Title Med-RM_Phy_SP-2_Ch-12-Thermodynamics.pdf ✅

Chapter Contents Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Chapter 12 Introduction Thermal Equilibrium Zeroth Law of Thermodynamics Heat, Internal Energy and Work First Law of Thermodynamics Specific Heat Capacity Thermodynamic State Variables and Equation of State Thermodynamic Processes Second Law of Thermodynamics Heat Engines Refrigerators and Heat Pumps Carnot Engine Additional Information Introduction The study of heat and its transformation to mechanical energy is called thermodynamics. In mechanics, our interest is in the motion of particles or bodies under the action of forces and torques. While in thermodynamics deal with the internal microscopic state of the body without considering with the motion of the system as a whole. In this chapter we shall study the laws of thermodynamics, various thermodynamic processes, basic theory of heat engines, refrigerators and Carnot engine. THERMAL EQUILIBRIUM In mechanics the term ‘Equilibrium’ means, that the net external force and torque on a system are zero. But in thermodynamics, the term thermal equilibrium means, temperature of two systems are equal, and there is no exchange of heat if put in contact. ZEROTH LAW OF THERMODYNAMICS If two systems A and B are in thermal equilibrium with a third system C separately, then A and B will also be in thermal equilibrium with each other. Note : Two or more systems are said to be in thermal equilibrium, if there is no exchange of heat energy between them when they are brought in thermal contact. HEAT, INTERNAL ENERGY AND WORK Heat : It is the disordered form of energy, which flows naturally from a body at high temperature to a body at low temperature till thermal equilibrium is obtained. Internal Energy : It is sum of the kinetic energies and potential energies of all the constituent molecules of the system. It is denoted by ‘U’. Thermodynamics
150 Thermodynamics NEET Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Internal energy depends only on the state of the system. It is a state variable which is independent of the path taken to arrive at that state. If we neglect the small intermolecular forces in a gas, the internal energy of the gas is just the sum of the kinetic energies associated with various random motions of its molecules. Note : Internal energy of a system does not depend on the motion of a system as a whole i.e., the sum of the kinetic energy of only the constituent molecules due to their randomness inside the system is considered. Work done by gas : When the piston of a cylinder is pushed down or gas is compressed, the work done by the gas is taken to be negative. Similarly when the gas expands the work done by the gas is taken to be positive. Work done :   W P dV  To calculate work done graphically from (P – V) graph, calculate area under (P – V) graph up to volume axis. A statement like ‘a gas in a given state has a certain amount of heat.’ is as meaningless as the statement that ‘a gas in a given state has a certain amount of work’. While a statement that ‘a gas in a given state has a certain amount of inernal energy’ is a meaningful statement. Similarly we may say that ‘a certain amount of heat is supplied to the system’ or ‘a certain amount of work was done by the system’ are meaningful. Example 1 : The figure shows a P-V graph of the thermodynamic behaviour of an ideal gas. Find out from this graph (i) work done by the gas in the process A  B, B  C, C  D and D  A, (ii) work done by the gas in complete cycle A  B  C  D  A. 1.0 2.0 3.0 4.0 5.0 6.0 7.0 14 12 10 8 6 4 2 V (Litre) P(10 N/m ) 5 2 A' A B D C B Solution : (i) The work done in a thermodynamic process is equal to the area enclosed between the P–V curve and the volume-axis. Work done by the gas in the process A  B is W1 = area ABB' A' = AB × A'A = (6.0 –1.0 ) litre × (12 × 105) N/m2 = 5.0 × 10–3 m3 × (12 × 105) N/m2 = 6000 N-m = 6000 joule. Work done in the process B  C is zero since volume remains constant. Work done on the gas in the process C  D is W2 = area DCB'A' = DC × A' D = (–5.0 × 10–3) × (2 × 105) = –1000 joule. (Negative sign is taken because volume decreases) Work done in the process D  A is also zero, because volume remains constant. (ii) Work W1 is positive, while W2 is negative. Hence the net work done in the whole cycle is W = W1 – W2 = 6000 – 1000 = 5000 joule This net work is done by the gas.
NEET Thermodynamics 151 Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Example 2 : Consider the process on a system shown in figure below. During the process, the work done by the system. Pressure Volume 1 2 (1) Continuously increases (2) Continuously decreases (3) First increases then decreases (4) First decreases then increases Solution : Don’t get tempted to mark the answer (3). We said that the area under the P-V diagram is equal to the work done. So, if we move from 1 to 2 the area under the graph is continuously increasing so the correct response will be (1). Note : In this question the rate with which the work is being done by the system is first decreasing then increasing. Example 3 : Find the work done during the perfectly circular cyclic process as shown in the diagram. P P1 P2 V1 V2 V Solution : The area of a circle is R2. But here you have two radii. The horizontal radius   2 1 1 – 2 V V R  and the vertical radius   1 2 2 – 2 P P R  . You know that the area of an ellipse is R1R2, where R1 & R2 are semi-major and semi-minor axis respectively. If R1 = R2 the ellipse becomes a circle. So, here the work done = area = R1R2 =   21 12 – –   . 2 2 VV PP  . The +ve sign is due to the fact that the cycle is clockwise. FIRST LAW OF THERMODYNAMICS First law of thermodynamics is in accordance with law of conservation of energy. According to Clausius statement of first law of thermodynamics, the heat given to the system Q = U + W
152 Thermodynamics NEET Aakash Educational Services Limited - Regd. Office: Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 U = Change in the internal energy W = Work done by the system against external pressure Q UUU = – f i Heat given to system taken as positive Heat given by system taken as negative W Work done by the system taken as positive Work done on the system taken as negative Note : It should be remembered that (1) The units of dQ, dU and dW should be same while using this equation. (2) If the temperature of an ideal gas rises, then dU should be positive and if the temperature falls, then dU is negative. (3) If the volume of the system increases, then dW is positive while if the volume of the system, decreases, then dW is negative. If the volume is fixed, dW = 0. (4) dQ and dW depends not only on the initial and final states but also on the path followed (or thermodynamic process). (5) dU depends only on initial and final states, not on the path. For all processes  V dTnCdU for an ideal gas even when volume is not constant. But for non-ideal gas it is true only when volume is constant. (6) During melting, change in volume is neglected,  W = 0, hence  dUdQ , But during boiling dV  0, so dQ = dU + PdV > dU. SPECIFIC HEAT CAPACITY We have seen in last chapter that heat capacity of a substance is given by Q S T    If we divide S by mass of substance m in kg. S Q 1 s m mT     Here s is known as the specific heat capacity of substance. If the amount of substance is specified in terms of moles  (instead of mass m in kg), we can define heat capacity per mole of the substance by S Q 1 C T      Here C is known as molar specific heat capcity of substance. Both s and C are independent of the amount of the substance. Value of specific heat may varies from 0 to . Specific heat capacity depends on the process or the conditions under which heat transfer takes place. For gases, we mainly calculate two types of specific heats. Specific heat capacity at constant pressure ‘Cp’ and specific heat capacity at constant volume ‘Cv’. CC R p v  