# CBSE Class 11 Physics Chapter 12 Thermodynamics Study Materials

### Class 11 Physics Chapter 12 Thermodynamics

Topics and Subtopics in  Class 11 Physics Chapter 12 Thermodynamics:

 Section Name Topic Name 12 Thermodynamics 12.1 Introduction 12.2 Thermal equilibrium 12.3 Zeroth law of thermodynamics 12.4 Heat, internal energy and work 12.5 First law of thermodynamics 12.6 Specific heat capacity 12.7 Thermodynamic state  variables and equation of state 12.8 Thermodynamic processes 12.9 Heat engines 12.10 Refrigerators and heat pumps 12.11 Second law of thermodynamics 12.12 Reversible and irreversible processes 12.13 Carnot engine

### Thermodynamics Class 11 Notes Physics Chapter 12

• The branch of physics which deals with the study of transformation of heat into other forms of energy and vice-versa is called thermodynamics.
Thermodynamics is a macroscopic science. It deals with bulk systems and does not go into the
molecular constitution of matter.
• A collection of an extremely large number of atoms or molecules confined within certain boundaries
such that it has a certain values of pressure (P), volume (V) and temperature (T) is called a ; thermodynamic system.
• Thermal Equilibrium
A thermodynamic system is in an equilibrium state if the macroscopic variables such as pressure, volume, temperature, mass composition etc. that characterise the system do not change in time. In thermal equilibrium, the temperature of the two systems are equal.
• Zeroth Law of Thermodynamics
This law identifies thermal equilibrium and introduces temperature as a tool for identifying f equilibrium. According to this law “If two systems are in thermal equilibrium with a third system then those two systems themselves are in equilibrium.”
• Heat, Work and Internal Energy
— Energy that is transferred between a system and its surroundings whenever there is temperature difference between the system and its surroundings is called heat.
— Work is said to be done if a body or a system moves through a certain distance in the direction of the applied force. It is given as
dW = PdV
where P is the pressure of the gas in the cylinder.
— If we consider a bulk system consisting of a large number of molecules, then internal energy ; of the system is the sum of the kinetic energies and potential energies of these molecules.
This energy is possessed by a system due to its molecular motion and molecular configuration. The internal energy is denoted by U.
U = Ek + Ep
where Ek and Ep represent the kinetic and potential energies of the molecules of the system.
• Internal energy of a system is a macroscopic variable and it depends only on the state of the system. Its value depends only on the given state of the system and does not depend on the path taken to arrive that state.
• First Law of Thermodynamics
The first law of thermodynamics is simply the general law of conservation of energy applied to any system. According to this law, “the total heat energy change in any system is the sum of the internal energy change and the work done.”
When a certain quantity of heat dQ is subjected to a system, a part of it is used in increasing the internal energy by dU and a part is used in performing external work dW, hence
dQ = dU + dW
• For gases, the specific heat capacity depends on the process or the conditions under which heat capacity transfer takes place. There are mainly two principal specific heat capacities for a gas. These are specific heat capacity at constant volume and specific heat capacity at constant pressure.
• From First Law of Thermodynamics we find a relation between two principal specific heats of an ideal gas. According to the relation
Cp-Cυ = R
Here Cp and Cυ are molar specific heats under constant pressure and constant volume condition respectively.
The specific heat capacity of a gas at constant pressure is greater than the specific heat capacity of the gas at constant volume i.e. Cp > Cυ. Reason is that when heat supplied to a gas at constant volume, no work would be done by the gas against the external pressure and all the energy is used to raise the temperature of the gas. On the other hand when the heat is supplied to the gas at constant pressure, its volume increases and the heat energy supplied to it is used to increase the temperature of the gas as well as in doing the work against the external pressure.
The difference, between the two specific heats is the thermal equivalent of the work done in expanding the gas against the external pressure.
• Expression for the Relation between Cp and Cυ
Let P, V and T be the pressure, volume and absolute temperature initially of one mole of an ideal gas.
Case (i): The heat dQ is supplied to the gas at constant volume so that the temperature increases to T + dT.

• Thermodynamic State Variables
Thermodynamic state variables of a system are the parameters which describe equilibrium states of the system. For example, equilibrium state of gas is completely specified by the values of pressure, volume, temperature, mass and composition.
• Equation of State
The equation of state represents the connection between the state variables of a system. For example, the those equation of state of an ideal/perfect gas in represented as
PV = μRT
where g is number of moles of the gas and R is gas constant for one mole of the gas.
• Thermodynamic state variables are of two kinds, extensive and intensive. Extensive variables indicate the size of the system but intensive variables do not indicate the size. Volume, mass, internal energy of a system are extensive variables but pressure, temperature and density are intensive variables.
• Thermodynamic Processes
Any process in which the thermodynamic variables of a thermodynamic system change is known as thermodynamic process.
• Quasi-Static Processes
Processes that are sufficiently slow and do not involve accelerated motion of piston and/or large temperature gradient are quasi-static processes.
In this process, the change in pressure or change in volume or change in temperature of the system is very small.
• Isothermal Process
A change in pressure and volume of a gas without any change in its temperature, is called an isothermal change. In such a change, there is a free exchange of heat between the gas and its surroundings.
A process in which no exchange of heat energy takes place between the gas and the surroundings, is called an adiabatic process.
• The work done dW under isothermal change is given by

• P-V Diagram
A graph representing the variation of pressure with the variation of volume is called P-V diagram. The work done by the thermodynamic system is equal to the area under P-V diagram. It is given as

• Reversible Process
A process which can retrace so that the system passes through the same states is called a reversible process, otherwise it is irreversible.
Irreversibility arises mainly from two causes:
(i) Many processes like free expansion or an explosive chemical reaction take the system to non-equilibrium states.
(ii) Most processes involve friction, viscosity and other dissipative effects.
• Second Law of Thermodynamics
This principle which disallows certain phenomena consistent with the First law of thermodynamics is known as the second law of thermodynamics.
Following are the two statements of second law of thermodynamics.
Kelvin-Planck Statement: It is impossible to construct an engine, operating in a cycle, to extract
heat from hot body and convert it completely into work without leaving any change anywhere i.e., 100% conversion of heat into work is impossible.
Clausius Statement: It is impossible for a self acting machine, operating in a cycle, unaided by any external energy to transfer heat from a cold body to a hot body. In other words heat can not flow itself from a colder body to a hotter body.
• A heat engine is a device by which a system is made to undergo a cyclic process that results
in conversion of heat to work. Basically, a heat engine consists of: (i) a heat source, (ii) a heat sink, and (iii) a working substance.
• Carnot’s Engine. He proposed a hypothetical engine working on a cyclic/reversible process operating between two temperatures. Its efficiency is independent of the working substance and is given by, η=1-T2/T1 where T1 is the temperature of source and T2 is the temperature of sink.
• According to Carnot’s theorem: (a) working between two given temperatures T1 and T2 of the hot and cold reservoirs respectively, no engine can have efficiency more than that of Carnot’s engine, and (b) the efficiency of the Carnot engine is independent of the nature of the working substance.
• Refrigerator
The process of removing heat from bodies colder than their surroundings is called refrigeration and the device doing so is called refrigerator.
In the refrigerator, heat is absorbed at low temperature and rejected at higher temperature with the help of external mechanical work. Thus, a refrigerator is a heat engine working backward and hence it is also called heat pump.
Refrigerator works on the reverse process of Carnot engine. By the work done on the system, heat is extracted from low temperature sink T2 and passed on to high temperature source T1. The coefficient of performance is given by

• IMPORTANT TABLES

### CBSE Class 11 Physics Chapter-12 Important Questions

1 Marks Questions

1.If a air is a cylinder is suddenly compressed by a piston. What happens to the pressure of air?

