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, work is a mode of energy transfer in which the point of application of a force moves through a displacement and is 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 DT in a body, its heat capacity is defined as
Heat capacity  $C = {Q \over {\Delta T}}$
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.
$Q{\rm{ }} = {\rm{ }}mC\Delta T$
where C is called the specific heat of the substance.
Specific heat may be defined as the heat capacity per unit mass.
$C = {{Heat\,\,\,capacity} \over {mass}}$
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,
$Q = n{C_m}\Delta T$
where Cm is the molar specific heat, measured in J/mol-K  (or cal/mol-K)
Cm = Mc

The 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 Cp is different from its specific heat at constant volume Cv. For air, Cv = 0.17 cal/g-K and Cp = 0.24 cal/g-K For solids and liquids the difference is generally small, and in practice Cp is usually measured.

Example 1

In an industrial process 10 kg of water per hour is to be heated from 20oC to 80oC. To do this, steam at 150oC 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 90oC. How many kg of steam is required per hour?
[Specific heat of steam = 1 cal/g oC, and latent heat of steam = 540 cal/g]

Solution

Heat required by 10 kg water to increase its temperature from 20 to 80oC in one hour is given by
${Q_1} = {\rm{ }}{[mc\Delta T]_{water}} = (10 \times {10^3})\left( 1 \right)\left( {80-20} \right) = 600kcal$
If m gram of steam is condensed per hour, the heat released by steam in converting into water at 90oC
${Q_2} = \;m{c_s}\left( {150{\rm{ }}-{\rm{ }}100} \right) + m{L_v} + m{c_w}\left( {100-90} \right)$
${Q_2} = m[1 \times 50{\rm{ }} + {\rm{ }}540{\rm{ }} + {\rm{ }}1 \times 10] = 600{m_{}}cal$
[Q   cs = cw = 1 cal / g oC]
According to given problem  Q2 = Q1
$600m = {\rm{ }}600 \times {10^3}$

$m = {\rm{ }}1 \times {10^3}g{\rm{ }} = {\rm{ }}1{\rm{ }}kg$

Example 2

Ice at 0oC is added to 200 g of water initially at 70oC in a vacuum flask. When 50 g of ice has been added and has all melted, the temperature of the flask and contents is 40oC. When a further 80 g of ice has been added and has all melted, the temperature of the whole becomes 10oC. Neglecting heat lost to the surroundings, calculate the latent heat of fusion of ice? (Specific heat of water is 1 cal/g oC) 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
${m'}L{\rm{ }} + {\rm{ }}{m'}C\Delta T{\rm{ }} = {\rm{ }}\left( {m{\rm{ }} + {\rm{ }}W} \right)C\Delta {T_{water}}$
i.e. $50[L + {\rm{ }}1 \times \left( {40{\rm{ }}-{\rm{ }}0} \right)] = {\rm{ }}\left( {200 + W} \right) \times 1 \times \left( {70{\rm{ }}-{\rm{ }}40} \right)$
i.e.  5L = 3W + 400                                                            (i)
Now the system contains (200 + 50) g of water at 40oC so when further 80 g of ice is needed:
$80[L + {\rm{ }}1 \times \left( {10{\rm{ }}-{\rm{ }}0} \right)] = \left( {250 + W} \right) \times 1 \times (40{\rm{ }}-{\rm{ }}10)$
i.e. 8L = 3W + 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 quasistatic process the thermodynamics variables (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

$W = \int\limits_{{V_i}}^{{V_f}} {P\,\,\,dV}$

In figure the work is represented by the area under the curve. If Vf > Vi, the work done by the gas is positive. If the volume decreases, the work done by the gas is negative. This may be interpreted as positive work done on the gas by the environment. The work done 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 Pi, Vi, Ti to a final state Pf, Vf, Tf. At each step the work done and heat exchanged are measured. We know that both the total work done 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 DU of the system.

$\Delta U = Q - W$

In  the above statement, Q is positive when heat enters the system and W is positive when work is done by the system on its surroundings.
The above equation is the mathematical statement of the first law of thermodynamics. It states that 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 = 0, so from the first law we conclude
$\Delta U\; = {\rm{ }}0\;$ 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  ${P \over T}$ = 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
$\Delta U = n{C_0}\Delta T = {{nR} \over {\gamma �- 1}}\Delta T$
Using first law
$Q{\rm{ }} = {\rm{ }}W{\rm{ }} + \Delta U$
$Q = \Delta U = {{nR} \over {\gamma �- 1}}\Delta T = {{{P_f}{V_f} - {P_i}{V_i}} \over {\gamma �- 1}}$
Here Vf = Vi

(c)  Isobaric Process
In an isobaric process pressure of the system remains constant i.e.
p = constant.
The work done is given by
$W = \int {PdV} = {P_o}\int\limits_{{V_i}}^{{V_f}} {dV}$
or   W Po(Vf – Vi)
Using gas equation  PV = nRT
We get, =nR(Tf – Ti)

