Explanation Given: The number of moles of ideal gas is 1 mol . The volume is 0.500 L and the temperature is 298 K . Explanation: Calculate the pressure exerted by helium gas as follows: P = n R T V = 1 mol × 0.08206 L atm mol − 1 K − 1 × 298 K 0.500 L = 48.9 atm Calculate the pressure exerted by neon gas as follows: P = n R T V = 1 mol × 0.08206 L atm mol − 1 K − 1 × 298 K 0.500 L = 48.9 atm Calculate the pressure exerted by H 2 gas as follows: P = n R T V = 1 mol × 0.08206 L atm mol − 1 K − 1 × 298 K 0.500 L = 48.9 atm Calculate the pressure exerted by CH 4 as follows: P = n R T V = 1 mol × 0.08206 L atm mol − 1 K − 1 × 298 K 0.500 L = 48.9 atm Calculate the pressure exerted by CO 2 as follows: P = n R T V = 1 mol × 0.08206 L atm mol − 1 K − 1 × 298 K 0.500 L = 48.9 atm The pressure is same for all the mentioned gases when calculated by the ideal gas equation. Calculate the pressure exerted by helium gas by van der Waals gas equation as follows: P + a ( n V ) 2 × ( V − n b ) = n R T For helium, the value of a is 0.0341 atm L 2 mol − 2 and b is 0.0237 L mol − 1 . Substitute these values in the above expression as follows: ( P + 0.0341 atm L 2 mol − 2 ( 1.0 mol 0.500 L ) 2 ) × ( 0.500 L − ( 1 mol ) ( 0.0237 L mol − 1 ) ) = 1 mol × 0.08206 L atm mol − 1 K − 1 × 298 K ( P + 0.1364 atm ) ( 0.4763 L ) = 24.45 L atm P = 51.19 atm Calculate the pressure exerted by neon gas by van der Waals gas equation as follows: P + a ( n V ) 2 × ( V − n b ) = n R T For neon, the value of a is 0.2135 atm L 2 mol − 2 and b is 0.01709 L mol − 1 . Substitute these values in the above expression as follows: ( P + 0.2135 atm L 2 mol − 2 ( 1.0 mol 0.500 L ) 2 ) × ( 0.500 L − ( 1 mol ) ( 0.01709 L mol − 1 ) ) = 1 mol × 0

Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

4th Edition

ISBN: 9780134162485

Author: Nivaldo J. Tro

Publisher: PEARSON

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Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can b…

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Chapter 5, Problem 152QGW

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To calculate: The pressure exerted by an ideal gas.

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Chapter 5, Problem 151QGW

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Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

Ch. 5 - Prob. 1SAQ Ch. 5 - Prob. 2SAQ Ch. 5 - Prob. 3SAQ Ch. 5 - Prob. 4SAQ Ch. 5 - Prob. 5SAQ Ch. 5 - Prob. 6SAQ Ch. 5 - Prob. 7SAQ Ch. 5 - Prob. 8SAQ Ch. 5 - Q9. A gas sample at STP contains 1.15 g oxygen and...Ch. 5 - Prob. 10SAQ

