JEE MAIN CHAPTER OVERVIEW
Chapter Priority | 1 (HIGHEST) |
Number of problems in PYQ | HIGH |
Difficulty | Easy |
KEY CONCEPTS FOR JEE MAIN
Thermodynamics is a branch of Physics that deals with the conversion of heat into other forms of energy.
POSTULATES OF KTG
· Gases are made up of large number identical molecules
· Size of molecules is very small as compared to the distance between them
· Molecules undergo elastic collisions
· Force between molecules is negligible except during collision
· As a whole, molecules are moving randomly, i.e., they move in all directions with equal probability and with speeds varying from very low to very high values
IMPLICATIONS OF POSTULATES
· Volume of the molecules is negligible as compared to volume of the container
· Molecules can have varying speeds – very low, medium or very high.
· Proportion of molecules with a given speed does not change even after collisions. (This means that we can draw a graph of speed on x-axis and % of molecules on y-axis and this graph remains the same as time passes)
· Potential energy between molecules is ZERO as there is no force between them except during collision.
MEAN FREE PATH
Mean free path is the average distance a molecule can travel between two successive collisions.
Mean free path (lambda) is give by:
Using PV = nRT, Mean Free Path can also be written as
From the above we can conclude that, mean free path is proportional to ratio of (T/P)
TRY JEE MAIN PYQ
The mean free path of molecules of a certain gas at STP is 1500d, where d is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K is approximately :
[Hint: Mean free path is proportional to ratio T/P]
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VELOCITIES OF MOLECULES
We define three special velocities of a gas under Kinetic Theory.
When,
M0 = Molecular mass of the gas
T = Temperature of the gas
Average velocity of gas molecules is given by
RMS Velocity of gas molecules is given by
Most probable Velocity
Note: VMP < VAVG < VRMS
[This is often asked directly in JEE MAIN]
Pressure & RMS Velocity
Relation between Pressure and Vrms of a gas is given by:
MAXWELL'S DISTRIBUTION CURVE
[Not important for JEE MAIN but it is good to know where the above velocities were derived from]
Within a volume, gas molecules will have all possible speeds ranging from ZERO to infinity. The fraction of gas molecules with a given speed is the Maxwell distribution curve.
The x-axis represents the speed of molecules and the y-axis represents the Maxwell Distribution Function.
[Again, not important for JEE but technically, the fraction of molecules having speeds in the range dv is equal to the area of the shaded rectangle, Nvdv. The function Nv approaches zero as v approaches infinity. ]
Vmp, Vavg and Vrms are derived from this function.
TRY JEE MAIN PYQ:
Q. The temperature of an ideal gas is increased from 200 K to 800 K. If r.m.s. speed of gas at 200 K is V0. Then, r.m.s. speed of the gas in terms of V0 at 800 K will be?
[Hint: Vrms is proportional to square root of T]
Q. The rms speed of oxygen molecule in a vessel at particular temperature is
[(1 + 5/x)^0.5]*v
where is the average speed of the molecule. The value of x will be:
(Take pie = 22/7)
[Hint: Write down the formula for both Vrms and Vavg and use the above equation as well]
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MOLAR SPECIFIC HEAT
Molar specific heat is the heat energy required to raise the temperature of one mole of a substance by 1 degree celsius. It is represented by 'C'
We can also write:
dQ = nC(dT)
For gases, always remember, C depends upon the process.
Cadiabatic = 0 [No heat is transferred in adiabatic process]
Cisothermal = infinity [Temperature does not change in isothermal process]
For isobaric (constant pressure) process, Cp is given by:
For isochoric (constant volume) process, Cv is given by:
Other important formulae:
CP - CV = R
Ratio of Cp/Cv is called (gamma)
We can also write
DEGREE OF FREEDOM (f)
It represents the number of independent ways a gas molecule can move.
A molecule can either
Translate (Move in a straight line in x/y/z direction)
Rotate (Spin about an axis in x/y/z direction)
Vibrate (Move back and forth in x/y/z direction)
Not all molecules will have the ability rotate or vibrate across all (x/y/z) axis.
Depending upon the structure of the molecule, degrees of freedom (f) are calculated.
Remember, Cp and Cv are dependent on degrees of freedom:
Cv = f(R/2)
Cp = R + f(R/2)
Cp/Cv = 1 + (2/f)
Vibrational modes of freedom are not a part of JEE Syllabus
Given below are the degrees of freedom (f), Cp , Cv and Cp/Cv values for various molecular structures
[Note:Vibrational modes of freedom are not a part of JEE Syllabus]
Gas | f | Cv | Cp | Cp/Cv |
Monoatomic | 3 | 3R/2 | 5R/2 | 5/3 |
Diatomic | 5 | 5R/2 | 7R/2 | 7/5 |
Linear Triatomic | 5 | 5R/2 | 7R/2 | 7/5 |
Triatomic | 6 | 3R | 4R | 4/3 |
MIXTURE OF GASES
Let there be a mixture of two gases.