Since the change in internal energy is independent of the path followed, therefore
$\Delta U = n{C_0}\Delta T = {{nR} \over{\gamma - 1}}\Delta T = {{nR} \over {\gamma - 1}}\left( {{T_f} - {T_i}} \right)$
Using first law of thermodynamics,
$Q{\rm{ }} = {\rm{ }}W{\rm{ }} + \Delta U$
$Q = nR\left( {{T_f} - {T_i}} \right) + {{nR} \over {\gamma �- 1}}\left( {{T_f} - {T_i}} \right)$
By defination $Q = n{C_p}\Delta T = n{C_p}\left( {{T_f}-{\rm{ }}{T_i}} \right)$
${C_p} = {{\gamma R} \over {\gamma - 1}}$

Important
1.${C_p}-{C_v} = R\;$
2.${{{C_p}} \over {{C_v}}} = \gamma$

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 35oC, 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 $W = P\Delta V = nR\Delta T\;\;\;\left( {as\;\;PV = nRT} \right)$
$W = nR\Delta T = 2 \times 2 \times \left( {35{\rm{ }}-{\rm{ }}30} \right) = {\rm{ }}20{\rm{ }}cal$
(b) $\Delta U = {Q_v} = {Q_p} - W$
so  $\Delta U = 70-20{\rm{ }} = 50{\rm{ }}cal$

Example 4

A cylinder with a piston contains 0.2 kg of water at 100oC. What is the change in internal energy of the water when it is converted to steam at 100oC at a constant pressure of 1 atm? The density of water is ro = 103 kg/m3 and that of steam is
rs = 0.6 kg/m3. The latent heat of vaporization of water is Lv = 2.26 $\times$106 J/kg

Solution

The heat transfer to the water is
$Q\; = \;m{L_v}\; = {\rm{ }}\left( {0.2{\rm{ }}kg} \right){\rm{ }}(2.26 \times {10^6}\;J/kg){\rm{ }} = {\rm{ }}4.52 \times {10^5}\;J$
The work done by the water when it expands against the piston at constant pressure is
$W = P({V_s} - {V_w}) = P\left( {{m \over {{\rho _s}}} - {m \over {{\rho _w}}}} \right)$
$(1.01 \times {10^5}N/{m^2})\left( {{{0.2\,\,kg} \over {0.6\,kg/{m^3}}} - {{0.2\,\,kg} \over {1000\,kg/{m^3}}}} \right)$
$= {\rm{ }}3.36 \times {10^4}J$
The change in internal energy is
$\Delta U = Q-W = 452{\rm{ }}kJ - 33.6{\rm{ }}kJ{\rm{ }} = {\rm{ }}418.4{\rm{ }}kJ$

(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
$W = \int\limits_{{V_i}}^{{V_f}} {PdV} = nRT\int\limits_{{V_i}}^{{V_f}} {{{dV} \over V}}$
or  $W = nRT\ln \left| {{{{V_f}} \over {{V_i}}}} \right|$

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
PVg = constant
where g is the adiabatic exponent.
work done=$W = \int\limits_{{V_i}}^{{V_f}} {PdV}$
$W = {{{P_f}{V_f} - {P_i}{V_i}} \over {1 - \gamma }} = {{nR\left( {{T_f} - {T_i}} \right)} \over {1 - \gamma }}$
Change in internal energy:
$\Delta U = n{C_0}\Delta T = {{nR} \over {\gamma - 1}}\Delta T$
By definition,   Q = 0
Therefore, using first law $Q = W + \Delta U\;\; \Rightarrow \;0 = W + \Delta U$
$W = - \Delta U\;$
Work done by the system is equal to the decrease in internal energy.
or  $- W = \Delta U$
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 (CP/CV) of the gas.
[loge5 = 1.61 and R = 8.31 J/mol K-1].

Solution

According to first law of thermodynamics,
$Q = \Delta U + W$
For an isothermal change,
T = constant, U = constant, $\Delta U$ = 0

and $W = nRT\log \left| {{{{V_F}} \over {{V_l}}}} \right|$
i.e.  $W\; = 3 \times 8.3 \times 300 \times lo{g_e}\left( 5 \right)$
= 12.03 kJ
Qisothermal­  = 0 + 12.03 = 12.03 kJ
For isochoric change as V = constant,
$W = \int {pdV} = {\rm{ }}0$
$\Delta U = n{C_v}\Delta T{\rm{ }} = {\rm{ }}3{C_v}\Delta T(n = {\rm{ }}3)$
Applying gas equation between points A and C,
${{PV} \over {300}} = {{P\left( {5V} \right)} \over {{T_C}}}$    i.e.        TC = 1500 K
so that $\Delta T = {T_C}-{\rm{ }}{T_B} = {\rm{ }}1500-300 = 1200$
and hence,  (DQ)isochoric = 3.6 Cv + 0 = 3.6 Cv kJ               (ii)

According to given problem,
$\Delta {Q_{isothermal}}\; + D{Q_{isochoric}} = {\rm{ }}83.14{\rm{ }}kJ$
Using equation (i) and (ii), we get
12.03 + 3.6 Cv = 83.14
or         Cv = (71.11/3.6) = 19.75 J
Thus    CP = CV  + R = 19.75 + 8.3 = 28.05 J/mol - K
$\gamma = {{28.05} \over {19.75}} = 1.42$

• No discussion about Cp and CV

• Simran

Good notes

• Anonymous

nice notes got full marks in the class test