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Home , Adiabatic process, Isobaric process, Isochoric process, Isothermal process, Thermodynamic system

PY105 C1

1. Help for Final exam has been posted on WebAssign.

2. The Final exam will be on Wednesday December 15th from 6-8 pm. First Law of

3. You will take the exam in multiple rooms, divided as follows: SCI 107: Abbasi to Fasullo, as well as Khajah PHO 203: Flynn to Okuda, except for Khajah SCI B58: Ordonez to Zhang

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Heat and done by a Thermodynamics Initial:

Consider a cylinder of ideal Thermodynamics is the study of involving gas at room . Suppose the on top of in the form of and work. the cylinder is free to move vertically without friction. When the cylinder is placed in a container of hot water, heat Equilibrium: is transferred into the cylinder. Where does the heat energy go? Why does the increase?

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The First Law of Thermodynamics The First Law of Thermodynamics The First Law is often written as: Some of the heat energy goes into raising the temperature of the gas (which is equivalent to raising the of the gas). The rest of it does work by raising the piston. ΔEQWint =− Conservation of energy leads to: QEW=Δ + int (the first law of thermodynamics) This form of the First Law says that the change in internal energy of a is Q is the heat added to a system (or removed if it is negative) equal to the heat supplied to the system

Eint is the internal energy of the system (the energy minus the work done by the system (usually associated with the motion of the atoms and/or molecules). via expansion.) So, the First Law is a form of ΔEint is the change in the internal energy, and is proportional to the change in temperature. conservation of energy. W is the work done by the system.

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1 A P-V diagram question Another P-V diagram question

An initially in state 1 progresses to a final state by An ideal gas initially in state 1 progresses to a final state by one of three different processes (a, b, or c). Each of the one of three different processes (a, b, or c). Each of the possible final states has the same temperature. possible final states has the same temperature. For which For which process is the change in internal energy the process is more heat transferred into the ideal gas? largest? Answer: Q = ΔU + W. Since Tf Answer: Because the final is the same for all three, ΔU is temperature is the same, the the same for all three. So change in temperature is the whichever process involves same as well. So, the more work requires more heat. change in internal energy is Because W = the area under the same for all three the P-V curve, it is clear that processes. process c involves more work and thus it requires the most 7 heat. 8

Constant volume vs. constant Constant volume (isochoric) process We have two identical cylinders of ideal gas. Piston 1 is free to move. Piston 2 is fixed so cylinder 2 has a constant No work is done by the gas: W = 0. The P-V diagram is a volume. We put both systems into a reservoir of hot water vertical line, going up if heat is added, and going down if heat and let them come to equilibrium. Which statement is true? is removed. 1.Will the change in internal energy be the same for the two cylinders? If not, which will be bigger? Applying the first law: QE= Δ Ans. Since both systems undergo the same change in int Endothermic Exothermic Temperature and they contain the same amount of gas, For a monatomic ideal gas: they have the same change in internal energy. T2 3 T 2.Will the change in heat be the same? If not, which will be EnRT= 1 bigger? int 2 Ans. Work done by the gas W is nonzero in case 1 while 3 QE= Δ=int nRT Δ W=0 in case 2. By the First Law of Thermodynamics, ΔQ is 2 Notice that T2 > T1. Why? bigger in case 1. 9 10

Constant-volume vs constant- Constant pressure (isobaric) process pressure processes In this case the region on the P-V diagram is rectangular, so its area is easy to find. WPV=Δ From the above, we see that the amount of heat involved in a heating or cooling process (i.e., where the temperature of a system is changed) depends on the details of the process. For a monatomic ideal gas: For constant-volume processes, the heat involved is the QEW=Δint + minimum since no work by the system (PΔV = 0) is involved 3 Endothermic and so Q = U + W = U. =Δ+ΔnR T P V 2 3 For constant-pressure processes, the heat involved is bigger. =Δ+ΔnR T nR T Exothermic If the system is an idea gas, we can calculate what the work 2 done is by using W = PΔV = nRΔT. From that, we can 5 =ΔnR T determined the heat, Q by using Q = U + W. 2 T3 > T1. 11 12