Let (n1, Cp, Cv) be the number of moles & molar specific heats of Gas 1
Let (n2,Cp2,Cv2) be the number of moles & molar specific heats of Gas 2
Equivalent Molar Specific heat at constant pressure is
Equivalent Molar Specific heat at constant Volume is
Equivalent Gamma(Cp/Cv)
INTERNAL ENERGY OF A GAS
Energy associated with a gas due to its Temperature. It is represented by 'U'
It represents the Kinetic Energy of a gas which is a result of Temperature alone.
[Note: Remember, within internal energy there is no component of potential energy because we assume there is no force between molecules.]
Internal Energy (U) is given by:
U = nCv(T)
dU = nCv(dT)
Where, Cv is the molar specific heat at constant volume. T is the temperature.
dT>0 => dU increases
dT<0 => dU decreases
Remember, this U is solely a function of T. The above formula is true under any process (isochoric, isobaric, isothermal, adiabatic etc.).
Law of Equipartition of Energy
Each degree of freedom has equal kinetic energy (kT/2) associated with it.
k = Boltzmann Constant = (R/NA)
For example,
A monoatomic gas has 3 degrees of freedom. Thus,
Energy of each molecule = (3kT/2)
Energy of a mole (U) = (3kT/2)*NA = (3R/2)*T
Energy of 'n' moles = n*(3R/2)*T = n*(Cv)*T
[Note: Cv (monoatomic) = 3R/2]
For any gas,
Internal energy Ugas = (f/2)*nRT
TRY A JEE MAIN PYQ
Q. The temperature at which the kinetic energy of oxygen molecules becomes double than its value at 27°C is?
[Hint: Use Kelvin scale for temperature & Kinetic energy is internal energy dependent on Temperature alone. Oxygen is diatomic so use approriate Cv.]
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Q. A gas mixture consists of 2 moles of oxygen and 4 moles of neon at temperature T. Neglecting all vibrational modes, the total internal energy of the system will be?
[Hint: U(total) = U(O2) + U(He)]
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WORK DONE BY A GAS
When a gas expands the work done by gas (W) = P(dV)
dV = Change in volume
W>0 when dV>0 [when gas expands]
W<0 when dV<0 [when gas compressed]
W = Area under P-V curve
FIRST LAW OF THERMODYNAMICS
The form of the first law useful for JEE MAIN is -
"Heat Given to System leads to Change in Internal Energy of Gas and Work Done by the Gas"
It can also be written for small change in Q,W or U. Remember the following formulae:
dQ = dW + dU
dU = nCv(dT)
dW = P(dV)
dQ = nCdT
dU is always nCv(dT) irrespective of process. Cv is the molar specific at constant volume.
dQ is by definition nC(dT). Here C is the molar specific heat of the process involved.
dQ>0 | Heat supplied to the system |
dQ<0 | Heat taken from the system |
dU>0 | temperature of gas increases |
du<0 | temperature of gas decreases |
dW>0 | Volume of gas increases |
dW<0 | Volume of gas decreases |
ISOBARIC PROCESS
Pressure = Constant
PV = nRT
W = P(V2 - V1)
In isobaric process, Volume is proportional to to temperature
P-V graph is a straight line parallel to x-axis
ISOCHORIC PROCESS
Volume = Constant
PV = nRT
W = 0 (ZERO)
In Isochoric process, Pressure is proportional to temperature
PV curve is a straight line perpendicular to x-axis.
ISOTHERMAL PROCESS
Temperature = Constant
PV = nRT
PV = Constant
P1V1 = P2V2
Work done in isothermal process:
P-V graph of Isothermal Process
Since, Temperature is constant, internal energy (U) is constant.
dU = 0
dQ = dW
Heat supplied = Work done by Gas
ADIABATIC PROCESS
dQ = 0
No heat is exchanged in this process
dQ = 0
dU + dW = 0
dW = -dU
dU = nCv(dT)
dW = -nCv(dT)
Relation between Pressure and Volume in Adiabatic Process
Pressure-Volume graph slopes downwards
Relation between Temperature and Volume
Graph of temperature-volume in adiabatic process slopes downwards
Relation between Pressure and Temperature in adiabatic process
Graph of pressure and temperature in adiabatic process slopes upwards
Work done in adiabatic process
dW = -dU
dW = -nCv(dT)
Now, Cv can be written as:
Work done in adiabatic process:
Comparison P-V graphs of adiabatic and isothermal:
CYCLIC PROCESS
Processes in which gas starts and ends at the same state are called cyclic processes.
The P-V graph of a cyclic process forms a closed loop.
Work done in cyclic process = Area of the loop
if process clockwise, W>0 (positive)
if process anti-clockwise, W<0 (negative)
GAS UNDER FREE EXPANSION
When gas expands against a vacuum, Pressure of the gas is Zero.