Ans.Since the sudden compression causes heating and rise in temperature and if the piston is maintained at same Position then the pressure falls as temperature decreases.

2. What is the ratio of find volume to initial volume if the gas is compressed adiabatically till its temperature is doubled?

PVY= constant

Since PV = RT

P =

So, constant

Or TV y – 1 = constant T1, V1 = Initial temperature and Initial Volume

∴ T1 Vy – 1 = T2 V2 y – 1 T2, V2 = Final temperature and Final volume.

Since T2 = 2 T1(Given)

So,

Since  > 1,  is less than  .

3.What is the ratio of slopes of P-V graphs of adiabatic and isothermal process?

Ans.The slope of P-V graph is

For an isothermal process, (PV = constant)

We have ,

For an adiabatic process ( PVY = constant)

Divide 2) by 1)

So, the ratio of adiabatic slope to isothermal slope is Y.

4.What is the foundation of Thermodynamics?

Ans.The foundation of thermodynamics is the law of conservation of energy and the fact the heat flows from a hot body to a cold body.

5.Differentiate between isothermal and adiabatic process?

Ans.

 Isothermal process Adiabatic process 1) In this, temperature remains constant 1) In this, no heat is added or removed. 2) It occurs slowly 2) It occurs suddenly. 3) Here, system is thermally conducting to surroundings 3) Here, system is thermally insulated from surroundings. 4) State equation :→ PV = constant 4) State equation : → PVY = constant.

6.A Carnot engine develops 100 H.P. and operates between 270C and 2270C. Find 1) thermal efficiency; 2) heat supplied3) heat rejected?

Ans. Here, energy = W = 100 H. P.

= 100 × 746 W ( 1 H.P. = 746W)

High temperature, TH = 2270C = 227 + 273 = 500K

Low temperature, Th = 270C = 27 + 273 = 300K

1) Thermal efficiency,

2) The heat supplied QH is given by:-

3) The heat rejected QL is given by:-

or

7.Draw a p – v diagram for isothermal and adiabatic expansion?

Ans.

8.State zeroth law of thermodynamics?

Ans. Acc. to this, when the thermodynamic system A and B are separately in thermal equilibrium with a third thermodynamic system C, then the system A and B are in thermal equilibrium with each other also.

9.Can a gas be liquefied at any temperature by increase of pressure alone?

Ans. No, a gas can be liquefied by pressure alone, only when temperature of gas is below its critical temperature.

10.Can you design heat energy of 100% efficiency?

Ans. Since efficiency of heat engine = 1-, so, efficiency will be 100% or 1 if T2 = OK or T1 = α. Since both these conditions cannot be practically attained, so heat engine cannot have 100% efficiency.

11.If air is a bad conductor of heat, why do we not feel warm without clothes?

Ans.This is because when we are without clothes air carries away heat from our body due to convection and we feel cold.

12.A body with large reflectivity is a poor emitter why?

Ans.This is because a body with large reflectivity is a poor absorber of heat and poor absorbers are poor emitters.

13.Animals curl into a ball, when they feel very cold?

Ans.When animals curl, they decrease their surface area and since energy radiated varies directly to surface area hence loss of heat due to radiation is also reduced.

14.Why is the energy of thermal radiation less than that of visible light?

Ans.The energy of an electromagnetic ware is given by :- E = hf

h = Planck’s constant; f = frequency of wave. Since the frequency of thermal radiation is less than that of visible light, the energy associated with thermal radiation is less than associated with visible light.

15.Two rods A and B are of equal length. Each rod has its ends at temperature T1 and T2 (T1 > T2). What is the condition that will ensure equal rates of flow through the rods A and B?

Ans. Heat flow,

K = Thermal conductivity

A = Area

T1 = Temperature of hot body

T= Temperature of cold body

d = distance between hot and cold body.

Q = heat flow

When the rods have the same rate of conduction,

Q1 = Q2

K1, K2 → Thermal conductivity of first and second region

A1, A2 → Area of first and second region

or, K1 A1 = K2 A2

or

16.A Sphere is at a temperature of 6oo k. Its cooling rate is R in an external environment of 200k. If temperature falls to 400k. What is the cooling rate R1 in terms of R?

Ans.Acc. to Stefan’s law;

E = constant T4

Also, R1 = constant (T24 – T14)

R = constant (T34 – T14)

T= heat of hot junction = 400K

T1 = heat of cold junction = 200K

T3 = heat of hot junction = 600K

R1 = constant

R1 = constant

Divide  eq4 1) & 2)

Therefore,

17.If the temperature of the sun is doubled, the rate of energy received on each will increases by what factor?

Ans.By Stefan’s law : →

Rate of energy radiated α T4

T = Temperature

E1 = constant T14

E2 = constant T24

T1 = Initial temperature

T2 = Final temperature

T2 = 2T1

T24 = (2)4 T14

T24 = 16T14

E2 = constant (16 T14)

E2 = 16 (constant T14)

E2 = 16 E1

18. On a winter night, you feel warmer when clouds cover the sky than when sky is clear. Why?

Ans.We know that earth absorbs heat in day and radiates at night. When sky is covered, with clouds, the heat radiated by earth is reflected back and earth becomes warmer. But if sky is clear the heat radiated by earth escapes into space.

19. If a body is heated from 270 C to 9270C then what will be the ratio of energies of radiation emitted?

Ans.Since, By Stefan’s law:→

T = Temperature.

E1, T1  Initial energy and temperature

E2, T2  Final energy and temperature.

T1 = 270C = 27+273 = 300K

T2 = 9270C = 927+273K = 1200K.

E = constant T4

So, E1 = constant T14

Equating equation 1) &2)

or E1 : E2 = 1 : 256

20. Which has a higher specific heat ; water or sand?

Ans.Water has higher specific heat than sand as

, where T = Temperature, Q = Heat, m = Mass,

C = Specific heat; Since for water temperature increases less slowly than sand hence the result.

