Chapter 1 Solutions Engineering and Chemical Thermodynamics 2e Wyatt Tenhaeff Milo Koretsky

Transcription

1 Chapter 1 Solutions Engineering and Chemical Thermodynamics 2e Wyatt Tenhaeff Milo Koretsky School of Chemical, Biological, and Enironmental Engineering Oregon State Uniersity

2 1.1 (b) The olume of water contaminating the wine is equal to the olume of wine contaminating the water. 2

3 1.2 (b) The amount of pennies in the jar of mostly nickels is the equal to the amount of nickels in the jar of mostly pennies. After step (1) there are 10 pennies and 90 nickels in the first jar. We can reason this problem assuming the jars get well mixed. After step (2) there are 9 nickels and 1 penny transferred back to the other jar. That leaes 9 pennies in the first jar and 9 nickels in the second jar. Een if it is not perfectly mixed, the same type of reasoning leads to choice (b).

4 1. 4

5 1.5 Derie the following expressions by combining Equations 1.4 and 1.5: V 2 a kt m a V 2 b kt m b Therefore, 2 V a 2 Vb m m b a Since m b is larger than m a, the molecules of species A moe faster on aerage. 5

6 1.6 (a) After a short time, the temperature gradient in the copper block is changing (unsteady state), so the system is not in equilibrium. (b) After a long time, the temperature gradient in the copper block will become constant (steady state), but because the temperature is not uniform eerywhere, the system is not in equilibrium. (c) After a ery long time, the temperature of the reseroirs will equilibrate; The system is then homogenous in temperature. The system is in thermal equilibrium. 6

7 1.7 The amount of water in the liquid will decrease. The apor pressure of water increases with temperature and the energy deliered to the system will cause some of the water molecules to enter the apor phase. 7

8 1.9 The uration pressure is the pressure at which a pure substance boils at a gien temperature. Vapor pressure is a substance s contribution to the total pressure in a mixture at a gien temperature, assuming ideal gases. Saturation pressure is used to describe a pure substance while apor pressure is used to describe a substance s contribution to an ideal gas mixture. 8

9 1.10 It is not a mistake. 9

10 1.11 Assuming a closed system, as the temperature decreases the pressure decreases and the lid is forced down by atmospheric pressure. 10

11 1.12 We consider the air inside the soccer ball as the system. We can answer this question by looking at the ideal gas law: P RT If we assume it is a closed system, it will hae the same number of moles of air in the winter as the summer. Howeer, it is colder in the winter (T is lower), so the ideal gas law tells us that P will be lower and the balls will be under inflated. Alternatiely, we can argue that the higher pressure inside the ball causes air to leak out oer time. Thus we hae an open system and the number of moles decrease with time leading to the under inflation. 11

12 1.1 As defined in the problem statement, the relatie humidity can be calculated as follows mass of water Relatie Humidity mass of water at uration We can obtain the uration pressure at each temperature from the steam tables. At 10 [ o C], the uration pressure of water is 1.22 [kpa]. This alue is proportional to the mass of water in the apor at uration. For 90% relatie humidity, the partial pressure of water in the apor is: p Water [kPa] This alue represents the apor pressure of water in the air. At 0 [ o C], the uration pressure of water is 4.25 [kpa]. For 590% relatie humidity, the partial pressure of water in the apor is: p Water [kPa] Since the total pressure in each case is the same (atmospheric), the partial pressure is proportional to the mass of water in the apor. We conclude there is about twice the amount of water in the air in the latter case. 12

13 (a) When you hae extensie ariables, you do not need to know how many moles of each substance are present. The olumes can simply be summed. V a V b V + 1 (b) The molar olume, 1, is equal to the total olume diided by the total number of moles. b a b a tot n n V V n V We can rewrite the aboe equation to include molar olumes for species a and b. b b a b a b a a b a b b a a n n n n n n n n n n b b a a x x + 1 (c) The relationship deeloped in Part (a) holds true for all extensie ariables. K a K b K + 1 (d) b b a a k x k x k + 1

14 1.15 The molar density of the two systems would be the same. 14

15 1.16 The mass density of system II would be larger than system I because both systems would hae the same amount of moles, i.e., 15

16 1.17 You can breathe in the same olume of helium because the system (your lungs) is at the same temperature and pressure. Assuming number of moles is constant 16

