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Physics R. Sudha Rani

Thermodynamics 2 Marks Questions Q: Define thermal equilibrium. How does it lead to Zeroth law of thermodynamics? A: If temperature of two systems are equal, then they are said to be in thermal equilibrium. ★ Zeroth Law: If two systems (A, B) are in thermal equilibrium with third system (C) separately. Then two systems are in thermal equilibrium with each other. TA = TC TB = TC

A ⇒ TA = TB

B C

Q: Thermodynamic variables than can be defined by (a) Zeroth law (b) First law? A: (a) Temperature (b) Internal energy. Q: Why a heat engine with 100% efficiency can never be realised in practise? Q2 A: Efficiency η = 1 − ⎯ Q1 For Q2 = 0, η = 1 i.e. engine will have 100% efficiency in converting heat into work. Such engine is called ideal heat engine. But first law of thermodynamics does not rule ideal engine. Thus η = 1 is never possible. Q: In summer, when the valve of a bicycle tube is opened, the escaping air appears cold. Why? A: When the valve of a bicycle tube is opened, the air expands under adiabatic process, internal energy decreases, causes a fall in temperature. Hence escaping air appears cold. Q: A thermos flask containing a liquid is shaken vigorously, what happens to its temperature? A: Work is done by the liquid on the wall of flask, when it is shaken vigorously. Then temperature of liquid increases due to increase of internal energy. Q: Can a room be cooled by leaving the door of an electric refrigerator open? A: No, a room cannot be cooled by leaving the door of a refrigerator open rather it will get

Q: Define calorie. What is the relation between calorie and mechanical equivalent of heat? A: Calorie: The amount of heat required to raise the temperature of 1 gm of water from 14.5°C to 15.5°C. ★ 1 Cal = 4.186 J/Kg − K ∼ 4.2 J/Kg − K =

How much heat is wasted...? slightly heated, because it exhaust more heat into the room than it extracts from the body.

Problems Q: An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 J per second. At what rate is the internal energy increases? dQ Sol: dQ = 100 J × t ⇒ ⎯⎯ = 100 W = 100 J/s dt dW dU ⎯⎯ = 75 J/sec, ⎯⎯ = ? dt dt dQ = dU + dW dQ dU dW dU ⎯⎯ = ⎯⎯ + ⎯⎯ ⇒ 100 = ⎯⎯ + 75 dt dt dt dt dU ⇒ ⎯⎯ = 25 J/sec dt = 25 W Q: A refrigerator is to maintain eatables kept inside at 9°C. If the room temperature is 36°C, calculate the coefficient of performance. Sol: T1 = 36 + 273 = 309 K T2 = 9° + 273 = 282 K T2 282 282 α = ⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯ = ⎯⎯ = 10.4 T1 − T2 309 − 282 27 Q: A steam engine delivers 5.4 × 108 J of work per minute and requires 3.6 × 109 J of heat per minute from its boiler. What is the efficiency of engine? How much heat is wasted per minute? Sol: W = 5.4 × 108 J/min Q = 3.6 × 109 J/min W 5.4 × 108 η = ⎯ = ⎯⎯⎯⎯ = 0.15 Q 3.6 × 109 = 15% Heat wasted per minute = (Q − W) per min. = 3.6 × 109 − 5.4 × 108 = 108 (36 − 5.4) = 3.1 × 108 J

4 Marks Questions Q: Derive a relation between two specific heat capacities of gas on the basis of first law of thermodynamics? A: Consider one mole of an ideal gas contained in a cylinder provided with a frictionless piston. When gas is heated, it expands and piston is pushed up through a distance ‘dx’, then work done dw = F.dx = P A dx = P(dv) ........ (1) [... Adx = dv] When gas is heated at constant volume dU = Cv dT ...... (2) When gas is heated at constant pressure dQ = Cp dT ..... (3) From first law of thermodynamics dQ = dU + dW Cp dT = Cv dT + Pdv [Cp − Cv ] dT = P dv From ideal gas equation P(dv) = R(dT) ∴ [Cp − Cv ] dT = R(dT) Cp − Cv = R Q: Explain qualitatively the working of a heat engine. A: “A device used to convert heat energy into work” is called heat engine. It consists of three parts. ★ Source: It is a body at high temperature 'T1'. Heat 'Q1' is extracted from this body. ★ Working substance: A body of engine containing the working substance. In steam engine, the working substance is steam. ★ Sink: It is a body at low temperature 'T2'. Heat 'Q2' is rejected by the working substance into the sink. Work done by the working substance W = Q1 − Q2

