# Chapter Notes: Thermodynamics Physics Class 11

Notes for Thermodynamics chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.

Thermodynamics is concerned with the work done by a system and the heat it exchanges with its surroundings. We are concerned only with work done by a system on its surroundings or on the system by the surroundings. We are not concerned with internal work done by one part of a system on another.

### Heat and Work

** Heat **is the energy transferred between two bodies as a consequence of a temperature difference between them. In

*contrast*,

**is a mode of energy transfer in which the point of application of a force moves through a displacement and is**

*work**not*associated with a temperature difference.

Both *heat *and *work *are “*energy in transit*” from one body to another during the operation of some process, once the process stops, heat and work have no meaning.

**Mechanical Equivalent of Heat**

It has been concluded from Joule’s experiment that the mechanical work required to produce a given change in temperature is in fixed proportion to the heat required for same change in temperature. This constant factor is called the ** mechanical equivalent **of heat.

1 calorie = 4.186 J

Thus, a change in the state of a system produced by the addition of 1 calorie of heat may also be produced by the performance of 4.186 J of work on the system.

**Specific Heat and Heat Capacity**

If a quantity of heat *Q *produces a change in temperature *D**T* in a body, its heat capacity is defined as

Heat capacity

The SI unit of heat capacity is JK^{-1}.

The quantity of heat *Q *required to produce a change in temperature *DT *is also proportional to the mass *m *of the sample.

where *C *is called the specific heat of the substance.

** Specific heat **may be defined as the

*heat capacity*per

*unit mass*.

It is sometimes convenient, especially in the case with gases, to deal with the number of moles

*n*of a substance rather than its mass. Then,

where

*C*is the molar specific heat, measured in J/mol-K (or cal/mol-K)

_{m}*C*

_{m}= McThe specific heat of a substance usually *varies *with *temperature*.

The specific heat changes abruptly when the substance transforms from solid to liquid, or from liquid to gas. It also depends on the conditions under which the heat is supplied. For example, the specific heat of a gas kept at constant pressure *C _{p}* is different from its specific heat at constant volume

*C*. For air,

_{v}*C*= 0.17 cal/g-K and

_{v}*C*= 0.24 cal/g-K For solids and liquids the difference is generally small, and in practice

_{p}*C*is usually measured.

_{p}*Example 1*

*In an industrial process 10 kg of water per hour is to be heated from 20 ^{o}C to 80^{o}C. To do this, steam at 150^{o}C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90^{o}C. How many kg of steam is required per hour?*

*[Specific heat of steam = 1 cal/g*^{o}C, and latent heat of steam = 540 cal/g]*Solution*

Heat required by 10 kg water to increase its temperature from 20 to 80^{o}C in one hour is given by

If *m *gram of steam is condensed per hour, the heat released by steam in converting into water at 90^{o}C

[Q *c _{s} *=

*c*= 1 cal / g

_{w}^{o}C]

According to given problem

*Q*

_{2}=

*Q*

_{1 }

*Example 2*

*Ice at 0 ^{o}C is added to 200 g of water initially at 70^{o}C in a vacuum flask. When 50 g of ice has been added and has all melted, the temperature of the flask and contents is 40^{o}C. When a further 80 g of ice has been added and has all melted, the temperature of the whole becomes 10^{o}C. Neglecting heat lost to the surroundings, calculate the latent heat of fusion of ice? (Specific heat of water is 1 cal/g ^{o}C) and water equivalent of flask.*

*Solution*

If *L* is the latent heat of ice and *W* is the water equivalent of flask, according to principle of calorimetry

i.e. heat gained = heat lost

*i.e. *

*i.e. *5*L* = 3*W *+ 400 (i)

Now the system contains (200 + 50) g of water at 40^{o}C so when further 80 g of ice is needed:

*i.e. *8*L* = 3*W *+ 670 (ii)

Solving equations (i) and (ii)

