All S.Qs of Chapter#04: Entropy and 2nd law of Thermodynamics of Quantum Mechanics-II. We have arranged all S.Qs for BS/MSc Physics students.
Q.1 Is it true that heat energy of the universe is steadily growing less available, if so why?
Answer: Yes, it is true that heat energy of the universe is steadily growing less available because thermal pollution is going on increasing which results to increase in entropy of the universe.
Q.2 How would you increase the efficiency of a Carnot engine? By increasing TL only? By decreasing only?
Answer: The efficiency of an ideal heat engine is given as $e\ =\ 1-T_L/T_H$ According to this formula efficiency increases when $T_H$ is increased and $T_L$is kept constant. Similarly, efficiency increases when $T_L$ is decreased and $T_H$ is kept constant.
Q.3 What factors reduce the efficiency of a heat engine from its ideal value?
Answer: The friction and heat losses to environment reduce the efficiency of a heat engine from its ideal value.
Q.4 Comment on the statement; A heat engine converts disordered mechanical motion into organized mechanical motion?
Answer: The heat engine is a device that converts heat energy into mechanical energy. The hot system has disordered mechanical motion while mechanical system has organized mechanical motion.
Q.5 The efficiency of a Carnot engine is always less than 100%. Why?
Answer: The efficiency of Carnot heat engine is given as $=1-\frac{Q_2}{Q_1} .$ . The efficiency will be 100% if $Q_2$ is zero. It is against second, law of thermodynamics. Hence, efficiency of Carnot heat engine must be less than 100%.
Q.6 What is meant by heat death of the universe?
Answer: The disorderness in the universe is increasing. The time at which the entropy would reach maximum value and every thing is at same temperature in the universe is called heat death.
Q.7 Explain why a room cannot be cooled by leaving open the door of a kitchen refrigerator.
Answer: The room cannot be cooled by leaving the door of a kitchen refrigerator opened because heat absorbed from the room is exhausted in the same room. The net heat transfer is zero. When LTR and HTR are placed in the same room, net heat transfer is zero.
Q.8 Explain why the specific heat at constant pressure is greater than the specific heat at constant volume?
Answer: The amount of heat energy needed to raise the temperature of one mole of a substance through one Kelvin is called molar heat capacity. When heat is added into a gas enclosed in cylinder at constant volume; the work done is zero so all heat supplied changes the internal energy that is function of temperature.
When heat is added into a gas enclosed in cylinder, at constant pressure, the heat supplied changes the internal energy which is function of temperature and partially performs work. The heat supplied is divided into parts. Therefore, to get same temperature specific heat at constant pressure is greater than the specific heat at constant volume and both are related as $C_P-C_V=R$
Q.9 Why do we call heat energy as energy in transit? How is it sometimes related. to thermal equilibrium?
Answer: Heat energy flows from hot body to a cold body when both bodies have thermal contact until thermal equilibrium is reached. The temperature determines the direction of heat flow so heat is called energy in transit.
Q.10 Under what conditions would an ideal heat engine be 100 % efficient?
Answer: The efficiency of ideal heat engine is given as $e=1-\frac{T_2}{T_1}$. The efficiency of an ideal heat engine will be 100 % when temperature of LTR is 0 K.
Q.11 Give some examples of irreversible processes in nature?
Ans. Chemical explosion is irreversible process. The dissipation of energy through conduction, convection and radiation is irreversible process. The isochoric process and isobaric process are also irreversible.
Q.12 Are there any natural processes that are reversible?
Answer: There is no any natural process which is reversible, however liquefaction and evaporation of a substance performed slowly are considered reversible. Similarly slow compression of a gas in a cylinder is also considered reversible.
Q.13 Can we calculate work done during an irreversible process in terms of area on PV diagram?
Answer: We cannot calculate work done during an irreversible process in terms of area on PV diagram because pressure and volume are not well defined in this process. Therefore, we cannot plot a line as continuous line on PV diagram. Yes, work done can be calculated during reversible process on PV diagram.
Q.14 Give examples in which the entropy of a system decreases and explain why the second law of thermodynamics is not violated.
Answer: The entropy of a system decreases when its temperature is reduced. For example formation of ice from water. The second law of thermodynamics is not violated because heat flows from cold body to hot body with the help of work.
Q.15 Is there a change in entropy in purely mechanical motion?
Answer: The change in entropy ‘takes place in purely mechanical motion because temperature rises and disorderness increases in pure mechanical motion.
Q.16 Show that total entropy increases when work is converted into heat by friction between sliding surfaces.
Answer: The temperature increases when work is converted into heat by friction between sliding surfaces, which results increase in disorderness. Therefore, entropy of system increases.
