hap First Law of hermodynamics Isochoric process: Isochoric rocess: hange in the state at constant volume,, ƌq /d (U / ) ; m mol J K - mol - kg J K -

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1 hap First Law of hermodynamics.6 hange in State at onstant olume : : heat capacity : ƌq / d (ƌ: inexact differentials, ƌq ) U,, du,, :,, U f (, ) : du (U/ ) d + (U/) d, du ƌq ex d (for -work only, ƌw - ex d ) ƌq du + ex d (U/ ) d + [ ex + (U / ) ] d X : X (, ) X (, ) ƌq (U/ ) d

2 hap First Law of hermodynamics Isochoric process: Isochoric rocess: hange in the state at constant volume,, ƌq /d (U / ) ; m mol J K - mol - kg J K - kg - U d q, U ( )

3 hap First Law of hermodynamics U ( ) H ( ) Fig.. he heat q absorbed by a substance when it is heated is equal to the initial temperature to the final temperature. If an amount n is heated at constant volume, n, and if an amount n is heated at constant pressure, n

hap First Law of hermodynamics Internal Energy hange vs.,,. he internal energy of a system varies with volume and temperature, perhaps as shown here by the surface.

4 hap First Law of hermodynamics Internal Energy hange vs.,,. he internal energy of a system varies with volume and temperature, perhaps as shown here by the surface. he variation of the internal energy with temperature at one particular constant volume is illustrated by the curve drawn parallel to. he slope of this curve at any point is the partial derivative (U/ ) v,for (U/ ) too

5 hap First Law of hermodynamics he Heat apacity he internal energy of a system increases as the temperature is raised; this graph shows its variation as the system is heated at constant volume. he slope of the graph at any temperature (as shown by the tangents at A and B) is the heat capacity at constant volume at that temperature. Note that, for the system illustrated, the heat capacity is greater at B than at A

6 hap First Law of hermodynamics Sum of the hanges o achieve a change of state from one temperature and volume to another temperature and volume, we may consider the overall change as composed of two steps. In the first step, the system expands at constant temperature; there is no change in internal energy if the system consists of a perfect gas. In the second step, the temperature of the system is increased at constant volume. he overall change in internal energy is the sum of the changes for the two steps

James rescott Joule James rescott Joule (88-889) is the scientist established the First Law of hermodynamics. When he was 4, he began his weekly chemistry lessons with John Dalton.

7 James rescott Joule James rescott Joule (88-889) is the scientist established the First Law of hermodynamics. When he was 4, he began his weekly chemistry lessons with John Dalton. By 838, Joule had published his first paper. His determination of the principle of conservation of energy is considered to be the chief contribution of the nineteenth century to physical science

8 hap First Law of hermodynamics (Gay-Lussac) 807 (Joule) 843, ex 0 dq0 dw0 dudq U U du d 0, since d 0 0 ( )

9 hap First Law of hermodynamics U 0 Fig.. Joule s experiment in which a gas expands into a vacuum. Joule found there was no discernible change in temperature and concluded that (U/) 0. We now know that this applies to ideal gases but not to real gases

10 hap First Law of hermodynamics Example.5 Heat absorbed in expansion of ideal gas alculate the heat absorbed and the change in internal energy for the two expansions of ideal gas described in Example.3 Answer: (a) According to Joule s experiment, U 0 for the reversible isothermal expansion of an ideal gas. herefore, q rev U w rev 0 (-3988 J) 3988 J (b) q irrev U w irrev 0 (-98 J) 98 J hus, more heat is absorbed by the gas in the reversible isothermal expansion

hap First Law of hermodynamics onstant ressure Expansion When a system is subjected to constant pressure and is free to change its volume, some of the energy

11 hap First Law of hermodynamics onstant ressure Expansion When a system is subjected to constant pressure and is free to change its volume, some of the energy supplied as heat may escape back into the surroundings as work. In such a case, the change in internal energy is smaller than the energy supplied as heat

12 hap First Law of hermodynamics.7 Enthalpy and hange in State at onstant ressure isobaric process):, work., w -. U q p, q p p v q p,. U -U U q p ( - ) q p (U + ) - (U + )

13 hap First Law of hermodynamics Definition of a new state function as enthalpy (H). Enthalpy: H U + ( : pressure of system) Now q p H H H H is an extensive quantity of heat supplied to system at constant pressure process ( ex ). H is path independent, the heat absorbed in a process at constant pressure is equal to the change in enthalpy

14 hap First Law of hermodynamics For an infinitesimal change at constant pressure ƌq dh For chemically inert system and fixed mass, it is most convenient to take enthalpy H as a function of temperature and pressure H f (, ) dh (H/ ) p d +(H/ ) d, also X : X(, ) X(, ) ƌq p (H/ ) d dh For -work only, ƌq / d (H/ ) Define the heat capacity at constant pressure: (H/ ) and H d q

