Explanation The volume of the vessel is given as 5.0 m × 5.0 m × 3.0 m . It is given that the temperature is ( T ) is 25 ° C . The conversion of 25 ° C to K is done as, 1 ° C = 273 K Therefore, the conversion of 25 ° C to K is done as, 25 ° C = 25 + 273 K = 298 K Therefore, the temperature is ( T ) is 298 K . For water: It is given that the vapour pressure of water at this temperature is 3.2 kPa . The conversion of kPa to N m -2 is done as, 1 kPa = 1 × 10 3 N m -2 Therefore, the conversion of 3.2 kPa to N m -2 is done as, 3.2 kPa = 3.2 × 10 3 N m -2 = 3200 N m -2 Therefore, the vapour pressure of water is 3200 N m -2 . The mole of water is calculated by the following formula. p V = n R T (1) Where, p is the vapour pressure. V is the volume of the container. R is the gas constant. T is the temperature. n is the number of moles. The value of gas constant is 8.314 J K − 1 mol -1 . Substitute the value of T , p , R and V in equation (1). p V = n R T 3200 N m -2 × 5.0 m × 5.0 m × 3.0 m = n × 8.314 J K − 1 mol -1 × 298 K 240000 N m = n × 2477.572 J mol -1 (2) The conversion of J to N m is done as, 1 J = 1 N m Rearrange the equation (2) as, n = 240000 N m 2477.572 J mol -1 = 96.86 mol Therefore, the moles of water is 96.86 mol . The molar mass of water is 18 g mol -1 . The mass of water is calculated by the following formula

11th Edition

ISBN: 9780198814740

Author: ATKINS, P. W. (peter William), 1940- (author.)

Publisher: Oxford University Press,

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Chapter 4, Problem 4B.12AE

Interpretation Introduction

Interpretation:

The mass of water and benzene at equilibrium in the air, if there is no ventilation is to be calculated.

Concept introduction:

In the absence of any ventilation in the container, the moles of the liquid will remain same at equilibrium. The value of moles is calculated by the ideal gas equation. The ideal gas equation is shown below.

pV=nRT

Expert Solution & Answer

Chapter 4 Solutions

Atkins' Physical chemistry

Ch. 4 - Prob. 4B.1ST Ch. 4 - Prob. 4B.2ST Ch. 4 - Prob. 4A.1DQ Ch. 4 - Prob. 4A.2DQ Ch. 4 - Prob. 4A.3DQ Ch. 4 - Prob. 4A.1AE Ch. 4 - Prob. 4A.1BE Ch. 4 - Prob. 4A.2AE Ch. 4 - Prob. 4A.2BE Ch. 4 - Prob. 4A.3AE

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