Presentation on theme: "Thermal & Kinetic Lecture 14 Thermodynamics: The 0 th and 1 st laws Recap…. The 1 st law: heat and work Ideal gas scale. LECTURE 14 OVERVIEW Gas thermometers."— Presentation transcript:
Thermal & Kinetic Lecture 14 Thermodynamics: The 0 th and 1 st laws Recap…. The 1 st law: heat and work Ideal gas scale. LECTURE 14 OVERVIEW Gas thermometers Is work a function of state?
Last time…. Calculation of entropy and ‘abuses’ of the 2 nd law. The 0 th law and thermal equilibrium
This is the second year that the Thermal & Kinetic module has run in this form (10 credits). Large amount (but not complete) overlap with ‘Heat and Properties of Matter’ 5 credit module which ran until last year. Relevant exam questions from ‘Heat and Properties of Matter’ papers: 2002Q2, Q4, Q5 (all elements of Q5 not covered yet) 2001Q1, Q5 (again, not all elements of Q5 covered) 2000Q1, Q2, Q4 1999Q2, Q4 and some of Q5 Previous exam questions
Website All notes and slides used up to this point in the module will be available on the website from the beginning of next week.(Slides up to Lecture 10 already on website).
Temperature scales and gas thermometers Figure taken from Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed. Figure 12.03. Environment to be measured To pressure measurement Fixed volume of gas Gas thermometers always agree at all points on the temperature scale: they are independent of the ‘material’ used in the thermometer.
Temperature scales and gas thermometers Volume of gas kept constant by adjusting height of mercury column until meniscus is at ‘0’ on the scale. Bulb of gas immersed in system whose temperature is to be measured. A ‘dead space’ or ‘nuisance volume’ is sometimes referred to when discussing gas thermometers. From where does this ‘nuisance volume’ arise?? ? Assuming that the temperature at the triple point of water is 273.16 K, write down an equation that relates T and P of the gas to T and P at the triple point. ANS: T gas = 273.16 (P gas /P TP )
Temperature scales and gas thermometers When the amount of working gas is reduced to the smallest possible value, all gas thermometers give the same temperature for a given system irrespective of the gas used.? Why !? ANS: …..because we’ve reached the limit of an ideal gas Define ideal gas scale as: The limit is performed by carrying out an extrapolation……………
Temperature scales and gas thermometers Absolute zero defined as -273.15°C The value of 273.16 K for the triple point was chosen so as to ensure that there are 100 K between the experimentally measured temperatures of the ice and steam points. Celsius scale is given by: T (°C) = T(K) – 273.15 NB The temperature defined according to the ideal gas scale is identical to the absolute thermodynamic temperature scale (as used, for example, in exp (- E/kT)).
Heat and work, engines and entropy We have some big questions left to answer: How can we use thermal processes to do mechanical work? What distinguishes heat from work? What is the maximum efficiency we can expect from a heat engine? How is the maximum efficiency of a heat engine related to entropy? How do we know the ideal gas scale is identical to the absolute temperature scale?
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 = - PdVreversible process. To get total work done we need to integrate: dx
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 V1V1 V2V2 P2P2 P1P1 Given that PV = nRT write down an expression for the work done by a gas when it expands from V 1 to V 2.? ANS: W = - nRT ln(V 2 /V 1 )! Negative quantity because gas does work on surroundings. NB For reversible process. Chinese proverb “I hear, and I forget I see, and I remember I do, and I understand.”
Work done by expanding gases: path dependence Isothermal expansion (1 2 on PV diagram) for ideal gas: W = - nRT ln (V 2 /V 1 ) P Isotherm at T 1 2 V1V1 V2V2 P2P2 P1P1 3 Write down an expression for the work done in an isobaric expansion of an ideal gas (3 2 on PV diagram). ? ANS: W = P 2 (V 1 – V 2 )
Work done by expanding gases: path dependence P Isotherm at T 1 2 V1V1 V2V2 P2P2 P1P1 3 Isothermal expansion (1 2) for ideal gas:W = - nRT ln (V 2 /V 1 ) Isobaric expansion (3 2) for ideal gas: W = P 2 (V 1 – V 2 ) 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 1 2 V1V1 V2V2 P2P2 P1P1 3 Shade in the region of the PV diagram that corresponds to the quantity of work done in an isobaric expansion from V 1 to V 2. ?? Now, shade in the region of the PV diagram that corresponds to the quantity of work done in an isothermal expansion from V 1 to V 2. Unlike P, T, and V, work is not a function of state. 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. (…but there is an exception which we’ll come to soon).