2 for monatomic ideal gas For and liquids: Q = mcΔT, where m is the of the specimen and c is the specific heat per For a monatomic ideal gas kg. ⎛⎞33 Constant volume: QnCTn=Δ=VV⎜⎟ R Δ T, so C = R For : QnCT =Δ , where n is number of moles ⎝⎠22 and C, the specific heat capacity per , depends on the process. ⎛⎞55 Constant pressure: QnCTn=Δ=PP⎜⎟ R Δ T, so C = R ⎝⎠22

where n is the number of moles of molecules contained in the gas. 13 14

Specific heat capacity for general Constant temperature (isothermal) process

ideal gas No change in internal energy: Δ=Eint 0 The P-V diagram follows the isotherm. For a general ideal gas: Applying the first law, and 1 ΔEint per molecule = (nf + 3/2)kT or ΔEint = (nf + 3/2)nRT per using a little calculus: mole, where nf is the number of additional degrees of ⎛⎞Vf freedoms the gas has besides translational degrees of QW== nRTln⎜⎟ freedom. ⎝⎠Vi 2 From this equation, Q > 0

So, CV = Q/n = ΔEint / n (endothermic) if Vf > Vi, but < 0 (exothermic) if V < V . What is Or, CV = (nf + 3/2)RT f i the physical explanation for this? and CP = CV + R Area under an

isotherm = nRTln(Vf/Vi) from W = PΔV = nRT 15 16

An Ideal Gas with Fixed Number of Atoms Zero heat (adiabatic) process A container of monatomic ideal gas contains just the right When Q = 0, the P-V diagram is an interesting line, given by: number of moles so that nR = 20 J/K. The gas is initially in state 1 where PV γ = constant P = 20 kPa, V = 100 x 10-3 m3 C 1 1 1 γ = P CV (a) What is the initial temperature T1 of the gas?

For a monatomic ideal gas: If Q = 2500 J of heat is added to the gas, which expands at constant pressure and reaches a new equilibrium state. C 5/25R γ ==P = 2 CRV 3/23 (b) What is the final temperature T2? (c) How much work is done by the gas? Applying the first law: (d) What is the final volume V ? Note that W = area under the P-V 2 Δ=−EWint curve is > 0 in the process 1→2, but

< 0 in 2→1. 17 18

3 An Ideal Gas with Fixed Number of Atoms An Ideal Gas with Fixed Number of Atoms

Solution (c) With constant pressure, (a) Use the , PV = nRT, so:

-3 3 WPVnRT= Δ= Δ=(20 J/K)( + 50 K) =+ 1000 J (20 kPa)(100 x 10 m ) = (20 J/K)T1 ⇒ T1 = 100 K n.b. The factor of 1000 in the kPa cancels the factor of 10-3 in (d) By the ideal gas law, the volume. nRT (20 J/K)× 150 K (b) With constant pressure for a monatomic ideal gas V ==2 =×150 10−33 m 2 P 20× 103 Pa 35 2 QEW=Δ + = nRTnRTnRT Δ + Δ = Δ int 22 2Q 1000 J ⇒ Δ=T = =50K 520nR J/K

TT21=+Δ= T100 K + 50 K = 150 K

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A Three-Step Process A Three-Step Process

A undergoes a three-step process. (a) nR = 100 J/K; P = P = 100 kPa; V = 0.5 m3; T = ? An adiabatic expansion takes it from state 1 to state 2; then 2 3 3 3 heat is added at constant pressure to move the system to Solution state 3; and finally, an isothermal compression returns the system to state 1. The system consists of a diatomic ideal By the Ideal Gas Law, gas with CV = 5R/2. PV (100×× 1033 Pa) 0.5 m T ==33 =500 K The number of moles is chosen so nR = 100 J/K. 3 nR 100 J/K The following information is known about states 2 and 3.

Pressure: P2 = P3 = 100 kPa 3 Volume: V3 = 0.5 m T = ? (a) What is the temperature of the Isothermal 3 Adiabatic compression system in state 3? 100kPa expansion

Isobaric 21 0.5 m3 22 expansion

A Three-Step Process A Three-Step Process (b) The system does 20000 J of work in the constant pressure Complete the table process that takes it from state 2 to state 3. What is the volume and temperature of the system in state 2? Given that the total work done by the system in the cycle is Solution –19400 J. Complete the table. For constant pressure, we can use: 100kPa Process Q ΔEint W WPVnRT=Δ= Δ 1 to 2 3 W +20000 J 3 T2 = ? 0.5 m Finding volume: Δ=V = =+0.2 m Isothermal P 100000 Pa 2 to 3 +20000 J Adiabatic compression 33 3 3 to 1 expansion Δ=VV32 − Vso V 2 = V 3 −Δ=+ V 0.5 m − 0.2 m = 0.3 m Entire -19400 J Isobaric Finding temperature: (use the ideal gas law, or …) cycle expansion T2 = 300 K W +20000 J T = 500 K Δ=T = =+200K First fill in all the terms that are zero. 3 nR 100 J/K Each row satisfies the First Law of Thermodynamics.

