Class XII, PHYSICS, Chapter # 13 "CURRENT ELECTRICITY"

CURRENT ELECTRICITY
Qs. Define Charge and Current.
CHARGE
Definition
Flow of electron is known as Charge.
It is denoted by Q.
Unit
Its unit is Coulomb.
1 Coulomb = 10(-6) μcoulomb
1 coulomb = 10 (-3) mili coulomb
1 coulomb = 10(-9) neno coulomb
CURRENT
Definition
The flow of charge per unit time is known as Current.
It is denoted by I.
Unit
The unit of current is coulomb/sec or Ampere.
AMPERE
If one coulomb charge passes through the conductor in 1 second then the current is 1 Ampere.
Mathematical Form
Mathematically,
I = Q/t
Qs. State and Explain Ohm’s Law.
OHM’S LAW
Introduction
A German scientist George Simon Ohm studied the relationship between voltage, current and resistance. On the basis of his experimental results, he proposed a law which is known as Ohm’s Law.
Statement
Ohm’s Law to metallic conductors can be stated as
The current through a conductor is directly proportional to the potential difference between the ends of the conductor provided that physical conditions are kept constant.
It can also be stated as
The ratio between voltage and current remains constant, if the physical conditions are kept constant.
Mathematical Form
Mathematically,
V ∞ I
V = IR
R = V/I
Where R is the constant of proportionality known as resistance of the conductor. Its unit is volt per ampere (Volt/Ampere) or Ohm (Ω).
Ohm (Ω)
If 1 ampere current passes through the conductor due to 1 volt potential difference then the resistance of conductor is 1 Ohm.
Resistance
Opposition offered in the flow of current.
Graphical Representation.
When graph is plotted between current and potential differences then straight line is obtained.
Limitations of the Law
Ohm’s Law is valid only for metallic resistance at a given temperature and for steady currents.
Qs. Define the term Resistivity or Coefficient of Resistor.
RESISTIVITY OR COEFFICIENT OF RESISTOR
Definition
It is the resistance of a unit conductor whose cross-sectional area is 1 sqm.
Unit
Its unit is Ohm meter.
Mathematical Form
The resistance of any conductor depends upon the following factors.
1. Length of the conductor
2. Cross-sectional area of the conductor.
3. Material of the conductor.
The resistance of the conductor is directly proportional to the length of the conductor and inversely proportional to the cross-sectional area.
Mathematically,
R ∞ L ——– (I)
R α 1/A —— (II)
Combining eq (I) and (II)
R α L/A
=> R = ρL/A
Where ρ is the constant of proportionality known as Resistivity or Coefficient of resistance.
ρ = RA/L
Qs. Explain the effect of temperature on resistance or temperature coefficient of resistance.
EFFECT OF TEMPERATURE ON RESISTANCE
It is observed that if we increase the temperature then resistance of a conductor will increase.
Consideration
Let Ro be the initial resistance of a conductor at 4°C. If we increase the temperature from t1°C to t2°C, then resistance will increase. This increment in resistance is denoted by ΔR. The increment in resistance depends upon the following two factors.
1. Original Resistance (Ro)
2. Difference in temperature Δt.
Mathematical Verification
The increment in resistance is directly proportional to the original resistance and temperature difference.
Mathematically,
ΔR ∞ Ro —– (I)
ΔR ∞ Δt —– (II)
Combining eq (I) and eq (II) we get
ΔR ∞ RoΔt
=> ΔR = αRoΔt
Where α is the temperature coefficient of resistance. It is defined as
It is the increment in resistance per unit resistance per degree rise in temperature.
Its unit is 1/°C or °C. If RT is the total resistance, then
RT = Ro + ΔR
=> RT = Ro + αRo Δt
=> RT = Ro (1 + αΔt)
As we know that resistance is directly proportional to resistivity therefore,
ρT = ρo (1 + αΔt)
Qs. Define the term Power Decipation in Resistor.
POWER DECIPATION IN RESISTORS
Definition
When current flows in a conductor then a part of electrical energy appears in the form of heat energy which is known as Power Decipation in Resistor.
Units
Its unit is Joule per second (J/s). Most commonly used unit is Kwh.
1 Kwh = 36 x 10(5) Joules
Mathematical Form
Since,
P = Electrical Work / Time
Electrical Work = QV —— (I)
This electrical work produces heat energy in the resistor.
P = QV / t
P = Q / t . V
But,
I = Q / t
P = VI
From Ohm’s Law
V = IR
P = IIR
P = I2R
OR,
P = 12R2 / R
=> P = V2 / R
As we know that,
Energy = Power x time
=> E = P x t
=> E = Vit
=> E = I2Rt
And,
E = V2 / R . t
Qs. Define and explain Electromotive Force.