17 1.18 Increasing pressure will increase the boiling point temperature of the liquid, allowing you to cook food at a higher temperature. 17

18 1.19 The ariables of the equation are thermodynamic properties. Two independent, intensie properties constrain the state of a pure substance. The equation relates thermodynamic properties in a gien state. 18

19 1.20 To sole these problems, the steam tables were used. The alues gien for each part constrain the water to a certain state. In most cases we can look at the urated table, to determine the state. (a) (b) (c) (d) Subcooled liquid Explanation: the uration pressure at T 170 [ o C] is 0.79 [MPa] (see page 508); Since the pressure of this state, 10 [bar], is greater than the uration pressure, water is a liquid. Saturated apor-liquid mixture Explanation: the specific olume of the urated apor at T 70 [ o C] is 5.04 [m /kg] and the urated liquid is [m /kg] (see page 508); Since the olume of this state, [m /kg], is in between these alues we hae a urated aporliquid mixture. Superheated apor Explanation: the specific olume of the urated apor at P 60 [bar] 6 [MPa], is [m /kg] and the urated liquid is [m /kg] (see page 511); Since the olume of this state, 0.05 [m /kg], is greater than this alue, it is a apor. Superheated apor Explanation: the specific entropy of the urated apor at P 5 [bar] 0.5 [MPa], is [kj/(kg K)] (see page 510); Since the entropy of this state, [kj/(kg K)], is greater than this alue, it is a apor. In fact, if we go to the superheated water apor tables for P 500 [kpa], we see the state is constrained to T 200 [ o C]. 19

20 1.21 An approximate solution can be found if we combine Equations 1.4 and 1.5: 1 2 molecular m V e 2 k molecular 2 kt e k V kt m 26 Assume the temperature is 22 ºC. The mass of a single oxygen molecule is m kg. Substitute and sole: V [ m/s] The molecules are traeling really, fast (around the length of fie football fields eery second). Comment: We can get a better solution by using the Maxwell-Boltzmann distribution of speeds that is sketched in Figure 1.4. Looking up the quantitatie expression for this expression, we hae: f ( ) d 4π 2 m πkt / 2 exp 2 m kt 2 2 d where f() is the fraction of molecules within d of the speed. We can find the aerage speed by integrating the expression aboe V 0 0 f ( ) d f ( ) d 8kT πm 449 [ m/s] 20

21 1.22 We hae the following two points that relate the Reamur temperature scale to the Celsius scale: ( 0 º C, 0 º Reamur) and ( 100 º C, 80 º Reamur) Create an equation using the two points: At 22 ºC, ( º Reamur) 0.8 T( º Celsius) T T 17.6 º Reamur 21

22 1.2 We assume the temperature is constant at 0 ºC. The molecular weight of air is MW 29 g/mol kg/mol Find the pressure at the top of Mount Eerest: ( ) P 1atm exp P 0.0 atm.4 kpa Interpolate steam table data: ( kg/mol) ( 9.81[ m/s] )( 8848 m) J 8.14 ( K) mol K T 71.4 º C for P.4 kpa Therefore, the liquid boils at 71.4 ºC. Note: the barometric relationship gien assumes that the temperature remains constant. In reality the temperature decreases with height as we go up the mountain. Howeer, a solution in which T and P ary with height is not as straight-forward. 22

23 1.24 From the steam tables in Appendix B.1: m ˆ critical ( T ºC, P MPa) At 10 bar, we find in the steam tables m ˆ l m ˆ kg Because the total mass and olume of the closed, rigid system remain constant as the water condenses, we can deelop the following expression: critical ( 1 x) ˆ xˆ ˆ + l where x is the quality of the water. Substituting alues and soling for the quality, we obtain x or 1.05 % A ery small percentage of mass in the final state is apor. 2

