Chapter#01: Equilibrium Thermodynamics

Team Quanta presents all possible short questions of Thermal & Statistical Physics’ Chapter#01: Equilibrium Thermodynamics.

Q.1- Define statistical mechanics. What are postulates of statistical mechanics?

Answer: Statistics mechanics is a branch of, mathematical physics that studies, using
probability theory, end the average behavior of a mechanical system where the slate of the system is uncertain. Statistical mechanics is the formalism that connects thermodynamics to the microscopic world. Remember that statistics is a quantitative measure of some selection of objects an observation of the macroscopic world is necessarily an observation of some statics of the molecular behaviors. Basic postulates of statistical mechanics are given below:

All gases are composed of molecules which are in state of constant random motion.
The total number of molecules in a gas is constant.
The total energy of system is constant.
All elementary cells are of same spin.
All accessible microstates corresponding to possible macro states are equally probable.
The equilibrium state (macro state) of a system corresponds to a state of maximum probability.

Important postulates of statistical mechanics are principle of equal a priori probability and erodic hypothesis.

Q.2- Name three different types of statistics. State postulates of quantum statistics.

Answer: Three deferent types of statistical mechanics are:

Classical Statistics
Bose-Einstein Statistics
Fermi-Dirac Statistics

The main points of quantum statistical mechanics are:

Quantum statistics applies to system of identical particles which obeys laws of quantum statistical mechanics.

In quantum statistical mechanics the volume of elementary cell in phase space cannot be less than , where, is Plank’s constant having value $6.63\times{10}^{-34}\ Js$

As the number of cells available in phase space in quantum statistical mechanics is approximately equal to number of particles, the occupation index

$\frac{n_i}{g_i}\approx1$

In quantum statistical mechanics the particles of system are considered indistinguishable.

Q.3: What is meant by available volume in statistical mechanics?

Answer. Suppose the particles of a dynamical system occupy a given volume in ordinary space. The geometry of this determines minimum and maximum possible values for particles of system. The product of volume elements of position and momentum space is called available volume in phase space and is given by,

$dxdydzdp_{x}dp_{y}dp_{z}$

Q.4: What is meant by term thermodynamic probability of a macro state? How it is related to probability of occurrence of state?

Answer. The number of microstates corresponding to any given macrostate is called thermodynamics probability and is denoted by . The probability of occurrence of a macrostate is defined as number of microstates in it to total number of possible microstates. Probability of occurrence of a macrostate is related to thermodynamic probability by relation.

$p \propto W \ \ \ \Rightarrow p=k_{B} \ln W$

$K_{B}$ being Boltzmann constant.

Q.5: What is purpose of dividing phase space into cells?

Answer: The purpose of dividing phase space into cells is to study the distribution of particles of a dynamical system among these compartments and cells and therefore to develop various relations in statistical thermodynamics for most probable macro states of system when the system is in equilibrium.

Q.6: What is distinguishing feature of Maxwell-Boltzmann and Fremi-Dirac
statistics?

Answer: In Maxwell-Boltzmann statistics, the particles are distinguishable and any number of particles can occupy a single cell in phase space where as in Fermi-Dirac statistics, the particles are indistinguishable and there cannot be more than one particle ina single cell in phase space i.e. particles obey Pauli Exclusion principle.

Q.7: Name the statistics obeyed by photons. What was the main difficulty with Maxwell-Boltzmann statistics?

Answer: Photons obey Bose-Einstein statistics. Maxwell-Boltzmann explained the energy and velocity distribution of molecules of an ideal gas to a fair degree of accuracy but is failed to explain the observed energy distribution of electrons in so called electron gas and that of photons in photon gas.

Q.8: Define chemical potential. Name the statistics obeyed by molecules of N2at N.T.P.?

Answer: Chemical potential is given by,

$\mathbf{\mu=-T\left(\frac{\partial S}{\partial N}\right)<em>{U,V}=\left(\frac{\partial G}{\partial N}\right)</em>{T,p}}$

S is entropy, G is Gibb’s free energy and N is number of particles. Chemical potential is defined as, “It is the energy required to add a particle to the system”. Molecules ofN2 at N.T.P. obey Maxwell-Boltzmann statistics.

Q.9: Name the statistics obeyed by free electrons in metals. State specific heat at constant volume and constant pressure.

Answer: Free electrons in metals obey Fermi-Dirac statistics.

Specific heat at constant volume: –The specific heat at constant volume is amount of heat required to raise temperature of a unit mass of gas through keeping thevolume constant.

$\mathbf{C_\nu=\left(\frac{dQ}{dT}\right)_V}$

If is amount of heat required to raise the temperature of unit mass of gas through , keeping the volume constant, then specific heat at constant volume is,

Specific heat at constant pressure: –The specific heat of a gas at constant pressure is amount of heat required to raise the temperature of a unit gas through 1^oC keeping its pressure constant.