Efficiency η Work done by engine (W) = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Amount of heat absorbed (Q1) W Q1 − Q2 Q2 η = ⎯⎯ = ⎯⎯⎯⎯⎯ = 1 − ⎯ Q1 Q1 Q1 Q1

Q2

Source ⎯→⎯⎯ working ⎯→⎯⎯ Sink T2 T1 substance

↓W=Q −Q 1 2 Q: State second law of thermodynamics. Explain the working of a refrigerator. A: Second law of thermodynamics gives direction of flow of heat. It consists of two statements. ★ Clausius Statement: It is impossible for a self acting machine unaided by any external agency to transfer heat from a body at low temperature to a body at high temperature. (OR) Heat on its own can not flow from a cold body to a hot body. ★ Kelvin Statement: No heat engine can convert whole of the heat energy supplied to it into useful work. ★ Refrigerator: It is a heat pump, which is the reverse of a heat engine. Q1

Q2

Source ⎯←⎯⎯ working ⎯←⎯⎯ Sink T2 T1 substance

↑ In refrigerator working substance extracts (Q2) an amount of heat from sink of lower temperature and an external work (W) is done on the working substance, finally (Q1) heat is transferred to the source. Q2 The coefficient of performance α = ⎯⎯⎯⎯ Q1 − Q2 'α' can be greater than one.

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8 Marks Question Q: Explain reversible and irreversible processes. Describe the working of a Carnot engine. Obtain an expression for the efficiency. ★ Reversible Process: A process that can be retraced back in the opposite direction in such a way that system passes through the same states as in the direct process and finally the system and surroundings return to their original states. A reversible process is only idealised process. Examples: 1. Slow isothermal and slow adiabatic processor 2. Fusion of ice and vapourisation of water. ★ Irreversible Process: A process that cannot be retraced back in the opposite direction is called as irreversible process. ✪ Examples: 1. Work done against friction. 2. Diffusion of gases. ★ Carnot engine: A reversible engine operating between two temperatures is called Carnot engine. The cycle operating it is known as ‘’Carnot’s cycle’’. The ideal gas acts as working substance, it is taken through a cycle of two isothermals and two adiabatics. They are

1

★ Step 1→2: Isothermal expansion of the gas taking its state from (P1, V1, T1) to (P2, V2, T2) shown in curve (a). Heat absorbed by the gas (Q1) from source at temperature T1, equals to work done by the gas. Then V2 W1→2 = Q1 = nRT1 log ⎯ ........... (1) V1 →3: Adiabatic expansion of the gas ★ Step 2→ from (P2, V2, T1) to (P3, V3, T2) shown by the curve (b). Work done by the gas nR W2→3 = ⎯⎯ [T1 − T2] ........ (2) r−1 →4: Isothermal compression of the ★ Step 3→ gas from (P3, V3, T2) to (P4, V4, T2) is shown by curve (c). Heat released Q2 by the gas to the sink at temperature T2 equals to work done on the gas. V4 W3→4 = n RT2 log ⎯ V3

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V3 = −nRT2 log ⎯ ........... (3) V4 →1: Adiabatic compression of the ★ Step 4→ gas from (P4, V4, T4) to (P1, V1, T1) is shown by the curve (d) work done on the gas is − n R (T2 − T1) W4 →1 = ⎯⎯⎯⎯⎯⎯⎯ ....... (4) (r − 1) W = W1→ →2 + W2→ →3 + W3→ →4 + W4→ →1

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V2 V3 W3→4 = n RT1 log ⎯ − n RT2 log ⎯ V1 V4 W = Q1 − Q2

W Q1 − Q2 Efficiency η = ⎯ = ⎯⎯⎯⎯ Q1 Q1

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V2 V3 ⎯ − T2 log ⎯ V1 V4 = ⎯⎯⎯⎯⎯⎯⎯⎯ V2 T1 log ⎯ V1 T1 log

From adiabatic relation TVr-1 = constant Apply for 2→3 step, 4→1 step V2 V3 which gives ⎯ = ⎯ V1 V4 T1 − T2 T2 ∴η = ⎯⎯⎯ = 1 − ⎯ T1 T1

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