21. Why is latent heat of vaporization of a material greater than that of latent heat of fusion?

Ans .When a liquid changes into a gas, there is large increase in the volume and a large amount of work has to be done against the surrounding atmosphere and heat associated with change from solid to gas is latent heat of vaporization and hence the answer.

22. Draw a P – V diagram for Liquid and gas at various temperatures showing critical point?

Ans.

23. Why is temperature gradient required for flow of heat from one body to another?

Ans. Heat flows from higher temperature to lower temperature. Therefore, temperature gradient (i.e. temperature difference) is required for the heat to flow one part of solid to another.

24. Why are Calorimeters made up of metal only?

Ans. Calorimeters are made up of metal only because they are good conductor of heat and hence the heat exchange is quick which the basic requirement for the working of calorimeter.

25. If a body has infinite heat capacity? What does it signify?

Ans. Infinite heat capacity means that there will be no change in temperature whether heat is taken out or given to the substance.

26. Define triple point of water?

Ans. Triple point of water represents the values of pressure and temperature at which water co-exists in equilibrium in all the three states of matter.

27. State Dulong and petit law?

Ans.Acc. to this law, the specific heat of all the solids is constant at room temperature and is equal to 3R.

28. Why the clock pendulums are made of invar, a material of low value of coefficient of linear expansion?

Ans.The clock pendulums are made of Inver because it has low value of α (co-efficient of linear expansion) i.e. for a small change in temperature, the length of pendulum will not change much.

29. Why does the density of solid | liquid decreases with rise in temperature?

Ans.Let P = Density of solid | liquid at temperature T

P1 = Density of solid | liquid at Temperature T+∆T

Since Density =

So, P = →(1) P1 = (2)

V1 = Volume of solid at temperature T + ∆T

V = Volume of solid at temperature T

Since on increasing the temperature, solids | liquids expand that is their volumes increases, so by equation

i) & 2) Density is inversely proportional to volumes, so if volume increases on increasing the temperature, Density will decrease.

30. Two bodies at different temperatures T1, and T2 are brought in thermal contact do not necessarily settle down to the mean temperature of Tand T2?

Ans.Two bodies at diff temperatures T1 and T2 when in thermal contact do not settle always at their mean temperature because the thermal capacities of two bodies may not be always equal.

31. The resistance of certain platinum resistance thermometer is found to be 2.56 Ω at 00c and 3.56 Ω at 1000c. When the thermometer is immersed in a given liquid, its resistance is observed to 5.06 Ω. Determine the temperature of liquid?

Ans.Ro = Resistance at00c = 2.56Ω

R= Resistance at temperature T = 1000c = 3.56Ω 100

Rt = Resistance at unknown temperature t ;

Rt = 5.06Ω

Since,

t =

t =  2500c

32. Calculate Cp for air, given that Cv =0.162 cal g-1 k-1 and density air at N.T. P is 0.001293 g|cm3?

Ans.Specific heat at constant pressure = Cp= ?

Specific heat at constant volume = C= 0.162 Cal g-1 k-1

Now, Cp – Cv =

Or CP – Cv =

Cp – Cv =

= 6.8×10-4+2

Cp – Cv = 0.068

Cp = 0.162+0.068

Cp = 0.23 Cal g-1 k-1

33. Develop a relation between the co-efficient of linear expansion, co-efficient superficial expansion and coefficient of cubical expansion of a solid?

Ans.Since, co-efficient of linear expansion = α =

∆L = change in length

L = length

∆T = change in temperature

Similarly, co-efficient of superficial expansion = β =

∆S = change in area

S = original area

∆T = change in temperature

Co-efficient of cubical expansion, = Y =

∆V = change in volume

V = original volume

∆T = change in temperature.

Now, ∆L=αL ∆T

L + ∆L = L + αL ∆T

L + ∆L = L (1+α∆T) → (1)

Similarly V+ ∆V = V (1+Y∆T) →(2)

And S+∆S=S (1+β∆T) → (3)

Also, (V+∆V) = (L+∆L)3

V+∆V =

V+∆V = L3

Since α2, α3 are negligible, so,

V+γ V∆T= V(1+3α∆T) [as L3=V]

So, V+γV∆T = V+V3α∆T

γV∆T = 3α∆T

Y = 3α

Similarly,  β = 2α  [using L2 = S (Area)]

So,

34. Calculate the amount of heat required to convert 1.00kg of ice at – 100c into steam at 1000c at normal pressure. Specific heat of ice = 2100J|kg|k. Latent heat of fusion of ice = 3.36×105J|kg, specific heat of water = 4200J|kg|k. Latent heat of vaporization of water = 2.25 x106J|kg?

Ans.(1) Here, heat is required to raise the temperature of ice from – 100c to 00c.

So, change in temperature = ∆T = T2-T1 = 0-(-10) = 100c

So, ∆Q1=cm∆T

C = specific heat of ice

M = Mass of ice

∆T = 100c

∆Q= 2100×1×10=21000J

(2) Heat required to melt the ice to 00c water:-

∆Q2 = mL

L = Latent heat of fusion of ice = 3.36×105J/kg

m = Mass of ice

∆Q2 = 1×3.36×105J/kg

∆Q2=3.36×105J

∆Q2 = 336000J

(3) Heat required to raise the temperature of water from 00c to 1000c:-

∆T = T2-T1 = 100-0=1000c

∆Q= cm∆Tc = specific heat of water

= 4200×1×100

= 420,000J

(4) Heat required to convert 1000c water to steam at  1000c

∆Q4 = mL   L = Latent heat of vapourisation  = 2.25×106J/kg

∆Q4 = 1×2.25×106J|kg

∆Q4 = 2250000J

∴ Total Heat required = ∆Q1+∆Q2+∆Q3+∆Q4

∆Q total = 21000+336000+420000+2250000

∆Q total = 3027000J

∆Q total = 3.027×106J

35. Why is mercury used in making thermometers?

Ans. Mercury is used in making thermometers because it has wide and useful temperature range and has a uniform rate of expansion.

36. How would a thermometer be different if glass expanded more with increasing temperature than mercury?

Ans. If glass expanded more with increasing temperature than mercury, the scale of the thermometer would be upside down.

37. Show the variation of specific heat at constant pressure with temperature?

Ans.

38. Two thermometers are constructed in the same way except that one has a spherical bulb and the other an elongated cylindrical bulb. Which one will response quickly to temperature change?

Ans. The thermometer with cylindrical bulb will respond quickly to temperature changes because the surface area of cylindrical bulb is greater than the of spherical bulb.

39. State Carnot’s Theorem?

Ans. According to Carnot’s Theorem, no engine working between two temperatures can be more efficient than a Carnot’s reversible engine working between the same temperatures.