If dQ is amount of heat required to raise the temperature of a unit mass of gas through dT^oC keeping the pressure constant, then speciflc heat at constant pressure is,

$\mathbf{C_p=\left(\frac{dQ}{dT}\right)_p}$

Q.10: What is the concept of a cell in a compartment?

Answer: A compartment is divided into a very large number of cells in such a way that each cell is of same size and therefore, all cells have same a priori probability. The sizeof cell is very small as compared to size of compartment so that number of cells in a compartment is very large. For a given size of cell, number of cells in compartment and hence total number of cells is fixed.

Q.11: What is the role played by most probable state in determining the behaviorof a system in equilibrium?

Answer: The most probable state of a system is that macro state which has the maximum probability of occurrence. The role played by most probable state in determining the behavior of a macroscopic system in equilibrium lies in fact that all properties of system can be deduced from knowledge of most probable state and we need not consider all other macro states which are extremely large in number.

Shifted at last

Q.12: What is probability of getting either a seven or a six when throwing two dice?

Answer: To get six we have five ways out of 36 possible outcomes, so

$\mathbf{p_6=\frac{5}{36}}$

To get seven we have six ways out of 36 possible outcomes, so

$\mathbf{p_7=\frac{6}{36}}$

So total probability is,

$\mathbf{p=p_6+p_7=\frac{5}{36}+\frac{6}{36}=\frac{11}{36}}$

Q.13: How many ways can one choose five objects out of twelve if either the order of the choice is important, or the order of choice is not important, only the objects chosen?

Answer: We can choose five objects out of twelve in following ways if order of choice is important:

$\mathbf{W=12\times11\times10\times9\times9=95040}$

We can choose in following ways if order of choice is not important:

$\mathbf{W=\frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1}=792}$

Q.14: If eight coins are thrown at random show that the probability of obtaining at least six heads is 37/256.

Answer. Here we can have eight head or seven heads and tail or six heads and two tails. Probability of getting eight heads is probability of getting seven headsis $\frac{1}{2^{8}} \times \frac{8!}{7! \times 1!}=\frac{28}{2^{8}}$ , Probability of getting six heads is $\frac{1}{2^{8}} \times \frac{8!}{6! \times 2!}=\frac{28}{2^{8}}$ , Thus total probability is,

$p=\frac{1}{2^{8}} +\frac{8}{2^{8}}+\frac{28}{2^{8}}=\frac{37}{256}$

Q.15: Three dices are thrown. What is likelihood that all faces are same? What is likelihood that sum of faces are 11?

Answer. The total number of ways arranging three dices is $6^{3}$ . So

$P=\frac{6}{6^{3}} =\frac{1}{36}$

If the sum of faces is 11, then there are several 27 ways to get this total so that,

$P=\frac{27}{6^{3}} =\frac{1}{8}$

Q.16: What is range of an event? Define accessible and inaccessible macro states?

Answer: The probability of an event lies between 0 & 1. It is zero for event which can never occur and 1 for event which is sure to occur. The probability of random event is 1/2.

The macro states which are allowed under a constraint are called accessible macro states and macro states which are not allowed under a constraint are called inaccessible macro states. For example, in distributing three particles in two compartments under constraints that no compartment will remain empty, the only accessible macro states are (2; 1) & (1; 2) and inaccessible macro states are (3; 0) & (0; 3)

Q.17: Calculate probability that is tossing a coin five times, we get three heads two tails.

Answer: Required probability is,

$\mathbf{P=\frac{\frac{1}{2^5}5!}{\left(5-2\right)!\times2!}=\frac{5}{16}}$

Q.18: Two six face dice, each marked 1 to 6 are thrown. Calculate the probability that one of dice shows 6 and the other shows 5.

Answer: The probability that first dice shows 5 or 6 is

$P_1=\frac{2}{6}$

The probability that second dice shows one out of 5 or 6 which the first has not shown is

$\mathbf{P}_\mathbf{2}=\frac{\mathbf{2}}{\mathbf{6}}$

The probability that one dice shows five and the other shows six is

$\mathbf{P=P_1\times P_2=\frac{2}{6} \times \frac{1}{6}=\frac{1}{18}}$

Q.19: Which law of thermodynamics is most pertinent to the statement “all kings’ horses and all kings’ men could not put Humpty Dumpty back together again”?

Answer: Second law of thermodynamics is most pertinent to this statement, according to which physical processes move in the direction Of increasing disorder.

Q.20: Comment on statement, “A heat engine converts disordered mechanical motion into organized mechanical motion.”

Answer: By an ordered motion, we mean such a motion of energy which can be put to some. use. If we have got two bodies, one being at a higher temperature than the other, then heat can move from body at a higher temperature to body at lower temperature if we just mix the two bodies, then the flow of heat will take place without producing any useful work. This is a disordered motion of heat energy.