ELECTROMOTIVE FORCE
Definition
It is the terminal voltage difference when no current draws from a cell or a battery.
OR
Work done per coulomb on the charges.
It is denoted by E.
Unit
Electromotive force or simply e.m.f is a scalar quantity it has the same dimension as that of voltage, therefore its unit is volt.
Explanation
When an electric current passes through a resistor, it dissipates energy, which is transformed into heat energy. Thus to sustain a current in conductor some source of energy is needed so that it could continuously supply power equal to that which is dissipated as heat in the resistor. The strength of this source is called Electromotive Force.
Consideration
Let consider a simple circuit in which a resistor “R” is connected by leads of negligible resistance to the terminals of a battery. The battery is made up of some electrolyte and electrode for the production of e.m.f and hence when this current flows from battery, it encounters some resistance by the electrolyte present in two electrodes. This resistance is known as internal resistance “r” of the battery.
Mathematical Form
According to Ohm’s Law
V = IR
I = V / R
Or,
I = E / R + r
Where E is e.m.f and r is internal resistance
=> E = IR + Ir
E = V + Ir

Class XII, PHYSICS, Chapter # 12 "ELECTROSTATICS"

ELECTROSTATICS
Qs. State and Explain Coulomb’s Law.
INTRODUCTION
In 1974, Sir Augusts de coulomb studied the relationship between localized charges. He carried out experiments using torsion balance. In the basis of his experimental results, he proposed a law known as Coulomb’s Law.
Statement
The electric force of interaction between two point charges is directly proportional to product of their charges and inversely proportional to the square of the distance from their centers.
Mathematical Expression
Consider two points charges q1 and q2 and let “1″ be the separation between them. According to Coulomb’s Law.
F ∞ q1 q2 ——– (i)
And, according to second part
F ∞ 1/r² ——— (ii)
Combining eq (i) and eq (ii), we get
F ∞ q1 q2 / r²
F = K q1 q2 / r²
Where K is the constant of proportionality and its value depends upon the medium between the two charges. In S.I system its value is equal to
K = 8.98 x 10(9) N-m² / c².
Coulomb’s Law can be expressed in terms of permitivity as follows.
F = 1 / 4 π Є . q1 q2 / r²
Where Єo is the permitivity of free space and its value is
Єo = 8.85 x 10(-12) col² / Nm²
if some other medium is used instead of air then
F = 1 / 4 π Єo Єr . q1 q2 / r²
Where Єr is the relative permitivity which is different for different dielectric.
Qs. Define electric field and electric intensity. Find the formula for the electric intensity due to a point charge.
ELECTRIC INTENSITY
DEFINITION
Electric intensity is the force experienced by a unit positive charge due to the presence of a charge body.
EXPLANATION
It is the measure of the strength of the electric field. Electric intensity is a vector quantity and its direction is same as that of force. If the charge is positive then electric intensity is directed from the charge and if the charge is negative, then it is directed towards the charge. The SI unit of electric intensity is Newton per Coulomb or volt per meter.
Mathematical Form
Mathematically electric intensity is given by
E = F / q°
Where E = Electric Intensity
F = Force
q° = small magnitude
ELECTRIC INTENSITY DUE TO A POINT CHARGE
CONSIDERATION
Consider a charged body q that has an electric field all around it. We want to find out electric intensity E at point P. For this purpose, we placed a point charge q1 at that point.
Qs. Define Electric flux. Find the formula for the electric flux due to point charge in a closed sphere.
ELECTRIC FLUX
DEFINITION
The total number of imaginary lines drawn in such a way that the attraction of tangent at any point is same as that of electric field crossing a surface normally is called electric flux or flux on the surface.
OR
The total number of lines of forces crossing a surface normally is called flux on that surface.
Mathematical Form
The flux at a surface is determined by the product of flux density i.e. electric field and the projection of its area perpendicular to the field or by the product of area and component of field normal to the area.
ELECTRIC FLUX DUE TO A POINT CHARGE IN A CLOSED SPHERE
Consider an isolated point charge +q. Th lines of forces from q will spread uniformly in space around it cutting the surface of an imaginary sphere. Now we want to find flux due to point charge. For this purpose, we divide the whole sphere into small patches. Each patch is denoted by ΔA.
Qs. State and prove Guass’s Law
The total electric flux diverging out from a closed surface is equal to the product of the sum of all charges present in that closed surface and 1/ Єo.
Find the Formula for Electric Intensity.