Very Short Answer Type Question (1 Marks)

1. Why spark is produced when two substances are struck hard against each other?
Ans. Work is converted into heat.
2. What is the Specific heat of a gas in an isothermal process.
Ans. Infinite
3. On what factors, does the efficiency of Carnot engine depend?
Ans.
4. What are two essential features of Carnot’s ideal heat engine.
Ans.
(i) Source and sink have infinite heat capacities.
(ii) Each process of the engine’s cycle is fully reversible
5. Plot a graph between internal energy U and Temperature (T) of an ideal gas.
Ans.
6. Refrigerator transfers heat from cold body to a hot body. Does this violate the second law of thermodynamics?
Ans. No, External Work is done
7. What is heat pump?
Ans. A heat pump is a device which uses mechanical work to remove heat.
8. Give two example of heat pump?
Ans. Refrigerator, Air Conditioner.
9. What is heat engine?
Ans. Heat engine is a device which convert heat energy into mechanical energy.
10. Why a gas is cooled when expanded?
Ans. Decrease in internal energy.
11. Can the temperature of an isolated system change?
Ans. Yes in an adiabatic process
12. Which one a solid, a liquid or a gas of the same mass and at the same temperature has the greatest internal energy.
Ans. Gas has greatest internal energy and solid has least internal energy.
13. Under what ideal condition the efficiency of a Carnot engine be 100%.
Ans. If the temperature of sink is OK.
14. Which thermodynamic variable is defined by the first law of thermodynamics?
Ans. Internal energy.
15. Give an example where heat be added to a system without increasing its temperature.
Ans. Melting.
16. What is the efficiency of carnot engine Working between ice point and steam point?
Ans.
17. Two blocks of the same metal having masses 5g and 10g collide against a target with the same velocity. If the total energy used in heating the balls which will attain higher temperature?
Ans. Both the balls will undergo the same rise in temperature.
18. What is the specific heat of a gas in an adiabatic process. Work is converted into heat.
Ans. Zero.

2 Marks Questions Part 1

1.A motor car tyre has a Pressure of four atmosphere at a room temperature of 270C. If the tyre suddenly bursts, calculate the temperature of escaping gas?

Ans. Since the tyre suddenly bursts, the change taking place is adiabatic, for adiabatic change:-

Or

Hence, T1 = 273 + 27 = 300K

P1 = Initial Pressure; P2 = final Pressure

So,

So, Putting the above values in eq4  i)

Taking 1.4 Power

W1=-150J→ (1)

Work done by the gas in the process B → C is : →

Net work done by the gas in the whole process is W = W1 + W2

W = 150 – 70 = – 22 OJ

T2 = 201.8 K

∴ T2 = 201.8 – 273 = – 71.20C

2.How does Carnot cycle operates?

Ans. A Carnot cycle operates a follows:-

1) It receives thermal energy isothermally from some hot reservoir maintained at a constant high temperature TH.

2) It rejects thermal energy isothermally to a constant low–temperature reservoir (T2).

3) The change in temperature is reversible adiabatic process.

Such a cycle, which consist of two isothermal processes bounded by two adiabatic processes, is called Carnot cycle.

3.Calculate the work done by the gas in going from the P-V graph of the thermodynamic behavior of a gas from point A to point B to point C?

Ans.Work done by the gas in the process A → B is

W= – (area under curve A B)

=  –

= –

A B = 500 Pa

= 5×105 N|m2

4.Why does absolute zero not correspond to zero energy?

Ans.The total energy of a gas is the sum of kinetic and potential energy of its molecules. Since the kinetic energy is a function of the temperature of the gas. Hence at absolute zero, the kinetic energy of the molecules ceases but potential energy is not zero. So, absolute zero temperature is not the temperature of zero energy.

5.State the Second law of thermodynamics and write 2 applications of it?

Ans.According to second law of thermodynamics, when a cold body and a hot body are brought into contact with each other, heat always from hot Body to the cold body. Also, that no heat engine that works in cycle completely converts heat into work.

Second law of thermodynamics is used in working of heat engine and of refrigerator.

6.At 00C and normal atmospheric pressure, the volume of 1g of water increases from 1cm3 to 1.091 cm3 on free zing. What will be the change in its internal energy? Normal atmospheric pressure is 1.013×105 N|m2 and the latent heat of melting of ice is 80 cal/g?

Ans.Since, heat is given out by 1 g of water in freezing is

m = Mass of water = 1 g

Q = – (mLf) Lf = Latent heat of melting of ice = 80 cal|g

During freezing, the water expands against atmospheric pressure. Hence, external work done (W) by water is :- W = P × ∆ V

P = 1.013×105 N|m2;  ∆ V = 1.091 – 1 = 0.091 cm3 = o.o91 × 10-6 m3

∆ V = V2 – V1; V2 = final volume = 1.91 cm3

V1 = Initial volume = 1 cm3

So, W =

W = 0.0092 J

Since, 1 cal = 4.2J so,

W =

Since the work has been done by ice, it will be taken positive.

Acc. to first law of thermodynamics,

Q = ∆∪ + W ∆∪ = change in internal energy

So, ∆∪ = Q – W

=

∆∪ = – 80.0022 cal

7.Two different adiabatic paths for the same gas intersect two thermals at T1 and T2 as shown in P-V diagram. How does Compare with ?

Ans.Now, A B and C D are isothermals at temperature T1 and T2 respectively and BC and AD are adiabatic.

Since points A and D lie on the same adiabatic.

T1 VA Y-1 = T2 VDY-1

Also, points B and C lie on the same adiabatic,

or T1VB Y-1 = T2VCY-1

From equation 1) & 2)

8.The internal energy of a compressed gas is less than that of the rarified gas at the same temperature. Why?

Ans.The internal energy of a compressed gas is less than that of rarified gas at the same temperature because in compressed gas, the mutual attraction between the molecules increases as the molecules comes close. Therefore, potential energy is added to internal energy and since potential energy is negative, total internal energy decreases.

9.Consider the cyclic process A B C A on a sample 2 mol of an ideal gas as shown. The temperature of the gas at A and B are300 K and 500K respected. Total of 1200 J of heat is with drawn from the sample. Find the work done by the gas in part BC?

Ans.The change in internal energy during the cyclic process is zero. Therefore, heat supplied to the gas is equal to work done by it,

∴ WAB + WBC + WCA = – 1200J →(1)

(- ve because the cyclic process is traced anticlockwise the net work done by the system is negative)

The work done during the process AB is

WAB = PA (VB-VA) = nR(TB-TA

WAB = 2×8.3(500-300)  = 3320J →2)

R = Universal gas constant

N = No. of volume

Since in this process, the volume increases, the work done by the gas is positive.

Now, WCA = O (volume of gas remains constant)

∴ 3320 + WBC + O = – 1200 (Using equation 1) & 2)

WBC = – 1200 – 3320

WBC = – 4520J

10.A refrigerator placed in a room at 300 K has inside temperature 264K. How many calories of heat shall be delivered to the room for each 1 K cal of energy consumed by the refrigerator, ideally?

Ans.High temperature, TH = 300K

Low temperature, Th = 264K

Energy = 1K cal.