Ch. 5 - Prob. 11SAQ Ch. 5 - Prob. 12SAQ Ch. 5 - Prob. 13SAQ Ch. 5 - Prob. 14SAQ Ch. 5 - Prob. 15SAQ Ch. 5 - Prob. 1E Ch. 5 - Prob. 2E Ch. 5 - Prob. 3E Ch. 5 - Prob. 4E Ch. 5 - Prob. 5E Ch. 5 - Prob. 6E Ch. 5 - Prob. 7E Ch. 5 - Prob. 8E Ch. 5 - Prob. 9E Ch. 5 - Prob. 10E Ch. 5 - Prob. 11E Ch. 5 - Prob. 12E Ch. 5 - Prob. 13E Ch. 5 - Prob. 14E Ch. 5 - Prob. 15E Ch. 5 - Prob. 16E Ch. 5 - Prob. 17E Ch. 5 - Prob. 18E Ch. 5 - Prob. 19E Ch. 5 - Prob. 20E Ch. 5 - Prob. 21E Ch. 5 - Prob. 22E Ch. 5 - Prob. 23E Ch. 5 - Prob. 24E Ch. 5 - Prob. 25E Ch. 5 - Prob. 26E Ch. 5 - Prob. 27E Ch. 5 - Prob. 28E Ch. 5 - 29. Given a barometric pressure of 762.4 mmHg,...Ch. 5 - Prob. 30E Ch. 5 - Prob. 31E Ch. 5 - Prob. 32E Ch. 5 - Prob. 33E Ch. 5 - Prob. 34E Ch. 5 - Prob. 35E Ch. 5 - Prob. 36E Ch. 5 - Prob. 37E Ch. 5 - Prob. 38E Ch. 5 - Prob. 39E Ch. 5 - Prob. 40E Ch. 5 - Prob. 41E Ch. 5 - Prob. 42E Ch. 5 - Prob. 43E Ch. 5 - Prob. 44E Ch. 5 - Prob. 45E Ch. 5 - Prob. 46E Ch. 5 - Prob. 47E Ch. 5 - Prob. 48E Ch. 5 - Prob. 49E Ch. 5 - Prob. 50E Ch. 5 - Prob. 51E Ch. 5 - Prob. 52E Ch. 5 - Prob. 53E Ch. 5 - Prob. 54E Ch. 5 - Prob. 55E Ch. 5 - Prob. 56E Ch. 5 - Prob. 57E Ch. 5 - Prob. 58E Ch. 5 - Prob. 59E Ch. 5 - Prob. 60E Ch. 5 - Prob. 61E Ch. 5 - Prob. 62E Ch. 5 - Prob. 63E Ch. 5 - 64. A 275-mL flask contains pure helium at a...Ch. 5 - Prob. 65E Ch. 5 - Prob. 66E Ch. 5 - Prob. 67E Ch. 5 - Prob. 68E Ch. 5 - Prob. 69E Ch. 5 - Prob. 70E Ch. 5 - Prob. 71E Ch. 5 - Prob. 72E Ch. 5 - Prob. 73E Ch. 5 - Prob. 74E Ch. 5 - Prob. 75E Ch. 5 - Prob. 76E Ch. 5 - Prob. 77E Ch. 5 - 78. Ozone is depleted in the stratosphere by...Ch. 5 - Prob. 79E Ch. 5 - Prob. 80E Ch. 5 - Prob. 81E Ch. 5 - Prob. 82E Ch. 5 - Prob. 83E Ch. 5 - Prob. 84E Ch. 5 - Prob. 85E Ch. 5 - Prob. 86E Ch. 5 - Prob. 87E Ch. 5 - Prob. 88E Ch. 5 - Prob. 89E Ch. 5 - Prob. 90E Ch. 5 - Prob. 91E Ch. 5 - Prob. 92E Ch. 5 - Prob. 93E Ch. 5 - Prob. 94E Ch. 5 - 95. Modern pennies are composed of zinc coated...Ch. 5 - Prob. 96E Ch. 5 - Prob. 97E Ch. 5 - Prob. 98E Ch. 5 - Prob. 99E Ch. 5 - Prob. 100E Ch. 5 - Prob. 101E Ch. 5 - Prob. 102E Ch. 5 - Prob. 103E Ch. 5 - Prob. 104E Ch. 5 - Prob. 105E Ch. 5 - Prob. 106E Ch. 5 - Prob. 107E Ch. 5 - Prob. 108E Ch. 5 - Prob. 109E Ch. 5 - Prob. 110E Ch. 5 - Prob. 111E Ch. 5 - Prob. 112E Ch. 5 - Prob. 113E Ch. 5 - Prob. 114E Ch. 5 - Prob. 115E Ch. 5 - Prob. 116E Ch. 5 - Prob. 117E Ch. 5 - Prob. 118E Ch. 5 - Prob. 119E Ch. 5 - Prob. 120E Ch. 5 - Prob. 121E Ch. 5 - Prob. 122E Ch. 5 - Prob. 123E Ch. 5 - Prob. 124E Ch. 5 - Prob. 125E Ch. 5 - Prob. 126E Ch. 5 - Prob. 127E Ch. 5 - Prob. 128E Ch. 5 - Prob. 129E Ch. 5 - Prob. 130E Ch. 5 - Prob. 131E Ch. 5 - Prob. 132E Ch. 5 - Prob. 133E Ch. 5 - Prob. 134E Ch. 5 - Prob. 135E Ch. 5 - Prob. 136E Ch. 5 - Prob. 137E Ch. 5 - Prob. 138E Ch. 5 - Prob. 139E Ch. 5 - Prob. 140E Ch. 5 - Prob. 141E Ch. 5 - Prob. 142E Ch. 5 - Prob. 143E Ch. 5 - Prob. 144E Ch. 5 - Prob. 145E Ch. 5 - Prob. 146E Ch. 5 - Prob. 147E Ch. 5 - Prob. 148QGW Ch. 5 - Prob. 149QGW Ch. 5 - Prob. 150QGW Ch. 5 - Prob. 151QGW Ch. 5 - Prob. 152QGW Ch. 5 - 153. Air contains about 78.08% nitrogen gas,...

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