1. Due to Charge Sheet
2. Due to Two Charge Sheets
OR
What are the application of Guass’s Law.
Introduction
Guass’s Law can be used to calculate the electric field only in those cases of charge distribution which are so symmetrical that by proper choice of Guassian surface the flux on it may possibly be evaluated. With the help of guass’s law we can measure the intensity in following cases.
1. Electric Intensity due to charge sheet.
2. Electric Intensity due to two charge sheets.
1. Electric Intensity Due to Charge Sheet.
Consider a charge sheet in which unit positive charges are uniformly distributed.
As we know that charge density is the charge stored per unit area and is denoted by
σ = Q / A
=> Q = σ A
According to Guass’s Law
σ = Q / Є
=> φe = σ A / Є ——– (I)
Now consider a cylindrical shell, which is placed inside the charge sheet. It has three surfaces.
Upper Surface
Curve Surface
Bottom
Electric Intensity Due to Upper Surface
φ1 = E ΔA
φ1 = E ΔA cos θ
But θ = 0° and cos 0° = 1, therefore,
φ1 = E ΔA cos 0°
= φ1 = E ΔA
Electric Intensity Due to Curved Surface
φ2 = E ΔA
φ2 = E ΔA cos θ
Since the angle between the field vector and area vector of all elements of curved surface is 90°, therefore,
φ2 = E ΔA cos 90°
But cos 90° = 0,
φ2 = E ΔA (0)
φ2 = 0
Electric Intensity Due to Lower Surface
φ3 = E ΔA
φ3 = E ΔA cos θ
But 0 = 0° and cos 0°, = 1 therefore,
φ3 = E ΔA cos 0°
=> φ3 = E ΔA
Total Flux
Total Flux is given by
φe = φ1 + φ2 + φ3
=> φe = E ΔA + 0 + E ΔA
=> φe = 2E ΔA
For the whole charge sheet:
=> φe = 2E Σ ΔA
=> φe = 2 EA ———– (II)
Comparing equation (I) and (II)
=> σ A / Єo = 2AE
=> E = σ / 2 Єo
OR
E = σ / 2 Єo
2. Electric Intensity Due to Two Charge Sheets.
As we know that electric intensity due to a charge sheet is
E = σ / 2 Єo
For two sheets E will be,
=> E = σ / 2 Єo + σ / 2 Єo
=> E = σ + σ / 2 Єo
=> E = 2 σ / 2 Єo
=> E = σ / Єo
=> E = σ / Єo i
Qs. Define the term Capacitor and Capacitance of a Capacitor
CAPACITOR
Capacitor is a device which is use to storage charge. A simple capacitor consists of two parallel metallic plates. A plate is connected to the positive terminal of the battery and another plate is connected to the negative terminal of the battery.
CAPACITANCE
Definition
The capacity of a capacitor to store the charge is known as Capacitance.
Mathematical Explanation
If V is the voltage provided to the capacitor and Q is the amount of charge stored in the capacitor, then it is observed that if more is the voltage, then more will be the charge stored in the capacitor.
Mathematically,
Q ∞ V
=> Q = CV
Where C is the capacitance of the capacitor which may be defined as
The ratio of the charge on one of the plate (conductor) to the potential difference between them.
Unit of Capacitance
The unit of Capacitance is Farad. It may be defined as,
If one coulomb charge is stored due to 1 volt potential, then capacitance will be 1 Farad.
Factors on which Capacitance Depends.
The Capacitance of a capacitor depends upon following factors.
1. Cross-sectional area of plate
2. Separation between plates
3. Dielectric
On the basis of dielectric capacitors are classified into different types. For example Electrolyte Capacitor, Paper Capacitor, Meca Capacitor, Oil Capacitor.
Types of Capacitor
There are two main types of Capacitors.
1. Fixed Capacitor
2. Variable Capacitor
1. Fixed Capacitor
Those capacitor whose capacitance is constant are known as Fixed Capacitor. For example Paper Capacitor and Meca Capacitor.
2. Variable Capacitor
Those capacitor whose capacitance is not fixed are known as Variable Capacitor. For example Gang Capacitor.
Find the Formula for the capacitance of Parallel plate capacitor.
PARALLEL PLATE CAPACITOR
Definition
A parallel plate capacitor is a device used to store the charge. It consist of two parallel metallic plates. A plate is connected to the positive terminal of the battery and another plate is connected to the negative terminal of the battery.
These plates are separated by a very small distance compared to the dimension of the plates.
CAPACITANCE OF A PARALLEL PLATE CAPACITOR
Consideration
Consider a parallel plate capacitor in which is the distance between the plates the charge stored is denoted by Q where as potential difference between the plates is V.