The same motion can be put to order if we produce it through a heat engine. The heat ,in this case also, will flow from hotter body to. colder one but this motion will help in operating the heat engine thereby producing a useful work. Thus, the heat engine has converted a disordered motion into ordered motion.

Q.21: Under what conditions would an ideal heat engine be 100% efficient?

Answer: An ideal heat engine will be 100% efficient, if:

The condition of thermodynamic equilibrium is satisfied.
No dissipative effect is present.
Piston should be friction less.
Temperature of sink must be zero.

Q.22: Which law of thermodynamics would be violated if heat were to? Spontaneously flow between two objects which are in thermal equilibrium?

Answer: Second law of thermodynamics will be a violated according to which heat always flows from a HTR to a LTR. If heat were to flow spontaneously between bodies having equal temperature, the result would be bodies at different temperatures. These bodies could then be used to run a heat engine until they were again at the same temperature, after which the process could be repeated indefinitely.

Q.23: What factors reduce the efficiency of heat engine from its ideal value?

Answer: Efficiency of a heat engine is given by,

$\mathbf{e=1-\frac{T_2}{T_1}}$

This show that $\mathbf{e\neq1}$ unless $\mathbf{T_2=0}$ . Thus, all heat engines have efficiency less than ideal value due to presence of friction and heat losses byconduction and radiation.

Q.24: If the temperature in a laboratory is decreased, is the cost to freeze adozen ice cubes less than or equal to as it was before the laboratory was cooled?

Answer: If the laboratory is cooler, then less energy would be required. This is because heat extracted from the water can be expelled at a lower temperature, which means that less work must be done by the refrigerator’s engine.

Q.25: Show that total entropy increases when work in converted into heat by friction between sliding surfaces.

Answer: If work is done by friction, the work will be converted into heat. The heat produced due to the friction goes into surrounding i.e., air and becomes useless. No useful work can be performed by it due to unavailability of this energy; we can say that entropy will increase when work is done by friction. Hence total entropy increases due to friction.

Q.26: The volume of a mono atomic gas doubles in an isothermal expansion. By what factor does its pressure change?

Answer: An isothermal process obeys the relation

PV=Constant

$\mathbf{p_1V_1=p_2V_2\ \ \Rightarrow\frac{p_2}{p_1}=\frac{V_1}{2V_{1\ }}=\frac{1}{2}}$

Therefore, pressure reduces to half of its original value.

Q.27: An ideal gas expands at constant pressure. Does heat flow into or out of gas? If heat flows into gas, justify your answer. If heat flows out of gas, give a proof.

Answer: For constant pressure $\mathbf{Q=nC_p\Delta T\ \ \ \ p\Delta V=nRT\Delta T}$ .

$\mathbf{\Rightarrow Q=nC_p\left(\frac{p\Delta V}{nR}\right)=\left(\frac{C_p}{R}\right)p\Delta V}$

Since gas expands, $\mathbf{\Delta V>0\Rightarrow Q>0}$ , i.e., heat goes into gas.

Q.28: Explain why temperature drops in an adiabatic process.

Answer: Suppose one mole of a gas having a volume V₁, pressure p₁& temperature T₁expands adiabatically to a volume V₂, pressure p₂& temperature T₂, then

$\mathbf{T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1\ }\ \Rightarrow T_2=T_1\left(\frac{V_2}{V_1}\right)^\frac{\gamma-1}{\gamma}}$

As gas expands, its volume increases i.e.

$\mathbf{V_1<V_2\Rightarrow\left(\frac{V_1}{V_2}\right)^{\gamma-1}<1}$

So i.e., $\mathbf{T_2<T_1}$ temperature of gas drops in adiabatic process

Q.29: For an ideal gas in an isothermal process, there is no change in thermal energy. Suppose gas does work W during such a process. How much is energy transferred by heat?

Answer: From 1^stlaw of thermodynamics,

$\mathbf{\mathrm{\Delta U}=Q-W}$

For isothermal process, above equation takes the form

$\mathbf{0=Q-W\ \ \Rightarrow Q=W}$

Since work is done by gas, so

$\mathbf{Q=-W}$

i.e. heat transferred is equal to work done by gas.

Q.30: What is minimum size of phase space cell in classical and quantum mechanical statistics?

Answer: In quantum mechanics, the minimum size of cell is , being Plank’s constant.In classical mechanics, size of cell can be made as smallas please, it can even approachzero.

Q.31: What are main points of difference between classical and quantum statistics.

Answer. The main points of difference between classical and quantum statistics are given below

Classical Statistics:

Classical statistics applies to system of identical particles which obeys laws of
classical statistical mechanics.
In classical statistical mechanics the volume of elementary cell in phase space canbe made as small as we please.

