Lecture 2: Temperature, Heat & The First Law Of Thermodynamics PDF

Lecture 2: Temperature, heat & the first law of thermodynamics (Fundamentals of Physics, 10th edition)  18.5: A closer look at heat and work Let us take a system where a gas is conf...

Lecture 2: Temperature, heat & the first law of thermodynamics (Fundamentals of Physics, 10th edition)  18.5: A closer look at heat and work Let us take a system where a gas is confined inside a cylinder with a movable piston, as in Fig. The upward force on the piston due to the pressure of the confined gas is equal to the weight of lead shot loaded onto the top of the piston. The walls of the cylinder are made of insulating material that does not allow any transfer of energy as heat. The bottom of the cylinder rests on a reservoir for thermal energy, a thermal reservoir (perhaps a hot plate) whose temperature T you can control by turning a knob. The system (the gas) starts from an initial state i, described by a pressure pi , a volume Vi and a temperature Ti. You want to change the system to a final state f, described by a pressure pf ,a volume Vf ,and a temperature Tf.The procedure by which you change the system from its initial state to its final state is called a thermodynamic process. During such a process, energy may be transferred into the system from the thermal reservoir (positive heat) or vice versa (negative heat). Also, work can be done by the system to raise the loaded piston (positive work) or lower it (negative work).  Suppose that we remove a few lead shot from the piston of Fig , allowing the gas to push the piston and remaining shot upward Ԧ through a differential displacement d𝑠Ԧ with an upward force 𝐹. Since the displacement is tiny, we can assume that 𝐹Ԧ is constant during the displacement. Then 𝐹Ԧ has a magnitude that is equal to pA, where p is the pressure of the gas and A is the face area of the piston. [p = F/A] The differential work dW done by the gas during the displacement is Work is done by the system (gas) on the environment (piston) Ԧ 𝑠Ԧ = Fdscos00= Fds(+1)=(pA)ds = p(Ads)= + pdV dW = 𝐹.𝑑 [p=F/A or F=pA] Work is done on the system (gas) by the environment (piston) Ԧ 𝑠Ԧ = Fdscos1800= Fds(-1)= -(pA)ds = -p(Ads)= - pdV dW = 𝐹.𝑑 in which dV is the differential change in the volume of the gas due to the movement of the piston. When you have removed enough shot to allow the gas to change its volume from Vi to Vf , the total work done by the gas is V W = ‫׬ = 𝑊𝑑 ׬‬V f 𝑝 𝑑𝑉 [any gas] i 18.5: Path dependent work done in terms of p-V diagram: Work done is always area under the p-V curve: The amount of work depends on path. Work done Calculation from PV Diagram: Always, work done equals the area under the PV curve. At constant volume, ΔV=0, W=0. At constant pressure, W= p ΔV When both volume and pressure changes, we have to consider the area concept only. 43. In Fig., a gas sample expands from V0 to 4.0V0 while its pressure decreases from p0 to p0/4.0. If V0 = 1.0 m3 and p0 = 40 Pa, how much work is done by the gas if its pressure changes with volume via (a) path A, (b) path B, and (c) path C? Solution of 43 The volume increases through the three paths , so the work done by the gas is always positive. 