*L* = 90 cal/g and *W *= (50/3) g

**Thermodynamic Work**

Figure shows a gas confined to a cylinder by a weight on a movable piston. Our *system *is the *gas*, whereas the *cylinder *and the *piston *form the *environment*. If the piston is allowed to move upward, the gas expands and does work on it. To calculate the work done by the gas, we assume that the *process *is ** quasistatic**. In a

**the thermodynamics variables (**

*quasistatic process**P, V, T, n,*etc.) of the system and its surroundings change infinitely slowly. Thus, the system is always arbitrarily close to an equilibrium state, in which it has a well-defined volume, and the whole system is characterized by single value of the macroscopic variables. To ensure that the piston moves very slowly, there must be some force, for example, provided by a weight, directed opposite to that due to the pressure. If the piston were to move suddenly, the rapid expansion would involve turbulence and the pressure would not be uniquely defined.

When the piston rises by *dx*, the work *dW *done by the gas is *dW *= *F dx *= *(PA) dx* where *A *is the cross-sectional area of the piston. Since the change in volume of the gas is *dV *= *A dx*, the work may be expressed as

(Quasistatic) * dW = P dV *

As a quasistatic process evolves, *P* and *V *are always *uniquely *defined. This allows us to depict the process on a *PV *diagram such as figure. When the system is taken quasistatically from the equilibrium state *i* to another equilibrium state *f*, the total work done by the system is

In figure the work is represented by the *area under the curve*. If *V _{f} *>

*V*, the work done by the gas is

_{i}**. If the volume**

*positive**decreases*, the work done by the gas is

**. This may be interpreted as positive work done on the gas by the environment. The work done**

*negative**depends*not only on the

*initial*and

*final states*but also on the details of the

*process*, that is, the thermodynamic

*path*between the states. Therefore, we need to know how the pressure varies with the volume.

**FIRST LAW OF THERMODYNAMICS**** **

Consider a system that consists of a gas enclosed by a piston in a cylinder. Suppose the system is taken quasistatically from an initial state *P _{i}, V_{i}, T_{i} *to a final state

*P*. At each step the work done and heat exchanged are measured. We know that both the total work done

_{f}, V_{f}, T_{f}*W*and the total heat transfer

*Q*

*to*or

*from*the system depend on the thermodynamic

*path*. However, the difference

*Q – W*, is the same for

*all*paths between the given

*initial*and

*final*equilibrium states, and it is equal to the change in internal energy

*D*

*U*of the system.

In the above statement, *Q* is ** positive **when heat enters the system and

*W*is

**when**

*positive***is done by the system on its surroundings.**

*work*The above equation is the mathematical statement of the

**. It states that**

*first law of thermodynamics**the internal energy of a system changes when work is done on the system (or by it), and when it exchanges heat with the environment*.

**Note **that the first law is valid for all processes *quasistatic *or *not*. However, if friction is present, or the process is not quasistatic, the internal energy *U* is uniquely defined only at the *initial *and *final *equilibrium states.

The first law establishes the existence of internal energy *U* as a *state function *– one that depends only on the thermodynamic state of the system.

In the macroscopic approach of thermodynamics, there is no need to specify the physical nature of the internal energy. The experimental results are sufficient to proof that such a function exists. The internal energy is the *sum *of all possible kinds of energies stored in the system – *mechanical*, *electrical*, *magnetic*, *chemical*, *nuclear*, and so on. It does *not *include the kinetic and potential energies associated with the *centre of mass *of the system.

### Misconception between Heat and Internal Energy

Confusion between *heat *and *internal energy *arises from erroneous statements that refer to the “*heat content” *of a body. Even correct terms like “*the heat capacity of a body” *can mislead one to believe that heat is somehow stored in a system. This is *not *correct.

The physical quantity possessed by a system is ** internal energy**, which is the sum of all the kind of energy in the system. As the first law indicates,

*U*may be changed either by heat exchange or

*by work*. The internal energy is a state function that depends on the equilibrium state of a system, whereas Q and W depend on the thermodynamic path between two equilibrium states. That is,

*Q*and

*W*are associated with processes. The heat absorbed by a system will increase its internal energy, only some of which is average translatory kinetic energy. It is therefore incorrect to say that heat is the energy of the random motion.