Co – efficient of performance, is given by:-

Now, COP =

Q= heat rejected

QL = COP × W

QL =

The mechanical work done by the compressor of the refrigerator is:-

W = QH – QL

QH = W + QL

QH =

QH = 8.33 K cal

11.If the door of a refrigerator is kept open in a room, will it make the room warm or cool?

Ans.Since a refrigerator is a heat engine that operates in the reverse direction i.e. it extracts heat from a cold body and transforms it to hot body. Since it exhaust more heat into room than it extracts from it. Therefore, the net effect is an increase in temperature of the room.

12.The following figure shows a process A B C A per formed on an ideal gas, find the net heat given to the system during the process?

Ans .Since the process is cyclic, the change in internal energy is zero. Therefore, the heat given to the system is equal to work done by it. The net work done by the gas in the process ABCA is:-

W = WAB + WBC + WCA

Now WAB = O

During the path BC, temperature remains constant. So it is an isothermal process. So, WBC = nRTLoge

During the CA, Vα T so that is constant.

∴ Work done by the gas during the part CA is :-

WCA = P (V1 – V2)

= nR (T– T2)

= – nR (T2 – T1)  → Using equation 1)

W= O + nR T2 Loge  – nR (T2 – T1)

13.A certain gas at atmospheric pressure is compressed adiabatically so that its volume becomes half of its original volume. Calculate the resulting pressure?

Ans.Let the original volume, V1=V

∴ final volume V(∴volume become half)

Initial pressure P1 = o.76m of Hg column

Final pressure P2 after compression =?

As the change is adiabatic, so

Y = = 1.4 for air

P= P1

= 0.76 ×

P2 = 0.76 × (2)1.4

P2 = 2m of Hg column

P2 = h sg

P2 = 2.672 x 105 N|m2

P2 = 2× (13.6×103)× 9.8

h = height of Hg column

s = Density of air

g = Acceleration due to gravity

14.Why is conversion of heat into work not possible without a sink at lower temperature?

Ans. For converting heat energy in to work continuously a part of heat energy absorbed from the source has to be rejected. The heat energy can be rejected only to a body at lower temperature which is sink, so we require a sink to convert heat into work

15.Write the sign conventions for the heat and work done during a thermodynamic process?

Ans. 1) When heat is supplied to a system d Q is taken positive but when heat is supplied by a system, d Q is taken negative.

2) When a gas expands, d w is taken as positive but when a gas compresses, work done is taken as negative.

16.Does the working of an electric refrigerator defy second law of thermodynamics?

Ans. No, it is not against the second law; this is because external work is done by the compressor or for this transfer of heat.

17.A Carnot engine absorb 6×105 cal at 2270c calculate work done per cycle by the engine if it sink is at 1270c?

Ans. Here, heat abs or bed = Q= 6 × 105 cal.

Initial temperature = T1 = 2270c = 227+273 = 500K

Final temperature = T2 = 1270c = 127 + 273 = 400K

As, for Carnot engine;

Q2 =

Q2 = 4.8 × 105 cal

Q2 = Final  heat emitted

As w = Q1 – Q2 = 6 × 105 – 4.8 × 105

= 1.2 × 105 cal

Work = w = 1.2 ×10× 4.2 J

Dore = 5.04 × 105 J

18.How does second law of thermodynamics explain expansion of gas?

Ans. Since from second law,.

d S ≥ O   d S = change in entropy

During the expansion of gas, the thermodynamic probability of gas is larger and hence its entropy is also very large. Since form second law, entropy cannot decrease ∴ following the second law, gas molecules move from one partition to another.

19.Why is it hotter at the same distance over the top of the fire than in front of it?

Ans. At a point in front of fire, heat is received due to the process of radiation only, while at a point above the fire, heat reaches both due to radiation and convection. Hence the result.

20. A metal rod of length 20cm and diameter 2cm is covered with a non-conducting substance. One of it ends is maintained at 1000c while the other is at 00c. It is found that 25g of ice melts in 5 min calculate coefficient of thermal conductivity of metal?

Ans.Length of rod = ∆x = 20cm = 2 × 10-3m

Diameter = 2cm

R = 10-2m

Area of cross-section = π r2

= π (10-2)2

= 10-4 π sq. m

∆ T = T2 – T1 = 100 – 0 = 1000c

Mass of ice melted = m = 25g

Latent heat office = 80 cal/g

Heat conducted, ∆ Q = mL

= 25 x 80

= 2000 cal

= 2000 × 4.2J

∆ t = 5 min = 300s

So,

K =  =

K = 1.78J |s|m|0c

K = coefficient of thermal conductivity

21.Calculate the temperature in Kelvin at which a perfectly black body radiates at the rate of 5.67 w/cm2?

Ans. E = 5.67w|cm2 ; E = energy radiated

= 5.67 x 10erg | s | cm2

= Stefan’s constant = 5.67 × 10-5 ergs |s | cm2| K4, from Stefan’s law

E = σ T4

T =

T =

22.How do you explain the emission of long – wavelength by the object at low temperature?

Ans.Since by Wein’s law: →

i.e temperature is inversely proportional to the wavelength so, if temperature is less, then wavelength will be long. If temperature is high, then wavelength will be short.

23.If the radiation from the moon gives maxima at  = 4700 A0 and = 14×10-6m. What conclusion can be drawn from the above information?

Ans. Acc. to wien’s displacement law,

Now, according to the question, m = 4700 A0 = 4700×10-10m

T1 = Temperature of moon,

T1 =

b = 2.9 ×10-3 mK

Let the temperature corresponding to

So, T2 =

T2 =

24.Differentiate between conduction, convection and radiation?

Ans.

 Properties Conduction Convection Radiation 1. Material Medium Essential Essential Not Essential 2. Molecules Do not leave their mean position More bodily from one place to another. Medium does not play any part 3. Transfer of heat Can be in any direction along any part Only vertically upward In all direction in straight lines 4. Speed of transfer of heat Slow Rapid Fastest with the speed of light.

25.The tile floor feels colder than the wooden floor even though both floor materials are at same temperature. Why?

Ans. This happens because the tile is better heat conductor than wood. The heat conducted from our foot to the wood is not conducted away rapidly. So, the wood quickly heats up on its surface to the temperature of our foot. But the tile conducts the heat away rapidly and thus can take more heat from our foot, so its surface temperature drops.

2 Marks Questions Part 3

52.A thermometer has wrong calibration. It reads the melting point of ice as – 100C. It reads 600C in place of 500C. What is the temperature of boiling point of water on the scale?

Ans.Lower fixed point on the wrong scale = -100C.

Let ‘n’ = no. divisions between upper and lower fixed points on this scale. If Q = reading on this scale, then

Now, C = Incorrect Reading = 600C

Q = Correct Reading = 500C

So,

n = 140

Now,

On, the Celsius scale, Boiling point of water is 1000C

So,

Q = 1300C

1) High accuracy of measurement

2) Measurements of temperature can be made over a wide range of temperature i.e. from – 2600C to 12000C.