As number of cells available in phase space in classical statistical mechanics is very large as compared to number of particles, the occupation index $\frac{n!}{g_{i} \ll 1}$
In classical statistical mechanics the particles of system are considered distinguishable.

Quantum Statistics:

Quantum statistics applies to system of identical particles which obeys laws of quantum statistical mechanics.
In quantum statistical mechanics the volume of elementary cell in phase space cannot be less than $h^{3}$ where $h$ is Plank’s Constant having value $6.63 \times 10^{-34} Js$ .
As number of cells available in phase space in quantum statistical mechanics is approximately equal to number of particles, the occupation index $\frac{n_{i}}{g_{i}} \approx 1$

In quantum statistical mechanics the particles of system are considered indistinguishable.

Q.32: Obtain relationship between volume in phase space and volume in position and momentum spaces.

Answer: Volume element in position space is dxdydz volume element in momentum space is dpxdpydpz . Volume element in phase space is,

dT = dxdydzdpxdpydpz

Q.33: Name the state functions used in statistical mechanics.

Answer: State functions used in statistical mechanics are; pressure P ,volume V, temperature T, internal energy U , entropy S, enthalpy H ,Helmholtz function F and Gibb’s function G and entropy σ .

Q.34: Define macro system and micro system, micro state and macro-state.

Answer: A system which can be observed with naked eye is known as macroscopic system. A state of system specified by macroscopically observable quantities is called macroscopic state. So, a macroscopic system containing billion of atoms can be observed with naked eye. While studying the macroscopic system we are not interested in individual behavior of atoms of system but in collective behavior of system like volume, pressure etc. Examples of macroscopic system are gas in a container and liquid in a vessel etc.

Q.35: Is it true that heat energy of the universe is steadily growing less available? If so, why?

Answer:- All natural processes continue towards a state of disorder. When the entropy of the universe will reach its maximum value, everything will be at same temperature. Then there will be no way to convert heat energy into useful work, thus we might face a heat death. When all systems are taken together as the universe, the entropy of universe always increases.

Q.36: The internal energy of a certain gas increases when it is compressed isothermally. Is the gas ideal?

Answer: The internal energy of ideal gas is function of temperature. In isothermal process temperature remains constant, so internal energy is constant. In this case internal energy increases without increasing temperature, so gas is not ideal.

Q.37: Why is heat energy named as energy in transit? How is it sometimes related to thermal equilibrium?

Answer: Joule established a relation between mechanical work W and heat produced Q in form

W =JQ

Where J is called mechanical equivalent of heat & its value is 4.2J/cal. This means that 4.2J of Mechanical work when performed produces one calorie of heat. Heat is a form of energy associated with molecular motion and hence is regarded as energy in transit. Thus, heat refers to energy that is transferred between two objects due to temperature difference between them. A thermal equilibrium is set up only when two objects attain a common temperature.

Q.38: The coefficient of performance of a refrigerator becomes infinite when temperatures of two bodies become equal. Explain.

Answer: The coefficient of performance of a refrigerator is given by,

$\mathbf{k=\frac{T_2}{T_1-T_2}}$

When $\mathbf{T_1=T_2. k\rightarrow\infty}$ . Also,

$\mathbf{k=\frac{T_2}{T_1-T_2}=\frac{Q_2}{Q_1-Q_2}\ ;Q_1-Q_2=0\ \ \ \Rightarrow Q_1=Q_2}$

But from $\boldsymbol{W=Q_1-Q_2}$ , we have for $\mathbf{Q_1-Q_2=0,\ \ W=0}$ i.e., no mechanical energy is drawn from outside and no external work is done on machine.

As no energy is drawn by the refrigerator, compressor may go on infinitely but no heat will be transferred from LTR to HTR in accordance with second law of thermodynamics.

Q.39: Name any two theories in physics based on statistical concepts.

Answer: Two quantities based on statistical concepts are explained below;.

Entropy: –According to statistical considerations equilibrium state of an isolated system is macro state of maximum probability. Hence in equilibrium state both entropy and probability has maximum value.

Distribution of molecular speeds: –With help of Maxwell-Boltzmann statistics the number of molecules per unit volume of a gas having speeds between v & v + dv has been calculated and hence value of most probable speed.

Q.40: Calculate the percentage error in using Stirling formula when .

Answer: By Stirling formula,

$\mathbf{\ln{n!}=n\ln{n-n}}$

Now,

$\mathbf{\ln{n!}=\ln{5!}=\ln{120}=4.7875}$ & $\mathbf{n\ln{n-n}=5\ln{5-5}=3.047}$

%age error = $\mathbf{\frac{4.7875-3.047}{4.7875}\times100}$ % = $\mathbf{\frac{1.74.5}{4.7875}\times}$ % =36.38%

Thermal & Satistical Physics