Derivation of the Formula
As we known that electric intensity due to two charge sheets is
E = σ / Єo
Since charge density is given by
σ = Q / A
Substituting the value in above equation
E = Q / A Єo ——— (I)
According to the definition of electrical potential.
ΔV = E Δ
Substituting the value of E from eq (I) in above equation
Q = C x Qd / A Єo
C = A Єo / d
If some other medium is used instead of air then,
C = A Єo Єr / d
Find the formula for equivalent capacitance when,
1. Capacitors are connected in Series
2. Capacitors are connected in Parallel
Introduction
Capacitors of some fixed values are used in a circuit. The capacitance of the desired value can however be obtained by suitable combination of capacitor. Capacitors can be combined in parallel, series or both.
WHEN CAPACITORS ARE CONNECTED IN SERIES OR SERIES COMBINATION
Consideration
Consider three capacitors having capacitance C1, C2, C3 connected in a series. These capacitors can be replace by an equivalent capacitor having capacitance Ce. When a cell is connected across the ends of system then a charge Q is transferred across the plates of capacitors. A charge upon one plate always attracts upon the other plate with a charge equal in magnitude and opposite in sign.
Let V be the potential difference across the combination. The potential difference across the individual capacitor is Vab, Vbc, Vcd.
Derivation of the Formula
As we know that in case of series combination the charge across the individual capacitor remains constant, where as potential difference varies such that the potential difference V is the sum of potential difference applied across individual capacitor.
Vad = Vab + Vbc + Vcd ———— (I)
We know that,
Q = CV
V = Q / C
Therfore,
Vad = Q / Ce
Vab = Q / C1
Vbc = Q / C2
Vcd = Q / C3
Substituting the values of Vad, Vab, Vbc and Vcd in eq (I)
Q/Ce = Q/C1 + Q/C2 + Q/C3
Q/Ce = Q [1/C1 + 1/C2 + 1/C3]
1/Ce = 1/C1 + 1/C2 + 1/C3
Conclusion
The reciprocal of the equivalent capacitance is equal to the sum of the reciprocals of the individual capacitance.
WHEN CAPACITORS ARE CONNECTED IN PARALLEL OR PARALLEL COMBINATION
Consideration
Consider three capacitors having C1, C2, C3 capacitance respectively are connected in parallel. We can replace them by an equivalent capacitor having capacitance Ce. A charge q given to a point divide it self reside on the plates of individual capacitor as Q1, Q2, Q3 respectively.
Derivation of the Formula
As we known that in case of parallel combination the potential difference across each capacitor is that of the source where as the charge across each capacitor varies, therefore the total charge Q is given by
Q = Q1 + Q2 + Q3 ——— (I)
As we known that,
Q = CV
Therefore,
Q = Ce Vab
=> Q1 = C1 Vab
=> Q2 = C2 Vab
=> Q3 = C3 Vab
Substituting the values of Q1, Q2, Q3 in eq (I)
Ce Vab = C1 Vab + C2 Vab + C3 Vab
Ce Vab + Vab (C1 + C2 + C3)
Ce = C1 + C2 + C3
Conclusion
The equivalent capacitance is equal to sum of the individual capacitances.
Qs. Define electric potential and absolute potential.
ELECTRIC POTENTIAL
Definition

Electric Potential is the amount of work done in order to bring a unit positive charge from one point to another point against the direction of electric field.
Explanation
In order to bring a unit positive charge from one point to another point, we have to do some work. This work is stored in the form of potential energy. This potential energy per unit charge is known as Electric Potential.
Mathematical Form
Electric Potential is denoted by V. It is defined as the potential energy per unit charge. Therefore mathematically,
ΔV = Δu / q ——- (I)
We know that
Δu = F. Δr
But E = F/q => F = Eq, therefore,
Δu = EqΔr
Substituting the value in eq (I)
ΔV = EqΔr / q
=> ΔV = E. Δr
Where,
ΔV = Electric Potential
E = Electric Intensity
Δr – Displacement
From above expression we know that
Electric Potential is the dot product of electric intensity and displacement.
It is a scalar quantity and its unit is Volt.
ABSOLUTE POTENTIAL
Definition
Absolute potential is the amount of work done in order to bring a unit positive charge from one point to infinite distance against the direction of magnetic field.
ELECTRIC POTENTIAL
Electric Potential is the amount of work done in order to move a unit positive charge from one point to another against the direction of electric.
