Answers to assigned problems from Chapter 1

Size: px

Start display at page:

Download "Answers to assigned problems from Chapter 1"

Opal Morgan
6 years ago
Views:

1 Answers to assigned probles fro Chapter a. A colun of ercury 1 in cross-sectional area and in height has a volue of and a ass of kg. Then 1 Hg kg s Pa By definition, 1 atosphere Pa. 1 Torr 1 (101 5 Pa) Pa 760 Thus Hg Torr The torr is now defined as 1 Hg. b. Fro P n, and since n N L P N L and the nuber density is N PL For P 10 6 Torr: N/ 10 6 Torr (1 at/760 Torr) J K 1 ol K (Pa at 1 ) ol For P Torr: 10 N/ 15 Torr (1 at/760 Torr) (J K 1 ol 1 ) K (Pa at 1 ) ol This is still a substantial nuber A colun of ercury 1 in cross-sectional area and exactly in height has a volue of and a ass of kg. Mass ties the gravitational acceleration is a weight or force. The colun s weight on the unit area gives a pressure kg s kg s Since 1 Pa 1 kg s, the pressure is kpa 1.9. The pressure of two different liquids of the sae volue at the sae teperature is proportional to their densities. Assue a 1 height of liquid of diaeter to accoodate 1 c. Then P DBP P Hg g c g c or P DBP 1 Torr Torr Thus, 1 of DBP is equivalent to Torr. This can be stated such that 1 Torr is equivalent to illietres of DBP:

2 Chapter 1 1 Torr/0.077 (Torr 1 ) 1.99 DBP P 1 1 P 697 Torr 0.51 d Torr 87( ) d P 1 1 P Pa 000. d Pa d T 1 T 0.00 d (550 K) d 0 K 1.1. Concentration n/ P/. Since P Pa 1 at a. b. 5 n Pa J K ol K ol since J/Pa ol d Nuber of olecules per unit volue (ol d ) ol d 4 n Pa J K ol K ol ol d d M ρpt P 1.9kg J ol K K Pa kg ol g ol Pa n/ P/ J K ol K ol ol d Thus ol g 1 ol 8.6 g olar ass

3 Chapter Fro Eq. 1.5 P t P H + P H O P H P t P H O The aount of each gas is kpa.17 kpa 96.8 kpa nn g/ g ol ol. The ole fractions are nco g / g ol. 7 ol x N ol 7. ol x CO ( ) ol ( ) ol The partial pressures are J K ol K PN ol bar.50 d P J K ol K. 7 ol. 51 bar.50 d CO The total pressure is bar 1.0. P of dry gas at 7.5 C is Torr 7.5 Torr 7.9 Torr P 1 1 /T 1 P /T gives 1 P 1 T /P T 1 (7.8 c ( 79. Torr)( K) ) ( Torr)( K) 6.6 c This can also be done using 1. Pa/Torr and using Pa to eliinate the pascal unit P P 0 e Mgz/ Assue that the teperature reains constant at 50 K. The value of M is kg ol 1. P 400 Torr e 0.017(9.807)8000/( ) 400 e Torr 1.7. Rate of effusion is proportional to 1/T; therefore, fro Graha s law v A 1 v B. 8 g ol 1 MW of A Molar ass of A 8 (.) g ol The tie for effusion is inversely proportional to the rate of effusion. (A high rate requires a short tie for diffusion.)

4 Chapter 1 4 rate (N ) rate (He) t He t N MW He MW N t He t N MW He MW N in 1.9. P ne k ; E k P n (Kinetic energy per ole of gas.) E k 1 (00 kpa) (8.0 d 0.5 ol ) 4800 J ol 1 For 0.5 ol, ne k 400 J 1.. The value of N A, the nuber of olecules fored for use in the ean-free-path equation, ust be calculated fro the ideal gas law. The nuber of olecules per unit volue is N A LP ( ol ) 101 5( Pa), ( J K ol )98. 15( K ) Since 1 Pa kg 1 s and J kg s λ 1/( πd N A ) 1/( ( π )( ) ( ) Z A πd u A N A π( ) 474.6( s 1 ) s Fro Eq Z AA 1 πd u A N A 1 N A Z A s 1 λ πd N 1 10 π( ) This is about a hundred ties greater than the distance between the earth and the nearest star (Proxia Centauri)! Ideal gas prediction: P n/ g /( 5.45 g ol ) ( J K ol ) 715. K P kpa

5 Chapter 1 5 an der Waals prediction: P n nb an n 0.468; a 6.49; b P ( ol) ( J ol K )7. 15 K ( ) ( ol)( ol ) ( ) ( ) Pa ( ) 10 Pa 650 k Pa P 1 1 P Z T Z 1 T bar 1.00 d K 00 bar K.77 d. P 1 1 P T T bar 1.00 d 7. K 00 bar. K 7.46 d. This represents an error of %..77 We conclude that using the copression factor gives the ore accurate result At the critical point, both the first and second derivatives of P with respect to at constant T are zero. P a + 0. b Rearrangeent gives a T. ( ) b (1) P a 0, ( b) T or a b ( ) Division of () by (1) gives. () /( b) a / or. /( b) a / b Fro the last expression, 1/ 1/( b). Therefore, b 0. As a result, the two derivatives cannot vanish siulataneously unless b 0. This is contrary to the stateent of the proble. Therefore, the gas does not have a critical point Using Figure 1. for the copressibility factor, find the reduced teperature and reduced pressure. T r T/T c ( )/(08.6) 1.0

6 Chapter 1 6 P r P/P c 90.0/ Fro Figure 1., the value of Z for these two values is 0.7. Fro Z P/ (Because pressure in the chart is based on atospheres, we ust use at d ol 1 K 1 for R.), we have: Z/P 0.7(0.081)(70.5) 90.0 Fro the ideal gas law: /P 0.081(70.5) d ol d ol Fro Table 1.4, b ol 1. For a sphere, (4/)πr. Note that the constant b is called the covolue, which is approxiately four ties the volue occupied by the individual olecules. (Section 1.1, p. 6 of the text book) Substituting b 4, the volue per olecule 4 b ol 9 πr L ol is given by Therefore, r [ /(4π/)] 1/ (The actual radius, i.e., the C-H bond distance, of ethane olecule is ) This equation reduces to P / 1 when [1 4(T c /T) 0. This occurs when T T B. Therefore, solve for T T B. 4(T c /T B ) 1 and T B T c a. Multiplying both sides of Eq. (1.101) by /(), we get P a b 1. b. Using the series expansion given, we get P a 1 b b a ( 1 b / ) c. Grouping together ters containing the sae power of, we get P Z 1+ b a 1 + b Coparing this to Eq. (1.118) with n 1, we see that B(T) b a/(), C(T) b. d. At the Boyle teperature, B(T B ) 0, i.e., b a/(r T B ) or T B a/(br).

Chemistry Department Al-kharj, October Prince Sattam Bin Abdulaziz University First semester (1437/1438)

Chemistry Department Al-kharj, October Prince Sattam Bin Abdulaziz University First semester (1437/1438) Exercise 1 Exercises- chapter-1- Properties of gases (Part-2- Real gases Express the van der Waals paraeters a = 1.32 at d 6 ol 2 and b = 0.0436 d 3 ol 1 in SI base units? * The SI unit of pressure is