→ Disadvantages of Platinum Resistance thermometer:-

1) High Cost

2) Requires additional equipment such as bridge circuit, Power supply etc.

54.If the volume of block of metal changes by 0.12% when it is heated through 200C. What is the co-efficient of linear expansion of the metal?

Ans.The co-efficient of cubical expansion y of the metal is given by:-

Here,

∴ Co-efficient of linear expansion of the metal is :-

55. The density of a solid at00C and 5000C is in the ratio 1.027 : 1. Find the co-efficient of linear expansion of the solid?

Ans .Density at 00C = SO

Density at 5000C = S500

Now, SO = S500

Where, Y = Co-efficient of volume expansion

∆T = Change in temperature

∆T = Change in temperature

∆T = Final Temperature – Initial temperature

∆T = 500 – 00C

∆T = 5000C

Or

Now, Co-efficient of linear expansion (α) is related to co-efficient of volume expansion (Y) as :-

56. If one Mole of a monatomic gas is mixed with 3 moles of a diatomic gas. What is the molecular specific heat of the mixture at constant volume?

Ans.For, a monatomic gas, Specific heat at consent volume = CV1 =  ; R = Universal Gas Constant

No. of moles of monatomic gas = n1 = 1 mole

No. of moles of diatomic gas = n2 = 3 moles.

For, diatomic gas, specific heat at constant volume

Applying, conservation of energy.

Let CV = Specific heat of the mixture;

R = Universal Gas constant

57. Calculate the difference between two principal specific heats of 1g of helium gas at N. T. P. Given Molecular weight of Helium = 4 and J = 4.186 J/cal and Universal Gas constant, R = 8.314J / mole / K?

Ans. Molecular weight of Helium = M = 4

Universal Gas Constant, R = 8.31J | mole | K

CP = specific heat at constant Pressure

CV = specific heat at constant Volume

Now,   for 1 mole of gas.

Where R = Universal Gas Constant = 8.31J | mole | K

J = 4.186 J | cal

M = Molecular weight of Helium = 4

58. Why does heat flow from a body at higher temperature to a body at lower temperature?

Ans.When a body at higher temperature is in contact with a body at lower temperature, molecule with more kinetic energy that are in contact with less energetic molecules give up some of their kinetic energy to the less energetic ones.

59. A one liter flask contains some mercury. IT is found that at different temperatures, then volume of air inside the flask remains the same. What is the volume of mercury in the flask? Given the co-efficient of linear expansion of glass = 9 × 10-6 / 0C and co-efficient of volume expansion of mercury = 1.8 × 10-4 / 0C

Ans.It is given that volume of air in the flask remains the same at different temperature. This is possible only when the expansion of glass is exactly equal to the expansion of mercury,

Co-efficient of cubical expansion of glass is :-

Co-efficient of cubical expansion of mercury is :→

Volume of flask, V = 1 liter = 1000 cm3.

Let Vm Cm3 be the volume of mercury in the flask.

Expansion of flask = Expansion of Mercury

∴ Volume of Mercury,

60. A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C, calculate the coefficient of performance.

Ans. Temperature inside the refrigerator, = 9°C = 282 K

Room temperature, = 36°C = 309 K

Coefficient of performance =

= 10.44

Therefore, the coefficient of performance of the given refrigerator is 10.44.

61. A steam engine delivers of work per minute and services of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

Ans. Work done by the steam engine per minute, W =

Heat supplied from the boiler, H =

Efficiency of the engine =

Hence, the percentage efficiency of the engine is 15 %.

Amount of heat wasted =

Therefore, the amount of heat wasted per minute is.

62. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13)

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F

Ans.Total work done by the gas from D to E to F = Area of ΔDEF

Area of ΔDEF =

Where,

DF = Change in pressure

= 300 N/

FE = Change in volume

= 3.0 m3

Area of ΔDEF = = 450 J

Therefore, the total work done by the gas from D to E to F is 450 J.

3 Marks Questions

1.Calculate the work done during the isothermal Process?

Ans.Let an ideal gas is allowed to expand very slowly at constant temperature. Let the expands from state A (P1, V1) to state B (P2, V2)

The work by the gas in expanding from state A to B is

For ideal gas, PV = N R T

or P =

Use 2) in i)

W =

Since n, R and T are constant so,

W =

W isothermal = nRT Loge V

W isothermal =

W isothermal – nRT Loge

W isothermal = 2.303 nRT Log 10

If M = Molecular Mass of gas then for 1 gram of ideal gas,

W isothermal = 2.303

W isothermal 2.303 r T Log 10

r = Gas constant for 1 gm of an ideal gas,

Since P1 V= P2 V2

So W isothermal = 2.303 r T log 10

2.Five moles of an ideal gas are taken in a Carnot engine working between 1000C and 300C. The useful work done in 1 cycle is 420J. Calculate the ratio of the volume of the gas at the end and beginning of the isothermal expansion?

Ans.High temperature, TH = 1000C = 100+ 273 = 373K

Low temperature, TL = 300C, = 30 +273 = 303K

Amount of the gas, n = 5 moles

Useful work done per cycle, W = QH – QL

Now, W = 420 J

So, QH – QL = 420J → 1)

Now,

Or QH =

QL = 1818J

or, QH-Q= 420J

QH – 1818 = 420J

QH = 420 + 1818 = 2238J

When the gas is carried through Carnot cycle, the heat absorbed QH during isothermal expansion is equal to the work done by gas.

V1 – Initial Volume

V2 = Final Volume,

In isothermal expansion,

QH = 2.303 nRTH Log 10

22 38 = 2.303 × 5 × 8.4 × 373 Log 10

Log 10

Log 10

3.Deduce the work done in the following complete cycle?

Ans.1) Work done during the process from A to B = WAB

WAB = area ABKLA (∴ because area under p-v curve gives work done)

= area of ∆ ABC + area of rectangle

BC = KL = 4-1 = 3l = 3×10-3m3

AC = 4-2 = 2N|m2

LC = 2-0 = 2N|m2

WAB

= 3×10-3+6×10-3

WAB = 9×10-3J

Since gas expands during this process, hence WAB = 9×10-3J

2) Work done during the process from B to C(compression) is WBC = -area BCLK

( – ve because gas compresses during BC)

= – KL × LC

WBC = -3×10-3×2

= – 6×10-3J

3) Work done during the process from C to A :-

As there is no change in volume of gas in this process, WCA = O

So, net work done during the complete cycle = WAB + WBC +WCA

= 9×10-3-6×10-3+0

Net work done = 3×10-3J

4.One kilogram molecule of a gas at 400k expands isothermally until its volume is doubled. Find the amount of work done and heat produced?

