Practical - Tutorial 4 Problem and Solution PDF

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Tutorial 4 Problem and Solution...

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ChE102 Chemistry for Engineers - Tutorial 4 Solutions Solve the following problems either individually or in small groups (3 students maximum). Submit one copy of the solutions per group. Ask the TA for help if you need assistance solving the problems

Problem 1 For 6.74 mol of N2 gas confined in a volume of 1.0 L at 151.4 K, (a) calculate the pressure exerted by the N 2 gas using the ideal gas law; (b) calculate the pressure exerted by the N 2 gas using the van der Waals equation; (c) calculate the compressibility factor of the N2 gas given that the actual pressure exerted by the gas is 46.1 atm. Data: In the van der Waals equation for N2 : a = 1.390 atm· L2·mol–2; b = 0.03913 L·mol–1 Solution (a) Ideal gas law: P=

nRT

=

V

l atm �(151.4 K) mol K

(6.74 mol)�0.08206

1.0 L

= 83.73697atm

(b) van der Waals equation: �P +

n2 a V2

nRT

� (V − nb) = nRT

P = �V−nb� − �

n2 a V2

�=�

l atm

(6.74 mol)�0.08206mol K�(151.4 K) L � mol

1.0 L−6.74 mol�0.03913

�−�

(6.74 mol)2 (1.390 atm L2 /mol2 (1.0 L)2

�=

50.58794 atm

(c) Compressibility factor: For ideal gases, P i V = nRT. For real gases, P r V = ZnRT. ∴Z=

Pr Pi

=

46.1 atm 84 atm

= 0.54881

Problem 2 Consider the following statements about gases. The statements are not related to one another. a) At a given temperature and pressure, the volume of 1 mol of a gas predicted by the ideal gas law is always larger than the volume predicted by van der Waals equation. b) According to the kinetic-molecular theory, at 100 °C, the root mean square speed of gaseous O 2 molecules is the same as the root mean square speed of gaseous N 2 molecules. c) Diffusion is caused by attractive forces between different types of molecule. d) The ideal gas law is a good model for the behaviour of most gases at very high pressures. Identify which statements are true or false and explain why.

Solution a) False. Volumes predicted by the ideal gas law may be larger or smaller than those predicted by the van der Waals equation. b) False. According to the kinetic theory of gases, u Nrms2 =

u Orms2 = N2 u rms

3RT and u Orms2 = MN2

3RT . Therefore, MO2

M N2 ≠ 1. M O2

c) False. Diffusion is caused by the random collision of molecules and is proportional to their concentration gradient. d) False. The ideal gas law is generally a good model for gas behaviour at low pressures, far away from condensation.

Problem 3 The following statements relate to a sample of 1 mol of CO 2 . A) If the volume of the sample is 100 L and the temperature is 273 K, the CO 2 behaves like an ideal gas because Z is approximately 1. B) If the volume of the sample is 1 L at 273 K, repulsive forces will dominate and Z will be greater than 1. C) If the volume of the sample is 100 L, it is expected that the CO 2 will behave like an ideal gas at all temperatures from 273 K up to 1000 K. D) If the volume of the sample is 0.1 L at 273 K, the real pressure will be about 14% of the calculated ideal pressure. E) If the volume of gas is 0.1 L then it is expected that the real CO 2 pressure will be less than the ideal pressure for all temperatures from 273 K up to 1000 K. Data: the van der Waal’s constants for CO 2 are a = 3.59 atm· L2·mol–2 and b = 0.0427 L·mol–1. Identify which statements are true or false and explain why. Solution A) If the volume of the sample is 100 L and the temperature is 273 K, the CO 2 behaves like an ideal gas because Z is approximately 1. P ideal = 0.22398 atm PVdW =

0.082 L atm/mol K× 273.15 K 3.59 atm L2/mol 2 (1 mol) 2 − = 0.22372 atm 100 L − (1 mol × 0.0427 L/mol) (0.1 L)2

Z = 0.22372/0.22398 = 0.999. True.

B) If the volume of the sample is 1 L at 273 K, repulsive forces will dominate and Z will be greater than 1. P ideal = 22.39 atm PVdW =

0.082 L atm/mol K × 273.15 K 3.59 atm L 2 /mol 2 (1 mol) 2 − = 19.80736 atm 1 L − (1 mol × 0.0427 L/mol) (1 L) 2

Therefore, Z < 1. False. C) If the volume of the sample is 100 L, it is expected that the CO 2 will behave like an ideal gas at all temperatures from 273 K up to 1000 K. PVdW =

0.082 L atm/mol K ×1000 K 3.59 atm L2/mol 2 (1 mol) 2 − = 81.99 atm 100 L − (1 mol × 0.0427 L/mol) (1 L)2

Z = 0.99999 True.

D) If the volume of the sample is 0.1 L at 273 K, the real pressure will be about 14% of the calculated ideal pressure. P ideal = 223.86 atm PVdW =

0.082 L atm/mol K × 273.15 K 3.59 atm L 2 /mol 2 (1 mol) 2 − = 31.89529 atm 0.1 L − (1 mol × 0.0427 L/mol) (0.1 L) 2

P real /P ideal = 0.14248 True.

E) If the volume of gas is 0.1 L then it is expected that the real CO 2 pressure will be less than the ideal pressure for all temperatures from 273 K up to 1000 K. P ideal = 820 atm PVdW = False.

0.082 L atm/mol K ×1000 K 3.59 atm L2 /mol 2 (1 mol) 2 − =1072.06457 atm 0.1 L − (1 mol × 0.0427 L/mol) (0.1 L) 2

Problem 4 A commercial airplane traveling at an altitude of 10 km is capable of reaching a speed of 1,000 km/h. Under these conditions, the airplane is exposed to a temperature of -58 oC. Using the kinetic theory of gases, what would the approximate molecular weight (g/mol) of a gas be if it has a root mean square speed equal to the airplane’s speed at that same temperature? Solution A gas has a root mean square speed of 1,000 km/h at of -58 oC 3RT M 3RT M= urms 2 urms =

The molecular weight in g/mol is 3

8.314 kg∙m2 (-58+ 273.15) K S2∙mol∙K

h2

(3600 s)2

1 km2

(1000)2 km2

1 h2

(103 m)2

=0.06958 kg/mol = 69.58 g/mol

Problem 5 Two containers of argon (Ar) gas are illustrated below. Both containers contain the same number of moles of argon and are at the same temperature.

Constant T Constant n

Read the following statements and determine which are true and which are false: i. The root-mean-square speed of the argon is the same in both containers ii. Every argon atom has the same kinetic energy in both containers iii. The frequency with which argon atoms collide with the walls is the same in both containers iv. The average force with which the argon atoms strike the wall is the same in both containers Identify which statements are true or false and explain why.

Solution i. True. The containers are kept at the same temperature. ii. False. The distribution of molecular speeds causes a distribution of kinetic energies. The average kinetic energy will be the same in both containers. iii. False. The containers have the same number of molecules at are at the same temperature; therefore, the container with smaller volume with have a higher pressure, thus a higher frequency of molecular collision with the walls. iv. True. The containers are at the same temperature; therefore, the average kinetic energy of their molecules is the same.

Numerical answers 1. (a) 83.73697atm, (b) 50.58794 atm, (c) 0.54881 4. 69.58 g/mol 12T...