24 1.25 The calculation methods will be shown for part (a), but not parts (b) and (c) (a) Use the following equation to estimate the specific olume: ( 1.9 MPa, 250 ºC) ˆ ( 1.8 MPa, 250 º C) + 0.5[ ˆ ( 2.0 MPa, 250 º C) ˆ ( 1.8 MPa, 250 º C) ] ˆ Substituting data from the steam tables, ˆ ( 1.9 MPa, 250 ºC) From the NIST website: ˆNIST ( 1.9 MPa, 250 ºC) m m Therefore, assuming the result from NIST is more accurate ˆ ˆ % Difference NIST 100 % % ˆ NIST (b) Linear interpolation: ˆ NIST website: Therefore, ( 1.9 MPa, 00 º C) ˆNIST ( 1.9 MPa, 00 ºC) % Difference (c) Linear interpolation: ˆ m % ( 1.9 MPa, 270 º C) m m

25 Note: Double interpolation is required to determine this alue. First, find the molar olumes at 270 ºC and 1.8 MPa and 2.0 MPa using interpolation. Then, interpolate between the results from the preious step to find the molar olume at 270 ºC and 1.9 MPa. NIST website: Therefore, ˆNIST ( 1.9 MPa, 270 º C) % Difference 0.17 % m With regards to parts (a), (b), and (c), the alues found using interpolation and the NIST website agree ery well. The discrepancies will not significantly affect the accuracy of any subsequent calculations. 25

26 1.26 For urated temperature data at 25 ºC in the steam tables, ˆ l m kg 1.00 L kg Determine the mass: m V ˆ l [ L] L kg kg Because molar olumes of liquids do not depend strongly on pressure, the mass of water at 25 ºC and atmospheric pressure in a one liter should approximately be equal to the mass calculated aboe unless the pressure is ery, ery large. 26

27 1.27 First, find the oerall specific olume of the water in the container: 1m ˆ 5 kg m 0.2 Examining the data in the urated steam tables, we find l ˆ ˆ < at P 2 bar ˆ < Therefore, the system contains urated water and the temperature is T T ºC Let m l represent the mass of the water in the container that is liquid, and m represent the mass of the water in the container that is gas. These two masses are constrained as follows: m l + m 5 kg We also hae the extensie olume of the system equal the extensie olume of each phase ml ˆ l + mˆ V m m / kg m m /kg 1 m l ( ) ( ) + Soling these two equations simultaneously, we obtain m l.88 kg m 1.12kg Thus, the quality is m x m 1.12 kg kg The internal energy relatie to the reference state in the urated steam table is U mluˆ l + muˆ From the steam tables: 27

28 Therefore, ˆ u l ˆ u [ kj/kg] [ kj/kg] (.88 kg) ( [ kj/kg] ) + ( 1.12 kg) ( [ kj/kg] ) U U kj 28

29 1.28 First, determine the total mass of water in the container. Since we know that 10 % of the mass is apor, we can write the following expression V m [( 0.9) ˆ l + ( 0.1) ˆ ] From the urated steam tables for P 1MPa ˆ l ˆ m m Therefore, V 1m m ( ) ˆ + ( ) ˆ l m + m m 48.9 kg and ( ) ˆ 0.1 m 0.1( 48.9 kg)( m / kg) V V m 29

30 1.29 In a spreadsheet, make two columns: one for the specific olume of water and another for the pressure. First, copy the specific olumes of liquid water from the steam tables and the corresponding uration pressures. After you hae finished tabulating pressure/olume data for liquid water, list the specific olumes and uration pressures of water apor. Eery data point is not required, but be sure to include extra points near the critical alues. The data when plotted on a logarithmic scale should look like the following plot. The Vapor-Liquid Dome for H 2 O Critical Point Pressure, P (MPa) Liquid Vapor-Liquid Vapor Specific Volume, (m /kg) 0

31 1.0 The ideal gas law can be rewritten as RT P For each part in the problem, the appropriate alues are substituted into this equation where L bar R mol K is used. The alues are then conerted to the units used in the steam table. (a) L 0.7 mol L 0.7 mol ˆ 10 MW kg mol m m 1.70 L kg For part (a), the calculation of the percent error will be demonstrated. The percent error will be based on percent error from steam table data, which should be more accurate than using the ideal gas law. The following equation is used: ˆ ˆ %Error IG ST ˆ ST 100 % where IG is the alue calculated using the ideal gas law and ST is the alue from the steam table. m m kg kg %Error % 1.62% m kg This result suggests that you would introduce an error of around 1.6% if you characterize boiling water at atmospheric pressure as an ideal gas. 1