Consideration
Consider a unit positive charge placed in a uniform electric field. We have to displace it from point O to point N. For this purpose we have to do some work. This work is known as electric potential. In order to determine electric potential from point O to point N we divide the whole distance into small equal patches because in a long distance intensity does not remain constant. This patch is denoted by Δr.

Class XII, PHYSICS, Chapter # 11, "Heat"

HEAT
DEFINITION
Total Kinetic energy of a body is known as HEAT.
OR
Transfer of energy from a hot body to a cold one is termed as Heat.
Heat is measured by using an measurement centimeter.
UNITS
Since heat is a force of energy therefore its unit is Joule (J).
TEMPERATURE
DEFINITION
The average kinetic energy of a body is known as Temperature.
OR
The quantitative determination of degree of hotness may be termed as Temperature.
SCALES OF TEMPERATURE
There are three main scales of temperature.
1. Celsius Scale
2. Fahrenheit Scale
3. Kelvin Scale
Celsius and Fahrenheit scales are also known as Scales of Graduation.
1. Celsius Scale
The melting point of ice and boiling point of water at standard pressure (76cm of Hg) taken to be two fixed points. On the Celsius (centigrade) scale the interval between these two fixed points is divided into hundred equal parts. Each part thus represents one degree Celsius (1°C). This scale was suggested by Celsius in 1742.
Mathematically,
°C = K – 273
OR
°C = 5/9 (°F – 32)
2. Fahrenheit Scale
The melting point of ice and boiling of water at standard pressure (76cm of Hg) are taken to be two fixed points. On Fahrenheit scale the lower fixed point is marked 32 and upper fixed point 212. The interval between them is equally divided into 180 parts. Each part represents one degree Fahrenheit (1°F).
Mathematically,
°F = 9/5 (°C + 32)
3. Kelvin Scale
The lowest temperature on Kelvin Scale is -273°C. Thus 0° on Celsius scale will be 273 on Kelvin scale written as 273K and 100 on Celsius scale will be 373K. The size of Celsius and Kelvin scales are same.
Mathematically,
K = °C + 273
THERMAL EQUILIBRIUM
Heat flows from hot body to cold body till the temperature of the bodies becomes same, then they are said to be in Thermal Equilibrium.
THERMAL EXPANSION
DEFINITION
The phenomenon due to which solid experience a change in its length, volume or area on heating is known as Thermal Expansion.
Explanation
If we supply some amount of heat to any substance then size or shape of the substance will increase. This increment is known as Thermal Expansion. Thermal expansion is due to the increment of the amplitudes of the molecules.
Types of Thermal Expansion
There are three types of Thermal Expansion.
1. Linear Expansion
2. Superficial Expansion
3. Volumetric Expansion.
1. Linear Expansion.
If we supply some amount of heat to any rod, then the length of the rod, then the length of the rod will increase. Such increment is known as Linear Expansion.
2. Superficial Expansion.
If we apply some amount of heat to any square or rectangle then area of the square or rectangle will increase. Such increment is known as Superficial Expansion.
3. Volumetric Expansion.
If we apply some amount of heat to any cube, then the volume of the cube will increase. Such increment is known as Volumetric Expansion.
COEFFICIENT OF LINEAR EXPANSION
CONSIDERATION
Let Lo be the initial length of rod at t1 °C. If we increase the temperature from t1 °C to t2 °C, then length of the rod will increase. This increment in length is denoted by ΔL. The increment in length depends upon the following two factors.
1. Original Length (Lo)
2. Difference in temperature Δt
Derivation
The increment in length is directly proportional to the original length and temperature difference.
Mathematically,
ΔL ∞ Lo —– (I)
ΔL ∞ Δt —– (II)
Combining eq (I) and (II), we get
ΔL ∞ LoΔt
=> ΔL = ∞LoΔt
Where α is the constant of proportionality and it is known as coefficient of Linear Expansion. It is defined as,
It is the increment in length per unit length per degree rise in temperature.
Its unit is 1/°C or °C. If Lt is the total length, then
Lt = Lo + ΔL
=> Lt = Lo + αLoΔt
=> Lt = Lo (1 + αΔt)
COEFFICIENT OF VOLUMETRIC EXPANSION
Consideration
Let Vo be the initial length of rod at t1 °C. If we increase the temperature from t1°C to t2°C then length of the rod will increase. This increment in length is denoted by ΔV. The increment in length depends upon the following two factors.
3. Original Volume (Lo)
4. Difference in temperature Δt
Derivation
The increment in volume is directly proportional to the original volume of temperature difference.
