Chemistry Department Al-kharj, October Prince Sattam Bin Abdulaziz University First semester (1437/1438)
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1 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 = d 3 ol 1 in SI base units? * The SI unit of pressure is Pa (1Pa = 1kg. -1 s -2, and that of volue is 3. We have: 1 at = Pa = kg. -1 s -2 1 d 3 = => 1 d 6 = a = 1.32 at d 6 ol 2 = 1.32 at d 6 ol kg. 1 s 2 a = kg. 5 s 2 ol 2 1 at d 6 * b = d 3 ol 1 We have 1 d 3 = b = d 3 ol d 3 = ol 1 Exercise 2 A gas at 350 K and 12 at has a olar volue 12 per cent larger than that calculated fro the perfect gas law. Calculate (a the copression factor under these conditions and (b the olar volue of the gas. Which are doinating in the saple, the attractive or the repulsive forces? (a Copression factor under the conditions of the exercise The copression factor, Z, of a gas is the ratio of its easured olar volue, =/n, to the olar volue of a perfect gas, 0, at the sae pressure and teperature: 1
2 Z = 0 At 350 K and 12 at, the gas has a olar volue 12 per cent larger than that calculated fro the perfect gas law, i.e, => Repulsive forces doinates = = = Z = 0 = = 1.12 (b The olar volue of the gas=? We have Z = 0 The olar volue of a perfect gas is equal to /p, thus Hence, Z = p = Z p = 1.12 (0.082 d3 at K 1 ol 1 (350 K 12 at = 2.7 d 3 ol 1 2
3 Exercise 3 Cylinders of copressed gas are typically filled to a pressure of 200 bar. For oxygen, what would be the olar volue at this pressure and 25 C based on (a the perfect gas equation, (b the van der Waals equation. For oxygen, a = d 6 at ol 2, b = d 3 ol 1. The cylinders pressure P = 200 bar The teperature, T = 25 C (a The olar volue of gas if it is considered as a perfect gas: The equation of a perfect gas: p = = p = (8.315 J K 1 ol 1 (298 K = 1.24 (200 bar 105 Pa ol 1 = d3ol 1 1 bar (b If the gas is real, and obey to the van der Waals equation. For oxygen, a = d 6 at ol 2, b = d 3 ol 1. The van der Waals equation is p = n nb a n2 2 This equation is often written in ters of the olar volue = /n as We arrange the equation to the for: p = b a 2 3 (b + p 2 + ( a p ab p = 0 This a cubic equation in The coefficient in the equation are: 3
4 b + p = d 3 ol 1 + (0.082 d 3 at K 1 ol 1 (298 K = (200 bar d3ol 1 Pa 1 bar 1 at Pa a p = d 6 at ol 2 (200 bar 105 Pa 1 bar 1 at Pa = d6 ol 2 ab p = (1.364 d6 at ol 2 ( d 3 ol 1 = d 9 ol bar 105 Pa 1 bar 1 at Pa the equation becoes: ( ( = 0 We use the calculator to solve the polynoial equation, we find = d 3 ol -1 The difference between the two volues calculated in (a and (b is 15% Exercise 4 At 300 K and 20 at, the copression factor of a gas is Calculate (a The volue occupied by 8.2 ol of the gas under these conditions (b An approxiate value of the second virial coefficient B at 300 K. The copression factor of a gas at 300 K and 20 at is 0.86 (a The volue occupied by 8.2 ol of the gas under these conditions We have Z = p and = n 4
5 Z = p Z => = n n p = ( (0.082 d3 at K 1 ol 1 (300 K 20 at (b An approxiate value of the second virial coefficient B at 300 K The virial equation of state of a gas: p = (1 + B + C 2 + = d 3 An approxiation value of the second virial coefficient B can be obtained by ignoring third, fourth virial coefficients. The above equation becoes: p = (1 + B => p = 1 + B Hence, p 1 = B B = ( p 1, where = n = Z p B = Z (Z 1 p B = 0.86 (0.082 d3 at K 1 ol 1 (300 K ( = 0.15 d 3 ol 1 20 at 5
6 Exercise 5 A vessel of volue 22.4 d 3 contains 1.5 ol H 2 and 2.5 ol N 2 at K. Calculate (a Mole fractions? (b Partial pressures? (c Total pressure? (a Mole fractions of H 2 and N 2? x H2 = n H 2 n t = x N2 = n N 2 n t = n H2 n H2 + n N2 = n N2 n H2 + n N2 = 1.5 ol 1.5 ol ol = ol 1.5 ol ol = 0.37 (b Partial pressures of H 2 and N 2 We consider gases as perfect gases, thus P H2 = n H2 => P H2 = n H2 = 1.5 ol (0.082 d3 at K 1 ol 1 ( K 22.4 d 3 => P H2 = 1.5 at P N2 = n N2 => P N2 = n N2 = 2.5 ol (0.082 d3 at K 1 ol 1 ( K 22.4 d 3 => P N2 = 2.5 at (c Total pressure? P t = P H2 + P N2 = 1.5 at at = 4 at 6
7 Exercise 6 The critical constants of ethane are p c = at, c = 148 c 3 ol 1, and T c = K. Calculate the van der Waals paraeters of the gas and estiate the radius of the olecules. an der Waals paraeters are a and b, which can be deterined fro the below critical points forula: * alue of b =? c = 3b; p c = a 27b 2 ; T c = 8a 27Rb c = 3b b = 1 3 c = c3 ol 1 = d 3 ol 1 * alue of a =? p c = a 27b 2 a = 27b 2 p c = at ( d 3 ol 1 2 = 3.16 at d 6 ol 2 * The radius of the olecules =? To calculate the excluded volue we note that the closest distance of two hard-sphere olecules of radius r, and volue olecule = 4 3 πr 3, is 2r, so the volue excluded is 4 3 π(2r 3, or 8 olecule. The volue excluded per olecule is one-half this volue, or 4 olecule, so b 4 olecule N A. b = 4 olecule N A = πr 3 N A 3b = 16πr 3 N A => r 3 = 3b => r = ( 3b 16πN A 16πN A 1/3 7
8 3 493 c 3 ol 1 r = ( ol 1 1/3 = c Exercise 7 Use the van der Waals paraeters for hydrogen sulfide H 2 S to calculate approxiate values of (a the Boyle teperature of the gas (b the radius of a H 2 S olecule regarded as a sphere. (a Boyle s teperature is the teperature at which The copression factor Z is: li ( dz d ( 1 = 0 And pressure, p of a real gas is: Hence, So, Z = p = Z = p b a 2 ( b a 2 dz d ( 1 = dz d d d ( 1 = b where a d d ( 1 2 = dz d ( 1 = 2 ( dz = 2 b d [ ( b 2 + a 2 2 ] = b ( b 2 a 8
9 Hence, Boyle s teperature is li ( dz d ( 1 = li T = ( 2 b ( b 2 a = b a = 0 b = a d 6 at ol 2 (0.082 d 3 at K 1 ol 1 ( d 3 ol 1 = 1259 K (b The radius of a H 2 S olecule regarded as a sphere. As in exercise 6, the radius of H 2 S is r = ( 3b 16πN A 1/ d 3 ol 1 r = ( ol 1 1/3 = d Exercise 8 A certain gas obeys the van der Waals equation with a = Pa ol 2. Its volue is found to be o l 1 at 288 K and 4.0 MPa. Fro this inforation calculate the van der Waals constant b. What is the copression factor for this gas at the prevailing teperature and pressure? The van der Waals equation is Which can be solved for b p = b a 2 9
10 The copression factor is b = p + a 2 = ol 1 Z = p =
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