Explanation It is given that the concentration of a solute is supposed to be followed a Gaussian distribution, c ( x ) = c 0 − β c 0 x . Where, c 0 is the concentration of solute at x = 0 with the length of the container. The given temperature is 25°C . The value of universal gas constant is 8.314 J mol − 1 K − 1 . The value of x = 8 cm . The conversion of cm into m is done below. 1 cm = 10 − 2 m The value of x = 8 × 10 − 2 m . The conversion of degrees Celsius into Kelvin is done below. 0 ° C = 273 K So, the temperature becomes 25 + 273 K = 298 K . The expression for the concentration of the solute following the Gaussian distribution with the length of a container is given below. c ( x ) = c 0 a x 2 (1) Where, a is the constant. c 0 is the initial concentration. x is the length of the container. The differential equation at constant temperature and pressure for the above expression is as follows. ( ∂ c ∂ x ) p , T = 2 c 0 a x The multiplication and division of right side of the equation by x is done below. ( ∂ c ∂ x ) p , T = 2 x ( c 0 a x 2 ) ( ∂ c ∂ x ) p , T = 2 c x (2) The formula of thermodynamic force is expressed as follows. F = − R T c ( ∂ c ∂ x ) p , T (3) Where, R is the universal gas constant. T is the temperature. Substitute ( ∂ c ∂ x ) p , T = 2 c x in the above expression. F = − R T c × 2 c x F = − 2 R T x (4) Substitute the values of R , T and x = 8 × 10 − 2 m in equation (4)

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Chapter 16, Problem 16C.3BE

Interpretation Introduction

Interpretation:

The thermodynamic forces over the solute at 25°C, at x=8 cm and x=16 cm have to be calculated.

Concept introduction:

A force that is based upon the system quantification of the different energy tendencies in thermodynamic is known as a thermodynamic force. This force is the energetic illustration of various driving force used in a process.

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Chapter 16 Solutions

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Ch. 16 - Prob. 16A.1ST Ch. 16 - Prob. 16B.1ST Ch. 16 - Prob. 16C.1ST Ch. 16 - Prob. 16A.1DQ Ch. 16 - Prob. 16A.2DQ Ch. 16 - Prob. 16A.1AE Ch. 16 - Prob. 16A.1BE Ch. 16 - Prob. 16A.2AE Ch. 16 - Prob. 16A.2BE Ch. 16 - Prob. 16A.3AE

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