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Thermal & Kinetic Lecture 17 Back to the 1 st law: PV cycles

Thermal & Kinetic Lecture 17 Back to the 1 st law: PV cycles. LECTURE 17 OVERVIEW. Heat and work. PV cycles. Functions of state. Last time…. Entropy and the 2 nd law Reversible and quasistatic processes. dS=dQ/T. Heat and work, engines and entropy.

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Thermal & Kinetic Lecture 17 Back to the 1 st law: PV cycles

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  1. Thermal & Kinetic Lecture 17 Back to the 1st law: PV cycles LECTURE 17 OVERVIEW Heat and work PV cycles Functions of state

  2. Last time… Entropy and the 2nd law Reversible and quasistatic processes. dS=dQ/T

  3. 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?

  4. To get total work done we need to integrate: 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

  5. P • Three different pathways: • Isothermal (constant temperature) • Isobaric (constant pressure) • Isochoric (constant volume) 1 P1 2 P2 Isotherm at T V1 V2 V ? 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. ! Negative quantity because gas does work on surroundings. Work done by expanding gases: path dependence T, V, and P are functions of state. Is W also a function of state? ANS: W = - nRT ln(V2/V1)

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

  7. ? Write down an expression for the work done in an isochoric process involving an ideal gas (e.g. 1  3 onPVdiagram). 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 (V2 – V1) P 1 P1 2 3 P2 Isotherm at T V1 V2 ANS: 0

  8. ? 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. 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 2 3 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.

  9. ? Therefore, from the first law, how are the values of Q and W related in an isothermal process involving an ideal gas? 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 ANS: Q= -W

  10. ? If Q is (3R/2)DT for the isochoric process shown above, by how much does the internal energy of the gas change? 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 ANS: Isochoric, so W = 0, hence DU is also (3R/2)DT

  11. ? Write down an expression for the work done by the gas. 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 DT Gas DQ ANS: W = -P(V2 – V1)

  12. ? Is it possible for the temperature of the gas to rise if there’s no heat input? ? For an adiabatic process, how are the change in internal energy and the work done on the gas related? ? If the temperature of an ideal gas changes by dT, by how much does its internal energy change? 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? X Gas DQ ANS: DU = W (because Q = 0) ANS: dU = CVdT

  13. 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)

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

  15. Rewriting this equation taking into account the properties of logs: Using PV = RT again: 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: Equation of an adiabatic

  16. PV curve for an adiabatic and an isothermal process P Adiabatic At each point (P, V), the adiabatic for an ideal gas has a slope gtimes that of an isotherm for an ideal gas. Isotherm V

  17. Water stirred by falling weights turning paddle wheel. • Water isolated from surroundings by adiabatic walls of container. http://lectureonline.cl.msu.edu/~mmp/kap11/cd295.htm 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.

  18. 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 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)

  19. Work done by a gas in expanding adiabatically from a state (P1, V1) to a state (P2, V2) Adiabatic work (c is a constant)

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