Heat, work, isothermal and Thermal & Kinetic Lecture 15 Heat, work, isothermal and adiabatic processes LECTURE 15 OVERVIEW Recap of previous lecture and overview of final 6 lectures. Isothermal and Adiabatic work Is work a function of state? What distinguishes heat from work? Using thermal processes to do work: Intro. to the Carnot cycle

Last time…. The zeroth law. Isotherms, equations of state. Temperature scales and gas thermometers

To be covered in remaining lectures Lecture 15: Isothermal and adiabatic processes; heat vs work. Lecture 16. The Carnot cycle, heat engines and efficiencies. Lecture 17. Efficiences, entropy, Maxwell’s demon. Lecture 18. Real gases, heat conduction in solids. Lecture 19. Problems class/ previous exam questions. Lecture 20. Problems class/ previous exam questions. NB See website for a “What am I expected to know for the exam?” handout. NB The coursework is an integral part of the module.

Work done by expanding gases: path dependence Back in Section 2 of the module (Section 2.8 of the notes), we wrote down the following expression for the work done by an expanding gas: dW = - PdV reversible process. dx To get total work done we need to integrate:

? ! Work done by expanding gases: path dependence T, V, and P are functions of state. Is W also a function of state? V Three different pathways: Isothermal (constant temperature) Isobaric (constant pressure) Isochoric (constant volume) P Isotherm at T 1 2 V1 V2 P2 P1 Chinese proverb “I hear, and I forget I see, and I remember I do, and I understand.” Given that PV = nRT write down an expression for the work done by a gas when it expands from V1 to V2. ? NB For reversible process. ANS: W = - nRT ln(V2/V1) ! Negative quantity because gas does work on surroundings.

Work done by expanding gases: path dependence Isotherm at T 1 2 V1 V2 P2 P1 3 Isothermal expansion (1  2 on PV diagram) for ideal gas: W = - nRT ln (V2/V1) Write down an expression for the work done in an isobaric expansion of an ideal gas (3  2 on PV diagram). ? ANS: W = P2(V1 – V2)

Work done by expanding gases: path dependence Isothermal expansion (1  2) for ideal gas:W = - nRT ln (V2/V1) Isobaric expansion (3  2) for ideal gas: W = P2 (V1 – V2) P 1 P1 3 2 P2 Isotherm at T V1 V2 Write down an expression for the work done in an isochoric process involving an ideal gas (e.g. 1  3 on PV diagram). ? ANS: 0

Work done by expanding gases: path dependence ? P Shade in the region of the PV diagram that corresponds to the quantity of work done in an isobaric expansion from V1 to V2. 1 P1 ? Now, shade in the region of the PV diagram that corresponds to the quantity of work done in an isothermal expansion from V1 to V2. 3 2 P2 V1 V2 Unlike P, T, and V, work is not a function of state. (…but there is an exception which we’ll come to soon). The amount of work depends on the path taken and there are an infinite number of paths connecting points 1 and 2 in the PV diagram above.

Heat, work and the 1st law The internal energy of an ideal gas is a function of temperature only. (Section 2a of the notes). ? What is DU for an isothermal compression of an ideal gas? What is the corresponding value for an isothermal expansion? ANS: 0 Therefore, from the first law, how are the values of Q and W related in an isothermal process involving an ideal gas? ? ANS: Q= -W

Heat, work and the 1st law: isochoric heat transfer Piston locked in place so V is constant. Consider a process involving isochoric heat transfer to 1 mole of an ideal gas. Q = CV DT = (3R/2) DT DT Gas DQ If Q is (3R/2)DT for the isochoric process shown above, by how much does the internal energy of the gas change? ? ANS: Isochoric, so W = 0, hence DU is also (3R/2)DT

Heat, work and the 1st law: Isobaric heat transfer ? Piston free to move so P is constant as gas expands from V1 to V2 Write down an expression for Q for this isobaric process. ANS: Q = CPDT = (5R/2) DT ? Write down an expression for the work done by the gas. DT Gas ANS: W = -P(V2 – V1) DQ