**Thermodynamic Processes**

We now apply the first law of thermodynamics to some simple situations.

**(a) Isolated System
**Consider first an

*isolated system*for which there is no

*heat exchange*and

*no work is*

*done*on the external environment. In this case

*Q*= 0 and

*W*= 0, so from the

*first law*we conclude

and U=constant

*The internal energy of an isolated system is constant*

**(b) Isochoric Process
**In case of an isochoric process volume of the system remains constant

i.e.

*V*= constant

or = constant

Since the boundary of the system does not displace because volume is constant, therefore,

*W*= 0

The change in internal energy is given by

Using first law

Here

*V*

_{f}= V_{i}**(c) Isobaric Process
**In an isobaric process pressure of the system remains constant

*i.e.*

*p*= constant.

The work done is given by

or

*W*=

*P*(

_{o}*V*)

_{f}– V_{i}Using gas equation

*PV = nRT*

*We get,*

*W*=

*nR*(

*T*)

_{f}– T_{i}Since the change in internal energy is independent of the path followed, therefore

Using first law of thermodynamics,

*By defination*

* Important
*1.

2.

*Example 3*

*If 70 calorie of heat is required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30 to 35 ^{o}C, calculate
*

*(a) the work done by the gas*

*(b) increase in internal energy of the gas and*

*[R = 2 cal/mol K]**Solution *

(a) At constant pressure

(b)

so

*Example 4*

*A cylinder with a piston contains 0.2 kg of water at 100 ^{o}C. What is the change in internal energy of the water when it is converted to steam at 100^{o}C at a constant pressure of 1 atm? The density of water is r_{o} = 10^{3} kg/m^{3} and that of steam is*

r_{s} = 0.6 kg/m^{3}. The latent heat of vaporization of water is L_{v} = 2.26 10^{6} J/kg

*Solution*

The heat transfer to the water is

The work done by the water when it expands against the piston at constant pressure is

The change in internal energy is

**(d) Isothermal Process
**In an isothermal process, temperature of the system remains constant. For an ideal gas the equation of the process is given by

*PV = nRT*= constant

Work done in an isothermal process is given by

or

**(e) Adiabatic Process
**In an adiabatic process, the system does not exchange heat with the surroundings, i.e.

*Q*= 0.

For an ideal gas the equation of the adiabatic process is

*PV*

*= constant*

^{g}where g is the adiabatic exponent.

work done=

Change in internal energy:

By definition,

*Q*= 0

Therefore, using first law

Work done by the system is equal to the decrease in internal energy.

or

Work done on the system is equal to the increase in internal energy

*Example 5*

*Three moles of an ideal gas at 300 K are isothermally expanded to five times its volume and heated at this constant volume so that the pressure is raised to its initial value before expansion. In the whole process 83.14 kJ heat is required. Calculate the ratio (C _{P}/C_{V}) of the gas.*

*[log*_{e}5 = 1.61 and R = 8.31 J/mol K^{-1}].*Solution*

According to *first law of thermodynamics*,

For an *isothermal change,
*

*T*= constant,

*U*= constant,

*= 0*

and

*i.e.*

= 12.03 kJ

For

Applying gas equation between points

so that

and hence, (D

= 12.03 kJ

*Q*= 0 + 12.03 = 12.03 kJ_{isothermal }For

*isochoric change*as*V*= constant,Applying gas equation between points

*A*and*C*,*i.e*.*T*= 1500 K_{C}so that

*Q*)

_{isochoric}= 3.6

*C*+ 0 = 3.6

_{v}*C*kJ (ii)

_{v}According to given problem,

Using equation (i) and (ii), we get

12.03 + 3.6

*C*= 83.14

_{v}or

*C*= (71.11/3.6) = 19.75 J

_{v}Thus

*C*=

_{P}*C*= 19.75 + 8.3 = 28.05 J/mol - K

_{V}+ R
No discussion about Cp and CV

Good notes

nice notes got full marks in the class test