32 (b) L 64. mol m ˆ.57 % Error 0.126% For a gien pressure, when the temperature is raised the gas behaes more like an ideal gas. (c) (d) L 0.64 mol m ˆ %Error 8.87% 1.06 L mol m ˆ %Error 0.82% The largest error occurs at high P and low T. 2

33 1.1 The room I am sitting in right now is approximately 16 ft long by 12 feet wide by 10 feet tall your answer should ary. The olume of the room is V ( 12 ft)( 16 ft)( 10 ft) 1920 ft 54.4 m The room is at atmospheric pressure and a temperature of 22 ºC. Calculate the number of moles using the ideal gas law: 5 ( Pa)( 54.4 m ) PV n RT J 8.14 mol K n 2246 mol ( K) Use the molecular weight of air to obtain the mass: ( ) ( mol)( kg/mol) 65.2 kg m n MW That s pretty heay.

34 1.2 First, find the total olume that one mole of gas occupies. Use the ideal gas law: RT P J 8.14 ( K) mol K m ( 1 10 Pa) mol Now calculate the olume occupied by the molecules: occupied occupied occupied Number of molecules Volume of 4 N A π r per mole one molcule molecules 4 π ( m) 1mol 6 m mol Determine the percentage of the total olume occupied by the molecules: occupied Percentage 100 % % A ery small amount of space, indeed. 4

35 1. Water condenses on your walls when the water is in liquid-apor equilibrium. To find the maximum allowable density at 70 ºF, we need to find the smallest density of water apor at 40 ºF which isfies equilibrium conditions. In other words, we must find the uration density of water apor at 40 ºF. From the steam tables, m ˆ (. water apor at 40 ºF 4.44 ºC) We must correct the specific olume for the day-time temperature of 70 ºF (21.1 ºC ) with the ideal gas law. ˆ, K ˆ 294. K T277.6 T294. ˆ Therefore, 294. K 1 ˆ ρ ˆ T T ( ˆ ) 294.,277.6 K kg m 294. K m m K If the density at 70 ºF is greater than or equal to [ kg/m ] wall. Howeer, if the density is less than [ kg/m ], the density at night when the temperature is 40 ºF will be greater than the uration density, so water will condense onto the, then uration conditions will not be obtained, and water will not condense onto the walls. 5

36 1.4 First, write an equation for the olume of the system in its initial state: l ml ˆ + m ˆ V Substitute m l ( 1 x)m and m xm : m [( 1 x) ˆ + xˆ ] V At the critical point mˆ critical V l Since the mass and olume don t change critical ( 1 x) ˆ xˆ ˆ + From the steam tables in Appendix B.1: l m ˆ critical ( T ºC, P MPa) At 0.1 MPa, we find in the steam tables ˆ l ˆ m m Therefore, soling for the quality, we get x or % % of the water is liquid. 6

37 1.5 (a) From the steam tables, this is superheated apor (steam) and the alue is ˆ m kg (b) This case is subcooled liquid water; howeer, the properties of a liquid do not ary much with pressure, so we can use the urated table at 100 o C: ˆ l m kg 7

38 1.6 We need to specify two independent intensie properties to constrain the state of the system. We can use T 90 o C and the quality x 0.8. From the steam tables, we get: P 70.1 kpa and ˆ kg m ( 1 x) ˆ + xˆ 1.89 l 8

39 1.7 We need two intensie properties. We can use the information in the problem statement to calculate the specific olume: ˆ V m m 0.25 kg From the steam tables, we see this corresponds to superheated steam at 00 o C. At P 1 m m MPa ˆ while at 1.2 MPa ˆ Thus we need to linearly kg kg interpolate. As discussed in the text, following the ideal gas relation, it is most accurate to interpolate in terms of 1 / ˆ 1/0.25-1/ P 1 MPa MPa 1.01 MPa 1/ /

40 1.8 The initial specific olume in the tank must match the alue at the critical point, m ˆ kg ( 1 x) ˆ + xˆ The alue of mass can be found as: V m ˆ 1.7 kg For water at 100 o C, l ˆ l m kg And ˆ m 1.67 kg Soling for x, we get x It is almost all liquid. 40

41 1.9 The final pressure is 500 kpa. Looking at the steam tables, the final temperature is o C and the final specific olume is m ˆ and the final olume is V mˆ m kg 41