Mathematically,
ΔV ∞ Vo —- (I)
ΔV ∞ Δt —- (II)
Combining eq (I) and (II), we get,
ΔV ∞ Vo Δt
=> ΔV = βVoΔt
Where β is the constant of proportionality and it is known as coefficient of Volumetric Expansion. It is defined as
It is the increment in volume per unit volume per degree rise in temperature.
Its unit is 1/°C or °C-1. If Vt is the total volume then
Vt = Vo + ΔV
=> Vt = Vo + αβVo Δt
=> Vt = Vo (1 + βΔt)
State and Explain Boyle’s Law and Charle’s Law.
INTRODUCTION
Gas Laws are the laws, which give relationship between Pressure, Volume, temperature and mass of the gas. There are two gas laws.
1. Boyle’s Law
2. Charle’s Law
BOYLE’S LAW
Statement 1
According to first statement of Boyle’s Law:
Volume of the known mass of gas is inversely proportional to the pressure, if temperature is kept constant.
Mathematical Form
Mathematically,
V ∞ 1/P
=> V = K 1/P
=> PV = K (Constant)
P1V1 = P2V2 = … = K
=> P1V1 = P2V2
The above equation is mathematical form of Boyle’s Law.
Statement II
According to second statement of Boyle’s Law.
The product of the pressure and volume of the known mass of the gas remain constant if the temperature is kept constant.
Statement III
According to third statement of Boyle’s Law.
The product of pressure and volume of a gas is directly proportional to the mass of a gas, provided that temperature is kept constant.
Mathematical Form
Mathematically,
PV ∞ m
=> PV = Km
=> PV/m = K
=> P1V1/m1 = P2V2/m2
Limitations of Boyle’s Law
Boyle’s Law does not hold good at high pressure, because at high pressure gases convert into liquid or solid.
Graphical Representation
The graph between pressure and volume is a curved line, which shows that volume and pressure are inversely proportional to each other.
CHARLE’S LAW
Statement I
According to first statement of Charle’s Law.
Volume of known mass of gas is directly proportional to the absolute temperature, if then pressure is kept constant.
Mathematical Form
Mathematically,
V ∞ T
=> V = KT
=> V/T = K
OR
=> V1/T1 = V2/T2
The above equation is mathematical form of Charles Law.
Statement II
According to second statement of Charles Law.
The ratio between volume and temperature of the known mass of a gas is always constant, if pressure is kept constant.
Limitations of the Law
This law does not hold good at low temperature because at low temperature gases convert into liquid or solid.
GENERAL GAS EQUATION
It is the combination of Boyle’s law, Charle’s Law and Avogadro’s Law. According to Boyle’s Law.
V ∞ 1/P —- (I)
According to Charle’s Law
V ∞ T —- (II)
According to Avogadro’s Law
V ∞ n —- (III)
Combining eq (I), eq (II) and eq (III)
V α nT/P
=> V = RnT/P
=> PV = RnT —- (A)
Where R is the universal gas constant, We Know that
R = R/NA
=> R = KNA
Where K is the Boltzman constant, Its value is
K = 1.38 x 10(-23) J/K
Substituting the value of R in eq (A)
=> PV = nKNAT
=> PV = nNAKT
But nNA = N1 (Total number of molecules), therefore,
PV = NtKT
=> P = Nt/V KT
Since Nt/V = N (Total Number of molecules in a given volume), therefore,
P = NKT
The above equation is other form of General Gas Equation.
Qs. What are the basic postulates of Kinetic Molecular Theory pf Gases?
INTRODUCTION
The properties of matter in bulk can however be predicted on molecular basis by a theory known as Kinetic Molecular theory of gases. The characteristic of this theory are described by some fundamental assumptions, which explained below:
BASIC POSTULATES OF KINETIC MOLECULAR THEORY OF GASES
1. Composition
All gases are composed of small, spherical solid particle called molecules.
2. Dimension of Molecules
The dimensions of the molecules is compared to the separation between the molecule is very small.
3. Number of Molecules
At standard condition, there are 3 x 10(23) molecules in a cubic meter.
4. Pressure of Gas
Gas molecules collide with each other as well as with the wall of the container and exert force on the walls of the container. This force per unit are is known as Pressure.
5. Collision Between the Molecules
The collision between the molecules is elastic in which momentum and Kinetic energy remains constant.
7. Kinetic Energy of Molecules
If we increase the temperature of gas molecules, then K.E will also increase. It means that average kinetic energy of the gas molecules is directly proportional to the absolute temperature.
8. Forces Of Interraction
There is no force of attraction or repulsion between the molecules.
9. Law of Mechanics
Newtonian mechanics is applicable to the motion of molecules.
