma state: (W.Crooks 1879):

Plasma is the fourth sate of matter which contains positive ions, electrons and neutral atoms.

How plasma is formed.

Characteristics of plasma:

(i) Plasma shows response to magnetic as well as electric field because it contains charged

particles

(ii) Plasma is good conductor of electricity because it contains free electrons.

(iii) Plasma is neutral because it contains equal no. of

and

particles.

Types of plasma
Natural plasma (i)The plasma produced naturally as a result of natural process. (ii) Natural plasma is stable at high temperature above e.g Plasma exists in sun and stars is natural plasma	Artificial plasma (i)Plasma produced by artificial means is called artificial Plasma. (ii) Artificial plasma is either stable at high temperature or inside a low temperature vacuum e.g. plasma in neon sighs and fluorescent tubes is artificial plasma stable inside low temperature vacuum .

Where is plasma found :

Plasma occurs in

1. Sun, stars.

2. Flames

3. Fluorescent tubes.

4. Auroras (lightning at northern pole

Canada

’ Siberia at night time)

5. Lightning bolts

Applications of plasma: Plasma is used in

1. Fluorescent light tube which light up our homes, offices.

2. Neon signs which produce colored light (Reddish light in case of neon).

3. For sterilization of food

4. In lasers which help in many medical treatment.

5. In particle accelerators which clean up the environment by removing dangerous chemicals.

6. For generation of electrical energy from fusion pollution control.

(Remaining Questions of exercise)

Exercise Q.No.(22): Calculate the number of molecules and the number of atoms in the

given amount of each gas.

Solution: (a) 20 cm ³ of CH₄ , at 0C^o and pressure of 700mm of mercury.

V = 20 cm³ or 0.020dm³ , T = 0C^o + 273 = 273 K , P = 700torrs = 0.92atm

760

PV = nRT

n = PV = 0.92atm x 0.020dm³ = 0.000822moles

RT 0.0821atm dm³ mol^-1 K^-1 x 273 K

No of CH₄ molecules = 0.000822moles x 6.022 x 10²³ = 4.9 x 10²⁰ molecules.

No of atoms in 4.9 x 10²⁰ CH₄ molecules = 5 x 4.9 x 10²⁰ = 24.7 x 10²⁰ atoms

Solution:(b) 1 cm ³ of NH₃ , at 100C^o and pressure of 1.5atm .

V = 1 cm³ or 0.001dm³ , T = 100C^o + 273 = 373 K , P = 1.5 atm

PV = nRT

n = PV = 1.5atm x 0.001dm³ = 0.000049 moles

RT 0.0821atm dm³ mol^-1 K^-1 x 373 K

No of NH₃ molecules = 0.000049 moles x 6.022 x 10²³ = 2.95 x 10¹⁹ molecules.

No of atoms in 2.95 x 10¹⁹ NH₃ molecules = 4 x 2.95 x 10¹⁹ = 11.8 x 10¹⁹ atoms

Exercise Q.No.(23): Calculate the masses of 10²⁰ molecules of each of H₂ ,O₂ and CO₂

At STP.What will happen to the masses of these gases, When the

temperature of these gases are increased by 100 C^o and the

pressure is decreased by 100 torr.

Solution: Molecules of H₂ = 10²⁰,Moles of H₂ = No of particles = 10²⁰ = 0.166 x 10^-3 moles

N_A6.022 x 10²³

Mass of H₂ = Moles x Molar mass = 0.166 x 10^-3 moles x 2g mol^-1 = 3.3 x 10^-4g.

Molecules of O₂ = 10²⁰,Moles of O₂ = No of particles = 10²⁰ = 0.166 x 10^-3 moles

N_A6.022 x 10²³

Mass of O₂ = Moles x Molar mass = 0.166 x 10^-3 moles x 32g mol^-1 = 5.31 x 10^-3g.

Molecules of CO₂ = 10²⁰,Moles of H₂ = No of particles = 10²⁰ = 0.166 x 10^-3 moles

N_A6.022 x 10²³

Mass of CO₂ = Moles x Molar mass = 0.166 x 10^-3 moles x 44g mol^-1 = 7.3 x 10^-3g.

Note: The masses of H₂,O₂ and CO₂ remains same, if temperature is increased by 100C^o and

Pressure by 100torrs.

ashakil

Tuesday, August 19, 2014

Chapter, 03: Gases Properties of Gases. 1st year

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