X ? ? ? Heat, work and the 1st law: Adiabatic compression of an ideal gas In what we’ve considered thus far, heat was transferred into or out of the gas. What happens if compression of a gas is carried out under adiabatic conditions? ? Is it possible for the temperature of the gas to rise if there’s no heat input? X Gas DQ For an adiabatic process, how are the change in internal energy and the work done on the gas related? ? ANS: DU = W (because Q = 0) If the temperature of an ideal gas changes by dT, by how much does its internal energy change? ? ANS: dU = CVdT

? Heat, work and the 1st law: Adiabatic compression of an ideal gas If we can assume that the process is quasistatic and reversible, write down an expression for the amount of work done, dW, at each quasistatic step. ANS: dW = -PdV Gas So,CVdT = -PdV However, we’re dealing with an ideal gas  PV = RT (for 1 mole)

? Heat, work and the 1st law: Adiabatic compression of an ideal gas Integrate the above expression. Constant of integration. We now divide through by CV to get:

Heat, work and the 1st law: Adiabatic compression and expansion of an ideal gas However, we’ve shown in Section 2 that CP = CV + R (Eqn. 2.45). Hence, we can write the equation above as: Rewriting this equation taking into account the properties of logs: Equation of an adiabatic Using PV = RT again:

PV curve for an adiabatic and an isothermal process At each point (P, V), the adiabatic for an ideal gas has a slope g times that of an isotherm for an ideal gas. Isotherm V In CW9 you’ll be asked to show that the statement above is correct.

Adiabatic work At the beginning of the 19th century it was assumed that heat was a substance called caloric which flowed between bodies. Prompted by measurements carried out by Benjamin Thompson, Joule wanted to determine the precise ‘form’ of heat. http://lectureonline.cl.msu.edu/~mmp/kap11/cd295.htm Water stirred by falling weights turning paddle wheel. Water isolated from surroundings by adiabatic walls of container.

Adiabatic work No matter how the adiabatic work was performed, it always took the same amount of work to take the water between the same two equilibrium states (whose temperatures differed by DT) If a thermally isolated system is brought from one equilibrium state to another, the work necessary to achieve this change is independent of the process used. (1st law, conservation of energy). This seems to contradict what was said earlier – it appears that the work done in this case is path independent? Adiabatic work is a function of state – the adiabatic work done is independent of the path DU = Q + W. If Q=0  DU = W

Adiabatic work (c is a constant) Work done by a gas in expanding adiabatically from a state (P1, V1) to a state (P2, V2) In CW9 you’ll again be asked to show that the statement above is correct.

The distinction between heat and work “If both heat and work increase the internal energy of a system, what is the distinction between the two at the microscopic level?” Heat changes the populations of the energy levels (so have change in entropy because there’s a change in the number of accessible microstates.) Works changes the energies, with the populations staying the same.

Using thermal processes to do work: heat engines Carnot noted that work is obtained from an engine because there are heat sources at different temperatures. Furthermore, he realised that heat could also flow from a hot to a cold body with no work being done. A temperature difference may be used to produce work OR it can be ‘squandered’ as heat. Engine How do we convert thermal energy transfer into useful work? (e.g. a steam engine) How efficient can we make this cycle? QH QL W

The most efficient process: the Carnot cycle In an ideal engine the temperature difference between the two reservoirs should yield the maximum amount of work possible. Carnot realised that this meant that all transfers of heat should be between bodies of nearly equal temperature. The Carnot engine involves reversible processes (these are the most efficient processes in terms of exploiting a temperature difference to do work). TH QH Engine W QL TL Heat supplied from high temp. reservoir: QH Heat rejected into lower temp. reservoir: QL A Carnot engine operates between only two reservoirs and is reversible. All the heat that is absorbed is absorbed at a constant high temperature (QH) and all the heat that is rejected is rejected at a constant lower temp. (QL).

The most efficient process: the Carnot cycle Carnot engine is an idealisation. We’ll use an ideal gas as our working substance. Carnot cycle may be constructed from a combination of adiabatic and isothermal compressions and expansions. P A B QH Adiabatic C W Isotherm D QL Animation V