𝑉𝑓 W =‫ = 𝑉𝑑𝑝 𝑖𝑉׬ = 𝑊𝑑 ׬‬Area under the curve of p-V (a) WA = WA1 + WA2 WA1 = 40 (4-1) = 120 J (constant pressure) WA2 = 0 (constant volume) WA = 120 +0 = 120 J or [W = p∆V = p(Vf – Vi)= (40-0)(4-1) W=40(3) = 120 J] (b) The work done by the gas is the area under the curve (yellow line) WB= ½ ×(4 - 1)(40 -10) + (4 -1)(10 - 0) = 75 J (c) WC= WC1 + WC2 WC1 = 0 (constant volume) WC2 = (4-1)(10-0) = 30 J WC = 0 + 30 = 30 J 18.5: The First Law of Thermodynamics When a system changes from a given initial state to a given final state, both the work W and the heat Q depend on the nature of the process but the quantity Q - W is the same for all processes. It depends only on the initial and final states and does not depend at all on how the system gets from one to the other. All other combinations of Q and W, including Q alone, W alone, Q +W, and Q - 2W, are path dependent; only the quantity Q - W is not. The quantity Q - W must represent a change in some intrinsic property of the system. We call this property the internal energy Eint and we write ∆𝑬𝒊𝒏𝒕 = 𝑬𝒊𝒏𝒕,𝒇 − 𝑬𝒊𝒏𝒕,𝒊 = Q - W The equation known as the first law of thermodynamics. If the thermodynamic system undergoes only a differential change, we can write the first law as 𝒅𝑬𝒊𝒏𝒕 = dQ - dW 15.5 Some Special Cases of the First Law of Thermodynamics 1. Adiabatic processes : An adiabatic process is one that occurs so rapidly or occurs in a system that is so well insulated that no transfer of energy as heat occurs between the system and its environment : Q = 0 1st law of thermodynamics, ∆𝐸𝑖𝑛𝑡 = Q – W = 0 - W ∆𝐸𝑖𝑛𝑡 = - W (adiabatic process). If work is done by the system (that is, if W is positive), the internal energy of the system decreases by the amount of work. ∆𝐸𝑖𝑛𝑡 = - (+W) = - W Conversely, if work is done on the system (that is, if W is negative), the internal energy of the system increases by that amount. ∆𝐸𝑖𝑛𝑡 = - (-W) = +W 2. Constant-volume processes: If the volume of a system is held constant, that system can do no work: W = p∆V = p(0) = 0 1st law of thermodynamics, ∆𝐸𝑖𝑛𝑡 = Q – W = Q - 0 ∆𝐸𝑖𝑛𝑡 = Q (constant-volume process). Thus, if heat is absorbed by a system (that is, if Q is positive), the internal energy of the system increases. ∆𝐸𝑖𝑛𝑡 = +Q Conversely, if heat is lost during the process (that is, if Q is negative), the internal energy of the system decreases. ∆𝐸𝑖𝑛𝑡 = - Q 3. Cyclical processes: There are processes in which, after certain interchanges of heat and work, the system is restored to its initial state. In that case, no intrinsic property of the system—including its internal energy—can possibly change. ∆𝐸𝑖𝑛𝑡 = Ef – Ei = Ef - Ef = 0 [Ei = Ef] 1st law of thermodynamics, ∆𝐸𝑖𝑛𝑡 = Q – W 0=Q-W Q=W (cyclical process). Thus, the net work done during the process must exactly equal the net amount of energy transferred as heat; the store of internal energy of the system remains unchanged. 4. Free expansions: These are adiabatic processes in which no transfer of heat occurs between the system and its environment and no work is done on or by the system. Q=W=0 1st law of thermodynamics, ∆𝐸𝑖𝑛𝑡 = Q – W = 0 – 0 = 0 ∆𝐸𝑖𝑛𝑡 = 0 (free expansion). 46. Suppose 200 J of work is done on a system and 70.0 cal is extracted from the system as heat. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of (a) W, (b) Q, and (c) ΔEint. Solution: (a) The work done is negative since work done on the system. W = - 200 J (b) Energy is extracted from the system, Q = - 70 cal = - 294 J [1 cal = 4.2 J] (c) Internal energy change, ∆𝐸𝑖𝑛𝑡 = Q – W = - 294 – (- 200) = - 294 + 200 = - 94 J 48. As a gas is held within a closed chamber, it passes through the cycle shown in Fig. Determine the energy transferred by the system as heat during constant-pressure process CA if the energy added as heat QAB during constant-volume process AB is 20.0 J, no energy is transferred as heat during adiabatic process BC, and the net work done during the cycle is 15.0 J. Solution:First law of thermodynamics, ∆𝐸𝑖𝑛𝑡 = Q – W For a cyclical process, ∆𝐸𝑖𝑛𝑡 = 0 0= Q – W Q = W QAB + QBC + QCA = W +20 + 0 + QCA = 15 QCA = - 5 J Homework: Check your concept A gas within a closed chamber undergoes the cycle shown in the p-V diagram of Fig. The horizontal scale is set by Vs = 4.0 m3. Calculate the net energy added to the system as heat during one complete cycle.