THERMODYNAMICS
DEFINITIONS
The branch of Physics that deals with the conversion of heat energy into mechanical energy or work or transformation of work into heat energy is known as Thermodynamics.
Laws of Thermodynamics
There are two laws of thermodynamics.
1. First Law of Thermodynamics
2. Second Law of Thermodynamics
State and explain first law of Thermodynamics. What are the application of first law of Thermodynamics?
FIRST LAW OF THERMODYNAMICS
First Statement
Whenever heat energy is converted into work or work is transformed into heat energy, the total amount of heat energy is directly proportional to the total amount of work done.
Mathematical Expression
Mathematically,
Q ∞ W
=> Q = JW
Where J is the mechanical equivalent of heat or joules constant. Its value is 4.2 joules.
Second Statement
If ΔQ is the amount of heat supplied to any system, then this heat will be utilized to increase the internal energy of the system in the work done in order to move the piston.
Mathematical Expression
Mathematically,
ΔQ – Au + Δw
The above equation is the mathematical form of first law of thermodynamics.
Where
Δu = Internal energy of the system.
Δw = Amount of work done.
ΔQ will be positive when heat is supplied to the system and it is negative when heat is rejected by the system.
Δw will be positive when work is done by the system and it will be negative when work is done on the system.
Third Statement
For a cyclic process, the heat energy supplied to a system and work done on the system is equal to the sum of heat energy rejected by the system.
Mathematical Expression
Mathematically,
Q(IN) + W(IN) = Q(OUT) + W(OUT)
Q(IN) – Q(OUT) = W(OUT) + W(IN)
ΔQ = ΔW
{dQ = {dW
{Shows cyclic process
Fourth Statement
For a system and surrounding the total amount of heat energy remains constant
APPLICATIONS OF THE LAW
There are four applications of first law of Thermodynamics.
1. Isometric or Isocohric Process.
2. Isobaric Process
3. Isothermal Process
4. Adiabatic Process
1. Isometric or Isocohric Process
The process in which volume of the system remains constant is known as Isometric Process.
In this process all supplied amount of heat is utilized to increase the internal energy of the system.
Mathematical Form
In this process first law of thermodynamics take the following form.
ΔQ = Δu + ΔW
But,
ΔW = 0
=> ΔQ = Δu = 0
=> ΔQ = Δu
2. Isobaric Process
The process in which pressure is kept constant is known as isobaric process.
In this process, all supplied amount of heat is utilized for the following two functions.
i. To increase the internal energy of the system.
ii. In work done in order to move the piston upward.
3. Isothermal Process
A process in which temperature is kept constant is known as Isothermal Process.
There are two parts of isothermal process.
i. Isothermal Expansion
ii. Isothermal Compression
i. Isothermal Expansion
In this process cyclinder is placed on a source and piston is allowed to move upward. When we do so temperature and pressure of the working substance will decrease while volume will increase. In order to keep the temperature constant, we have to supply required amount of heat from source to cylinder.
Since in this expansion, temperature is constant therefore it is known as Isothermal Expansion.
ii. Isothermal Compression
In this process, cylinder is placed on a sink and piston is allowed to move downward. When we do so temperature and pressure of working substance will increase while volume will decrease. In order to maintain the temperature, we have to reject required amount of heat from cylinder to the sink.
Since in this compression, temperature is kept constant therefore it is known as isothermal compression.
SECOND LAW OF THERMODYNAMICS
Introduction
It is inherit tendency of heat that it always flows from hot reservoir to cold reservoir. Rather than to flow in both the directions with equal probability. On the basis of this tendency of heat a law was proposed that is known as Second Law of Thermodynamics.
Statement
It is impossible to construct a process which reserves the natural tendency of heat.
This law is also known as Law of heat and can also be stated as
Efficiency of heat engine is always less than unity.
Explanation
Many statements of this law has been proposed to cover similar but different point of vies in which two are given below.
1. Lord Kelvin Statement
2. Clausius Statement
1. Lord Kelvin Statement
According to this statement,
It is impossible to construct a heat engine which extract all heat form the source and convert it into equal amount of work done and no heat is given to the sink.
Mathematically,
Q1 ≠ W
Q2 ≠ O
2. Clausius Statement
According to Clausius Statement,
Without the performance of external work heat cannot flow from cold reservoir towards, the hot reservoir.
Example
In case of refrigerator flow of heat is unnatural but this unnatural flow of heat is possible only when we apply electrical power on the pump of the refrigerator.
Qs. Define the term Entropy and Give its Uses
ENTROPY
Definition
It measures the disorderness of any system.