23 Also remember that CV = 5nR/2 24 Δ=TT32 − Tso T 2 = T 3 −Δ= T 500K − 200K = 300K

4 A Three-Step Process A Three-Step Process Complete the table Complete the table For the same system, complete the table. The total work For the same system, complete the table. The total work done by the system in the cycle is –19400 J. done by the system in the cycle is –19400 J.

Process Q ΔEint W Process Q ΔEint W 1 to 2 0 1 to 2 0 Isothermal Isothermal 2 to 3 +20000 J compression 2 to 3 +20000 J Adiabatic Adiabatic compression 3 to 1 0 expansion 3 to 1 0 expansion

Entire 0 -19400 J Isobaric Entire -19400 J 0 -19400 J Isobaric expansion cycle cycle expansion T2 = 300 K T2 = 300 K T3 = 500 K T = 500 K Q is zero for an . Even the last row has to satisfy the first law: 3 The change in internal energy is zero for an isothermal 25 QEW= Δ+ 26 process, and is always zero for a complete cycle. int

A Three-Step Process A Three-Step Process Complete the table Complete the table For the same system, complete the table. The total work For the same system, complete the table. The total work done by the system in the cycle is –19400 J. done by the system in the cycle is –19400 J.

Process Q ΔEint W Q ΔEint W 1 to 2 0 compression Adiabatic 1 to 2 0 -50000 J Isothermal expansion 2 to 3 +50000 J +20000 J Adiabatic compression 2 to 3 +70000 J +50000 J +20000 J Isobaric expansion 3 to 1 0 expansion 3 to 1 0 Isobaric

Entire -19400 J 0 -19400 J T2 = 300 K expansion T3 = 500 K Entire -19400 J 0 -19400 J cycle T2 = 300 K cycle T3 = 500 K Find the change in internal energy for the 2 Æ 3 process. Rows have to obey the first law. 55 Δ=EnCTnRT Δ=Δ=(100 J/K)( + 200K) =+ 50000 J 27 Columns have to sum to the value for the entire cycle. 28 int V 22

A Three-Step Process Three Ways of Adding The Same Heat, Q Complete the table An ideal gas is contained in a cylinder, which is sealed at the top by a piston that can move up or down, or can be fixed in For the same system, complete the table. The total work place to keep the volume constant. The weight of the piston can be adjusted to adjust the pressure as necessary. Starting done by the system in the cycle is –19400 J. with the same initial conditions, you do three experiments, each involving adding the same amount of heat, Q. Process Q ΔE W int A – Add the heat at constant pressure. 1 to 2 0 -50000 J +50000 J Isothermal Adiabatic compression B – Add the heat at constant temperature. 2 to 3 +70000 J +50000 J +20000 J expansion C – Add the heat at constant volume. 3 to 1 -89400 J 0 -89400 J Isobaric expansion (a)Rank the processes by the final temperature. Entire -19400 J 0 -19400 J T = 300 K 2 (b)Rank the processes by the work. cycle T3 = 500 K (c)Rank the processes by the final pressure. Rows have to obey the first law. Columns have to sum to the value for the entire cycle. 29 30

5 Three Ways of Adding The Same Heat, Q Three Ways of Adding The Same Heat, Q (b) Rank by work: Solution: In process C (constant V), no work is done.

In process A (constant P), W = Q – ΔEint ⇒ only some of the heat (a) Rank by final temperature: added goes to doing work.

In process B (constant T), W = Q – ΔEint = Q – CVΔT = Q ⇒ all the ΔU = nCV(Tf – Ti) = Q – W. Hence the bigger Q – W is, the heat added goes to doing work. bigger Tf would be. B > A > C In process B (constant T), there is no change in T. (c) Rank by final pressure: In process A (constant P), some of the heat added goes to doing work. Pf = nRTf/Vf In process C (constant V), the gas does no work and all the In process A (constant P), the pressure stays constant. heat added goes to increasing the temperature. In process B (constant T), W = Q > 0 ⇒ΔV > 0. With T constant, the pressure must decrease. C > A > B In process C (constant V), ΔEint = Q > 0 ⇒ΔT > 0. With V constant, the pressure must increase. 31 32 C > A > B

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