Ans.Initial volume, V1= V

Final volume, V2= 2V

Initial temperature T = 400k

Find temperature = 400k (∴ process is isothermal)

Gas constant, R = 8. 3kJ |mole |k=8.3×10-3J|mole|k

Work done during is thermal process=w=2.3026RT Log 10

W = 2.3026×8.3×10-3×400×log 10

W = 2.3026×8.3×10-3×400×Log 10 (2)

W = 2.3016J

If H is the amount of heat produced than,

5.Calculate difference in efficiency of a Carnot energy working between:-

1) 400K and 350K

2) 350K and 300K

Ans.Efficiency of heat engine = n = 1 –

T2 = final temperature

T1 = Initial temperature

1) 400K and 350K,

T2 = 350, T1 = 400

n = 1 –

n1 =

2) 350K and 300K

T2 = 300K; T1 = 350K

n1 = 1 –

= 1 –

n1 =

Change in efficiency = n2 – n1 = 14.3% – 12.5% = 1.8%

6.How do you derive Newton’s law of cooling from Stefan’s law?

Ans.Acc. to Newton’s law of cooling, the rate of loss of heat of a liquid is directly proportional to the difference in temperature of the liquid and the surrounding, provided the difference in temperature is very small.

Let a body be maintained at T K. Let To be the temperature of the surroundings. Let T ≫ To. There will be loss of heat be the body

Acc. to Stefan’s law, amount of heat energy lost per second per unit area of the body is

E = ϵ σ

σ = Stefan’s constant

ε = Emissivity of the body and surroundings

E =

Incase of Newton’s cooling, T ≈ To

E = ε σ

E = ε σ

Hence, → Hence the Newton’s law of cooling

E α

7.Define the terms reflectance, absorptance and transmittance. How are they related?

Ans.1) Reflectance – Ratio of amount of thermal radiations reflected by the body in a given time to total amount of thermal radiations incident on body It is represented by r, 2) Absorptance – is the ratio of the amount of thermal to the total amount of thermal radiations incident on it. It is represented by a

3) Transmittance – It is the ratio of the amount of thermal radiations transmitted by body in a given time to the total amount of thermal radiations incident on the body in a given time. It is represented by t.

Let Q = Amount of the radiations incident by the body in a given time

Q1 = Amount of thermal radiations reflected by the body in a given time.

Q= Amount of thermal radiations absorbed by the body in a given time.

Q3 = Amount of thermal radiations transmitted by the body in a given time,

∴ By definition,

New, r + a + t =

R + a + t =

R + a + t = 1

If t = 0

A = 1 – r

that is  good reflectors are bad absorbers

8. If half mole of helium is contained in a container at S. T. P. How much heat energy is needed to double the pressure of the gas, keeping the volume of the gas constant? Given specific heat of gas = 3J | g |K.

Ans.Number of moles of Helium gas =

Specific heat of Helium gas =

Molecular weight = M = 4

Temperature, T1 = 273 K.

∴ Molar specific heat at constant volume = CV = MCV

CV = 4 × 3

CV = 12 J | mol | K

Since, Volume is constant, Pα T or = Constant

∴

Or

P2 = Final Pressure = 2 P

P1 = Initial Pressure = P

Now, Heat required,

Heat required = 1638 J

9. Calculate the amount of heat necessary to raise the temperature of 2 moles of HE gas from 200C to 500C using:-

1) Constant – Volume Process   2) Constant Pressure Process

Here for, He; CV = 1.5 R and CP = 2.49R

Ans .1) The amount of heat required for constant – volume process is :-

Here, n = 2 moles, CV = 1.5 R = 1.5 X 8.314 J | mol | 0C

T2 = final Temperature

T1 = Initial Temperature

2) The amount of heat required for constant – Pressure process is :-

Here, n = 2 moles,

Since the temperature rise is same in both the cases, the change in internal energy is the same i.e. 748J. However, in constant – pressure Process excess heat is supplied which is used in the expansion of gas.

10.  An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing?

Ans.Heat is supplied to the system at a rate of 100 W.

∴Heat supplied, Q = 100 J/s

The system performs at a rate of 75 J/s.

∴Work done, W = 75 J/s

From the first law of thermodynamics, we have:

Q = U + W

Where,

U = Internal energy

U = Q–W

= 100 –75

= 25 J/s

= 25 W

Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

4 Marks Questions

1. A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is  J/g?

Ans.Water is flowing at a rate of 3.0 litre/min.

The geyser heats the water, raising the temperature from 27°C to 77°C.

Initial temperature, = 27°C

Final temperature, = 77°C

∴Rise in temperature, ΔT =

= 77–27= 50°C

Heat of combustion =

Specific heat of water, c = 4.2

Mass of flowing water, m = 3.0 litre/min = 3000 g/min

Total heat used, ΔQ = mc ΔT

Rate of consumption = = 15.75 g/min

2.  What amount of heat must be supplied to  kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of.)

Ans. Mass of nitrogen,

Rise in temperature, ΔT = 45°C

Molecular mass ofM = 28

Universal gas constant, R =

Number of moles,

Molar specific heat at constant pressure for nitrogen,

The total amount of heat to be supplied is given by the relation:

= 933.38 J

Therefore, the amount of heat to be supplied is 933.38 J.

3.  Explain why

(a) Two bodies at different temperatures and if brought in thermal contact do not necessarily settle to the mean temperature.

(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.

(c) Air pressure in a car tyre increases during driving.

(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Ans.(a) When two bodies at different temperatures  and  are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature only when the thermal capacities of both the bodies are equal.

(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.

(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.

(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

4.  A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Ans.The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.

Initial pressure inside the cylinder =

Final pressure inside the cylinder =

Initial volume inside the cylinder =

Final volume inside the cylinder =

Ratio of specific heats, Y = 1.4

For an adiabatic process, we have:

The final volume is compressed to half of its initial volume.

Hence, the pressure increases by a factor of 2.639.

5. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

Ans.The work done (W) on the system while the gas changes from state A to state B is 22.3 J.

This is an adiabatic process. Hence, change in heat is zero.

∴ ΔQ = 0

ΔW = –22.3 J (Since the work is done on the system)

From the first law of thermodynamics, we have:

ΔQ = ΔU + ΔW

Where,

ΔU = Change in the internal energy of the gas

∴ ΔU = ΔQ– ΔW = –(–22.3 J)

ΔU = + 22.3 J

When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

ΔQ = 9.35 cal = 9.35  4.19 = 39.1765 J

Heat absorbed, ΔQ = ΔU + ΔQ

∴ΔW = ΔQ – ΔU

= 39.1765 – 22.3

= 16.8765 J

Therefore, 16.88 J of work is done by the system.

6.  Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:

(a) What is the final pressure of the gas in A and B?

(b) What is the change in internal energy of the gas?

(c) What is the change in the temperature of the gas?

(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

Ans.(a) 0.5 atm

(b) Zero

(c) Zero

(d) No

Explanation:

(a) The volume available to the gas is doubled as soon as the stopcock between cylinders A and B is opened. Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5 atm.