Mathematically,
ΔS = ΔQ/T
Where Δs shows change in entropy.
Units
Joule per degree Kelvin – J/°K.
Explanation
As we know that incase of isometric process volume is constant. In case of Isothermal process temperature and pressure is constant, but in case of adiabatic process neither temperature, nor pressure or volume is constant but one thermal property is constant which is known as Entropy.
There are two types of Entropy.
1. Positive Entropy
2. Negative Entropy
1. Positive Entropy
If heat is supplied to the system the entropy will be positive.
2. Negative Entropy
When heat is rejected by the system the entropy will be negative.
Qs. What is carbot engine an carnot cycle?
CARNOT ENGINE
Definition
‘Carnot engine is an ideal heat engine which converts heat energy into mechanical energy.
Working of Carnot Engine
It consists of a cylinder and a piston. The walls of the cylinder are non-conducting while the bottom surface is the conducting one. The piston is also non-conducting and friction less. It works in four steps. Which are as follows.
1. Isothermal Expansion
2. Adiabatic Expansion
3. Isothermal Compression
4. Adiabatic Compression
1. Isothermal Expansion
First of all, cylinder is placed on a source and allow to move upward as a result temperature and pressure of the working substance decreases, while volume increases. In order to maintain temperature we have to supply more amount of heat from source to the cylinder. Since in this expansion temperature is kept constant.
2. Adiabatic Expansion
Secondly cylinder is placed on an insulator and piston is allow to move downward as a result temperature and pressure of working substance will decrease. While volume will increase but no heat is given or taken of the cylinder.
3. Isothermal Compression
In this state cylinder is placed on a sink and piston is allow to move downward as a result temperature and pressure of the working substance will increase while volume will decrease. In order to maintain temperature we have to reject extra heat from cylinder to the sink. Since in this compression temperature is constant.
4. Adiabatic Compression
Finally cylinder is placed on an insulator and piston is a flow to move downward, when we do so neither temperature nor pressure or volume is constant. But no heat is given or taken out of the cylinder.
CARNOT CYCLE
Definition
By combining the four processess Isothermal Expansion, Adiabatic Expansion, Isothermal Compression and Adiabatic Compression which are carried out in carnot engine, then we get a cycle knows as Carnot cycle.
Qs. How can we increase the efficiency of Heat Engine?
If we want to increase the efficiency of any heat engine then for this purpose we have to increase temperature of source as maximum as possible and reduce the temperature of sink as minimum as possible.
Qs. Define Specific Heat and Molar Specific Heat.
SPECIFIC HEAT
Definition
Specific heat is the amount of heat required to raise the temperature of a unit mass of a substance by one degree centigrade.
Different substances have different specific heat because number of molecules in one kg is different in different substances. It is denoted by c.
Mathematical Expression
Consider a substance having mass m at the temperature t1. The amount of heat supplied is ΔQ, which raises the temperature to t2. The change in temperature is Δt.
The quantity of heat is directly proportional to the mass of the substance.
ΔQ ∞ m
And the temperature difference
ΔQ ∞ Δt
Combining both the equations
ΔQ ∞ mΔt
=> ΔQ = cmΔt
=> c = ΔQ / mΔt —- (I)
Where c is the specific heat of the substance. Its unit is Joules / Kg°C.
MOLAR SPECIFIC HEAT
Definition
Molar specific heat is the amount of heat required to raise the temperature of one mole of a substance through one degree celsius.
Almost all the substances have the same amount of molar specific heat because the numbers of molecules in all substances are same in one mole. It is denoted by cM.
Mathematical Expression
Mathematically,
No. of Moles = Mass / Molecular Mass
=> n = m / M
=> nM = m
=> nM = ΔQ / nΔt
Where n is the number of moles. The unit of molar specific heat is J/Kg°C.
Qs. Define Molar Specific Heat at Constant volume and at Constant Pressure.
MOLAR SPECIFIC HEAT AT CONSTANT VOLUME
Definition
The amount of heat required to raise the temperature of one mole of any gas through one degree centigrade, at constant volume is known as molar specific heat volume.
It is denoted by Cv.
Mathematical Expression
Mathematically,
ΔQv = nCvΔt
Where ΔQv is the heat supplied at constant volume.
MOLAR SPECIFIC HEAT AT CONSTANT PRESSURE
Definition
The amount of heat required to raise the temperature of unit mass of a substance through one degree centigrade at constant pressure is known as Molar Specific Heat at Constant Pressure.
It is denoted by Cp.
Mathematical Expression
Mathematically,
ΔQp = nCpΔt
Where ΔQp is the heat supplied at constant volume.