Explanation The total pressure exerted by the mixture of neon and fluorine is 3.32 a t m . The 2.5 L of mixture is heated to 15 ° C . The unit conversion of temperature of mixture from ° C to K is shown below. T = 15 + 273 K = 288 K The total number of moles of neon and fluorine in the mixture at 15 ° C is calculated by the relation shown below. n = P V R T (1) Where, n is the number of moles. P is the pressure. V is the volume. R is the gas constant. T is the temperature. The value of P is 3.32 a t m . The value of V is 2.5 L . The value of R is 0.082057 L ⋅ a t m ⋅ m o l − 1 ⋅ K − 1 . The value of T is 288 K . Substitute the value of P , V , R and T in equation (1). n = ( 3.32 atm ) × ( 2.5 L ) ( 0.082057 L ⋅ atm ⋅ mol − 1 ⋅ K − 1 ) × ( 288 K ) = 8.3 L ⋅ atm 23 .6324 L ⋅ atm ⋅ mol − 1 = 0.35 mol Therefore, the number of moles of neon and fluorine in the mixture at 15 ° C is 0.35 mol . The entropy for the mixture is calculated by the relation shown below. Δ S = ∫ T 1 T 2 n C V d T T (2) Where, Δ S is the change in the entropy. C V is the molar heat capacity at constant volume. n is the number of moles. T 1 is the initial temperature. T 2 is the final temperature. The integral used to evaluate the above expression is shown below. ∫ d x x = ln x + constant The modified equation (2) on using the above integral is shown below. Δ S = C V ( ln [ T 2 ] − ln [ T 1 ] ) (3) The value of Δ S is 0 .345 J ⋅ K − 1 . The value of n is 0.35 mol . The value of T 1 is 273 K . The value of T 2 is 288 K . Substitute the value of Δ S , n , T 1 and T 2 in equation (3). 0 .345 J ⋅ K − 1 = ( 0.35 mol ) × C V × ( ln ( 288 K ) − ln ( 273 K ) ) 0 .345 J ⋅ K − 1 = C V × ( 0.35 mol ) × ( 5.6629 − 5.6094 ) C V = 0 .345 J ⋅ K − 1 ( 0.35 mol ) × ( 0.0535 ) = 18.42 J ⋅ K − 1 ⋅ mol − 1 Therefore, the molar heat capacity of the mixture at constant volume is 18.42 J ⋅ K − 1 ⋅ mol − 1 . The molar heat capacity at constant volume for monoatomic molecules like Ne is 3 2 R . However, for diatomic molecules like F 2 it is equal to 5 2 R . Suppose the mole fraction of neon be x . Therefore, the mole fraction of fluorine be ( 1 − x )

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM

7th Edition

ISBN: 9781319399849

Author: ATKINS

Publisher: MAC HIGHER

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Chapter 4, Problem 4H.10E

Interpretation Introduction

Interpretation:

The amount of neon and fluorine in the mixture has to be determined.

Concept Introduction:

The molar heat capacity is the ratio of the energy supplied in the form of heat to product of number of moles and the rise in the temperature of the system. The mathematical relation for the calculation of enthalpy in terms of molar heat capacity is shown below.

ΔH=nCΔT

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Chapter 4 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM

Ch. 4 - Prob. 4A.1AST Ch. 4 - Prob. 4A.1BST Ch. 4 - Prob. 4A.2AST Ch. 4 - Prob. 4A.2BST Ch. 4 - Prob. 4A.3AST Ch. 4 - Prob. 4A.3BST Ch. 4 - Prob. 4A.4AST Ch. 4 - Prob. 4A.4BST Ch. 4 - Prob. 4A.1E Ch. 4 - Prob. 4A.2E

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Solution Summary: The author explains the molar heat capacity, which is the ratio of the energy supplied in the form of heat to product of number of moles.

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Chapter 4 Solutions