(b) The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change.

(c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.

(d) The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non-equilibrium states, they do not lie on the PVT surface of the system.

5 Marks Questions

1.Derive the equation of state for adiabatic change?

Ans.Let P = pressure, V = volume and T = Temperature of the gas in a cylinder fitted with a perfectly

frictionless piston.

Suppose a small amount of heat d Q is given to the system. The heat is spent in two ways:-

1) In increasing the temperature of the gas by la small range d T, at constant volume

2) In expansion of gas by a small volume d v

So, d Q = CV d T + P d V

In adiabatic change, no heat is supplied from outside

So, d Q = O

CV d t + P d V = O  →(1)

Acc. to standard gas equation

PV = RT

Diff both sides

P d V + V d P = R d R

R d T = P d V +V d P (d R=O as  R is a constant)

d T =

Using this in equation i)

C v

CV P d V + CV V d P + R P d V = O

(C+ R) P d V + CV d P = O →2)

As, CP – CV = R

or CP = R + CV

So equation 2) becomes

CP P d V + CV V d P = O

Dividing above equation by CV PV

Integrating both sides

Loge V + Loge P = constant

Loge + Loge P = constant

Loge P  = constant

= antilog (constant)

K = another constant

2.Derive an expression for the work done during isothermal expansion?

Ans.Consider one gram mole of ideal gas initially with pressure, volume and temperature as P, V, T, Let the gas expand to a volume V2, when pressure reduces to P2 and at the same temperature T

If A = Area of cross – section of piston

Force = Pressure × Area

F = P x A

If we assume that piston moves a displacement d x,

the work done : → d w = F d x

d w = P × A × d x

d w = P × d v

Total work done in increasing the volume from Vto V2

W =

Since, PV = RT (from ideal gas equation)

P =

W =

W = RT

W = R T Loge

W = R T

W = R T Loge

W = 2.3026 R T Log 10

As P1 V1 = P2 V2

So W = 2.3026 R T Log 10

3.Briefly describe a Carnot cycle and derive an expression for the efficiency of Carnot cycle?

Ans.The construction of a heat engine following Carnot  cycle is :-

1) Source of heat :- It is maintained at higher temperature T1

2) Sink of heat – It is maintained at lower temperature T2

3) Working base :- A perfect ideal gas is the working substance.

Theory :- Carnot cycle consist of four stages:-

1) I so thermal expansion

3) I so thermal compression

4. Discuss briefly energy distribution of a black body radiation. Hence deduce wien’s displacement law?

Ans.For a black body, the monochromatic emittance  of the black body and the wavelength  of the radiation emitted.

So, at a given temperature of black body :→

a) The energy emitted is not distributed uniformly amongst all wavelengths.

b) The energy emitted in maximum corresponding to a certain wavelength and its falls on either side of it.

As temperature of black body is increased.

a) The total energy emitted rapidly increases for any given wavelength.

b) The wavelength corresponding to which energy emitted is maximum is shifted towards shorter wavelength side i. e,  m decreases with rise in temperature

or   m T = constant

Thus is the wein’s displacement law.

NUMERICALS

1. When a system is taken from state A to state B along the path ACB, 80 k cal of heat flows into the system and 30 kcal of work is done.
1. How much heat flows into the system along path ADB if the work done is 10 k Cal?
2. When the system is returned from B to A along the curved path the Work done is 20 kcal. Does the system absorb or librate heat.
3. If UA = 0 and UD = 40 kcal, find the heat absorbed in the process AD

Ans.

1. dWADB = +10 k cal
Internal energy is path independent
dQADB = duACB = 50 k cal
dQADB = 50 + 10 = 60 k cal.

2. = -50 – 20 = -70 k cal
3. UA = 0 UD = 40 k cal

dWDB = 0 since dV = 0
dQAD = 40 + 10 = 50 k cal
2. mole of helium is contained in a container at S.T.P. How much heat energy is needed to double the pressure of the gas, keeping the volume Constant? Heat capacity of gas is 3 J g-1 K-1.
Ans.
CV = MCV = 12 J/mole k         M → Molecular mass

3. The volume of steam produced by 1 g of water at 100°C is 1650 cm3. Calculate the change in internal energy during the change of state given J = 4.2 x 107 erg cal-1 g = 98 J cm/s2?
latent heat of steam = 540 Cal/g
Ans. Mass of water = 1g = 10-3 kg
Volume of water
= 1 cm3
Change in volume = 1650 – 1 = 1649 cm3
dQ = M L = 540 cal = 540 × 4.2 × 107 erg
P = 1 atm = 76 × 13.6 × 981
Du = dQ – pdv = 22.68 × 109 – 1.67 × 109
= 21.01 × 109 erg.
4. What is the coefficient of performance () of a carnot refrigerator working between 30°C and O°C?
Ans.
5. Calculate the fall in temperature when a gas initially at 72°C is expanded suddenly to eight times its original volume. (= 5.3)
Ans.

6. A steam engine intake steam at 200°C and after doing work exhausts it directly in air at 100°C calculate the percentage of heat used for doing work. Assume the engine to be an ideal engine?
Ans.

7. A perfect Carnot engine utilizes an ideal gas the source temperature is 500K and sink temperature is 375K. If the engine takes 600kcal per cycle from the Source, Calculate
1. The efficiency of engine
2. Work done per cycle
3. Heat rejected to sink per cycle

Ans.

1. T1 = 500 K T2 = 375 K
Q1 = Heat absorbed = 600 k cal

= 25%

2. w =Q1 – Q2 Q2 = Q1 – W = 600 – 150
= 450 k cal
8. Two carnot engines A and B are operated in series. The first one A receives heat at 900 K and reject to a reservoir at temperature T K. The second engine B receives the heat rejected by the first engine and in turn rejects to a heat reservoir at 400 K calculate the temperature T when
1. The efficiencies of the two engines are equal
2. The work output of the two engines are equal

Ans. WA = WB

T = 650 K

T2 = 900 × 400
= 600 K
T1 = 273 K T2 = 673 K
Mass of gas = 10 mole

= -8.4 × 104 J work being done on the gas
Du = -dw = 8.4 × 104 J

9. Ten mole of hydrogen at NTP is compressed adiabatically so that its temperature become 400°C How much work is done on the gas? what is the increase in the internal energy of the gas
R = 8, 4 J mol-1 K-1  = 1.4
10. The temperature T1 and T2 of the two heat reservoirs in an ideal carnot engine be 1500°C and 500°C respectively, which of these increasing T1 by 100°C or decreasing T2 by 100°C would result in a greater improvement in the efficiency of the engine.
Ans.
1. T1 is increased from 1500°C to 1600°C
T1 = 1873 K
T2 remain constant T2 = 773 k
2. T1 remain constant 1500°C
T1 = 1500 + 273 = 1773 k
T2 is decreased by 100 i.e. 400°C
T2 = 400 + 273 = 673 k

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