15 hap First Law of hermodynamics.8 Heat apacities,,. (able.. at 5 for ~00 substances), (able.3. at K) : p + + Enthalpy, : H H H d H H dh H α ( ) ( ) ( ) β

16 hap First Law of hermodynamics Example.6 he change in molar enthalpy on heating Using the table.3 to calculate the change in the molar enthalpy of methane in going from 500 to 000 K. Answer: o o 500 ( ) ( ) o o o o H 000 H H H H H ( ) kj / mol kj/mol

17 hap First Law of hermodynamics Enthalpy versus emperature he slope of a graph of the enthalpy of a system subjected to a constant pressure plotted against temperature is the constant-pressure heat capacity. he slope of the graph may change with temperature, in which case the heat capacity varies with temperature. hus, the heat capacities at A and B are different. For gases, the slope of the graph of enthalpy versus temperature is steeper than that of the graph of internal energy versus temperature, and p,m is larger than,m

18 hap he First Law: the concepts Example Evaluating an increase in enthalpy with temperature What is the change in molar enthalpy of N when it is heated from 5 to 00? Use the heat capacity information in able. Answer: For convenience, we denote the two temperatures (98 K) and (373 K). he integrals we require are H Now we use the integrals: to obtain H ( α + β ) + H d H d H (373 K)H (98 K)+.0 kj mol - dx x x dx ( ) ( ) ( ) ( ) ( 3 3 H α ) β 3 dx x + const x + const 3 x + const

19 he Relation Between p and v for a real Gas :, :..., du (U / ) d + (U / ) d dq ex d ƌq (U / ) d + [ ex + (U / ) ] d, ex. dq p d + [ + (U / ) ] d d (ƌq p /d ) hap First Law of hermodynamics U U q U q x d d d

20 : Here is the cubic expansion coefficient, :. is the internal pressure: herefore _ (p + ) ( / ) / Here is the isothermal compressibility, hap First Law of hermodynamics > + U U H U πα α α - κ - - U κ κ α π

21 hap First Law of hermodynamics For an ideal Gas _ [ + (U / ) ] ( / ) ( / ) nr. (U / ) ( / ) 0 _ nr _,m,m R nr, K, nr ( K ), K

22 hap First Law of hermodynamics Kinetic theory shows that molar translation energy of a monoatomic ideal gas is 3R/ For monoatomic ideal gas, the translation energy U t,m independent of pressure or molar mass. So and U t,m 3R/ and H t,m U t,m + 3R/ + R 5R/,m (U t,m / ) 3R/.47 J/K-mol,m (H t,m / ) 5R/ J/K-mol Dulong and etit law: At room temperature and above, the heat capacities of solids elements heavier than K are about 3R (~5 J/K-mol ) is

23 hap First Law of hermodynamics able he Heat apacities of arious ypes of Molecules ypes of Molecule ranslations Rotations ibrations v +R Momoatomic 3R/ 0 0 3R/ 5R/ Diatomic 3R/ R/ R 7R/ 9R/ riatomic linear 3R/ R/ 4R 3R/ 5R/ nonlinear 3R/ 3R/ 3R 6R 7R

24 hap First Law of hermodynamics Fig..3 he effect of temperature on the molar heat capacities of several gases at constant pressure. Starting with the highest curve at 00 K, the gases are H 4, NH 3, O, H O, N and He

25 hap First Law of hermodynamics Fig.6 Heat capacity data for several solid elements plotted versus / D, where D is the Debye temperature

26 hap First Law of hermodynamics.9 Joule-homson Expansion( ) Joule-homson effect (throttling process) 85-6 James. Joule and William homson (porous plate )

27 hap First Law of hermodynamics Joule-homson expansion. he transition from the top diagram to the bottom diagram, which represents the passage of a given amount of gas through the throttle, occurs without change of enthalpy

28 hap First Law of hermodynamics Joule-homson coefficient enthalpy

29 hap First Law of hermodynamics H µ J. For expansion, dp < 0 if µ > 0, d J < 0,if µ J < 0, d > 0 Fig.3 Joule-homson expansion

30 hap First Law of hermodynamics :, ( < ). ( > ), : w U - U (q 0 ) (U + ) (U + ) or H H (isenthalpic) :. Joule-homson coefficient: J H p J 0 lim µ H p H 0 H d d d + H J H H H µ

hap First Law of hermodynamics Isenthalps and inversion curve for nitrogen. Inside the boundary, the shaded area, sign of is positive and outside is negative.