Therefore Work done by gas = ZERO
TRY JEE MAIN PYQ
Q. A source supplies heat to a system at the rate of 1000 W. If the system performs work at a rate of 200 W. The rate at which internal energy of the system increase is?
[Hint: Write first law of thermodynamics and divide equation by time to get rate.]
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Q. The Thermodynamic process, in which internal energy of the system remains constant is?
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SUMMARY OF PROCESSES
PROCESS | PROPERTY |
ISOTHERMAL | Internal energy is constant. Heat supplied to gas equals work done by gas. |
ISOBARIC | The heat absorbed goes partly to increase internal energy and partly to do work |
ISOCHORIC | No work is done on or by the gas |
ADIABATIC | Work done by the gas decreases internal energy |
POLYTROPIC PROCESSES
Processes whose P-V curve is of the form:
For example:
Molar specific heat of a polytropic process is given by:
MIXING TWO GASES
This topic is learnt from an example:
Case 1: Same gas in two containers:
Same gas having characteristics P1, V1,T1 & P2, V2, T2 is placed in two chambers as shown. Find final temperature & pressure of mixture if knob is removed. Assume no heat is loss from the container.
dQ = 0
dW = 0 [since total volume remains constant, work done by system is ZERO]
Therefore, dU = 0 [from 1st Law of Thermodynamics]
Get n1 and n2 from PV = nRT
Equate initial internal energy and final internal energy
Case 2: Different gas in two containers:
Different gases having characteristics P1, V1,T1 & P2, V2, T2 is placed in two chambers as shown. Find final temperature & pressure of mixture if knob is removed. Assume no heat is loss from the container.
dQ = 0
dW = 0 [since total volume remains constant, work done by system is ZERO]
Therefore, dU = 0 [from 1st Law of Thermodynamics]
Get n1 and n2 from PV = nRT
Equate initial internal energy and final internal energy
Keep in mind, Use Cv_equivalent in the above equation:
SECOND LAW OF THERMODYNAMICS
KEVIN-PLANCK STATEMENT
"It is impossible to construct a heat engine which can give 100% efficiency"
HEAT ENGINES
Heat engine converts heat into mechanical energy.
For example, when you boil water in a kettle and keep a lid on top, after a while the pressure from steam causes the lid to move a little. Heat from burning of gas is converted into motion of lid.
In a heat engine there are three parts:
Source: This part provides the heat energy to the system. This is at a high temperature (T1).
Working Substance: This is the gas which will do Work.
Sink: This is the part to which residual heat is lost. It is at a low temperature (T2).
In the above boiling water example, the burning of cooking gas provides heat (it is the source), steam/water is the working substance and air around the kettle is the sink.
Q1: Heat supplied by source
W: Work done by working substance
Q2: Heat rejected to sink
(These represent magnitudes only, take signs accordingly)
From 1st law:
Q1 = W + Q2
Efficiency = (Work Done/Heat Supplied)*100
CARNOT CYCLE
Carnot cyle is the most efficient heat engine. It represent a cyclic process with 4 stages:
Stage 1: Isothermal Expansion (AB)
Stage 2: Adiabatic Expansion (BC)
Stage 3: Isothermal Compression (CD)
Stage 4: Adiabatic Compression (DA)
Assume:
A (P1,V1), B (P2,V2), C (P3,V3), D (P4,V4)
A & B have same Temperature (T1)
C & D have same Temperature (T2)
The following equations hold true for Carnot cycle:
V2*V4 = V1*V3
For a Carnot cycle, from the above volume relation we can derive the following:
Efficiency of Carnot engine can be written as:
TRY A JEE MAIN PYQ:
Work done by a Carnot engine operating between temperatures 127°C and 27°C is 2 kJ. The amount of heat transferred to the engine by the reservoir is?
[Hint: Write Temperature in Kelvin]
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A carnot engine with efficiency 50% takes heat from a source at 600K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be?
[Hint: Write efficiency formula in terms Temperature]
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2nd Law of Thermodynamics
Clausius Statement:
"It is impossible to Construct a Refrigerator unaided by any external agent, which would transfer heat from a body at a low Temperature to Another Body at a High Temperature."
Refrigerator takes heat from a cold body and throws away this heat to a hot body. For example, the refrigerator in your home would take heat from inside (where temperature is low) and throw that heat into the atmosphere (where temperature is high)
The refrigerator takes heat Q2 from sink with the help of work W done by external agent and throws out heat Q1 to the atmosphere.
We define a coefficient of performance
Entropy
It is the degree of randomness of a system
It is represented by S.
Change in entropy equals the (Heat supplied to the system/Temperature)
Change in entropy: dS = dQ/T
dQ = T(dS)
[This can be used in dQ = dW + dU]
Area under T-S curve gives us the heat supplied to a system (integral of TdS).
Hope you can REVISE KTG & THERMODYNAMICS FOR JEE MAIN PHYSICS from this guide to answer all varieties of JEE MAIN PYQs. Thanks a lot!
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