31 hap First Law of hermodynamics Isenthalps and inversion curve for nitrogen. Inside the boundary, the shaded area, sign of is positive and outside is negative. he temperature corresponding to the boundary at a given pressure is the `inversion temperature' ( I ) of the gas at that pressure. Reduction of p under adiabatic conditions moves the system along one of the isenthalps curves. he I curve runs through the points of the isenthalps where their slopes change from negative to positive

32 hap First Law of hermodynamics, J >0,., H, He. J < 0,, J >0, J <0,. -, J (/) H, (Inversion urve)., J >0.., J <0,.,, J 0. :, J I 43 K,?

hap First Law of hermodynamics he inversion temperatures, I he inversion temperatures for three real gases, nitrogen (607 K), hydrogen (04 K), and helium (43 K).

33 hap First Law of hermodynamics he inversion temperatures, I he inversion temperatures for three real gases, nitrogen (607 K), hydrogen (04 K), and helium (43 K). For a given p, the must be below a certain value if cooling is required but, if it becomes too low, the boundary is crossed again and heating occurs

34 Example Joule-homson coefficient of an ideal gas Show that the Joule-homson coefficient for ideal gas is zero. Answer: hap First Law of hermodynamics H H H µ J ( ) 0 / nr H

35 hap First Law of hermodynamics.0 Adiabatic rocesses with Gases (q 0) : du q + w w ( q 0) du - ext d (d <0) (du >0, d >0) (d (du <0, d <0), du - ext d U U -U w

hap First Law of hermodynamics Adiabatic Expansion An adiabat depicts the variation of pressure with volume when a gas expands reversibly and adiabatically.

36 hap First Law of hermodynamics Adiabatic Expansion An adiabat depicts the variation of pressure with volume when a gas expands reversibly and adiabatically. (a) An adiabat for a perfect gas. (b) Note that the pressure declines more steeply for an adiabat than it does for an isotherm because the temperature decreases in the former

37 hap First Law of hermodynamics, U : du,m d, du,m d w U U -U,m ( - ),m ex,m d - d -(R/) d,m d/ -R d/,m ln ( / ) -R ln ( / ) R ln ( / ), d R d R,m

38 , - R ln ( / ) R ln ( / ) ( )ln( / ) hap First Law of hermodynamics p p - p p ln ln + - p p ln ln ln,m R p p x p p

39 , hap First Law of hermodynamics ( ) R then R where where,m R

40 Justification.3 he initial and final states of a perfect gas satisfy the perfect gas law regardless of how the change of state takes place, so we can use p nr to write However, we know that for a reversible adiabatic change the temperature changes so as to satisfy where we use the definition of heat capacity ratio. hen we combine the twp expressions to obtain It follows that constant c - i f c i f f i f f i i p p f i f f i i p p p i i c,m /R.

41 hap First Law of hermodynamics AB : p A A p B B A : p A A p p / v > i, f, B >

42 hap First Law of hermodynamics Example constant polytropic processes.? Answer: constant d const for, d const d const const ( d const d ) ( )

43 hap he First Law: the concepts Illustration.6 When a sample of argon (for which 5/3) at 00 ka expands reversibly and adiabatically to twice its initial volume, the final pressure will be f i i f i f i f 5 3 ( 00 ka) 3.5ka For an isothermal doubling of volume, the final pressure would be 50 ka

44 hap First Law of hermodynamics Fig..5 Isothermal and reversible adiabatic expansions of one mole of an ideal monoatomic gas. :(AB) :(A)

45 hap First Law of hermodynamics Example.7 Reversible adiabatic expansion of a monatomic ideal gas When one mole of an ideal gas is allowed to expand adiabatically and reversibly from.7 L/mol at bar and 0 (at point A on the graph) to a volume of 45.4 L/mol (at point ), the pressure drop to 0.35 bar. onfirm this pressure and calculate the temperature at. How much work is done in the adiabatic expansion? Answer:

46 hap First Law of hermodynamics Example.7 Answer: R R L mol bar L mol -.7 L mol ( 73.5K) L mol 7.07 K (or 0.08 ) 3 w d R ( ) bar 3 ( ) 7.7 K 73.5K -6J mol

47 hap First Law of hermodynamics c,m /R. Adiabatic Expansions he variation of temperature as a perfect gas is expanded reversibly and adiabatically. he curves are labelled with different values of c,m /R. Note that the temperature falls most steeply for gases with low molar heat capacity

3 (s05q6) The diagram shows the velocity-time graph for a lift moving between floors in a building. The graph consists of straight line segments. In t

3 (s05q6) The diagram shows the velocity-time graph for a lift moving between floors in a building. The graph consists of straight line segments. In t Mechnics (9709) M1 Topic 1 : s-t and v-t graph(9) Name: Score: Time: 1 (s03q3) The diagram shows the velocity-time graphs for the motion of two cyclists P and Q, whotravel in the same direction along a