Explanation The expression given is written below as, C p − C V = T ( ∂ p ∂ T ) V ( ∂ V ∂ T ) p (1) Since volume is a function of temperature and pressure, therefore, it can be written as, V = f ( T , p ) ∂ V = ( ∂ V ∂ T ) p ∂ T + ( ∂ V ∂ p ) T ∂ p At constant volume, the value of ∂ V becomes zero and the above equation can be rearranged as, 0 = ( ∂ V ∂ T ) p ( ∂ T ) V + ( ∂ V ∂ p ) T ( ∂ p ) V ( ∂ V ∂ T ) p ( ∂ T ) V = − ( ∂ V ∂ p ) T ( ∂ p ) V ( ∂ p ∂ T ) V = − ( ∂ V ∂ T ) p ( ∂ V ∂ p ) T Substitute the above equation in equation (1) as, C p − C V = T ( ∂ p ∂ T ) V ( ∂ V ∂ T ) p = T ( − ( ∂ V ∂ T ) p ( ∂ V ∂ p ) T ) ( ∂ V ∂ T ) p = − T ( ∂ V ∂ T ) p 2 ( ∂ V ∂ p ) T (2) Thus, the above relation matches with the given relation. The perfect gas equation is given by the formula as, p V = n R T (3) Where, • p is the total pressure

11th Edition

ISBN: 9780198834700

Author: ATKINS

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Question

Chapter 2, Problem 2D.2P

Interpretation Introduction

Interpretation: The given relation has to be shown using the relation between partial derivatives. The expression to evaluate difference in heat capacities at constant pressure and constant volume for a perfect gas has to be stated.

Concept introduction: The expansion coefficient is defined as the fraction change in volume with respect to change in temperature. Thus, it defines the change in size on changing temperature. This is represented by the formula given below as,

α=1V(∂V∂T)p

The isothermal compressibility is defined as the fraction change in volume with respect to change in pressure. This is represented by the formula given below as,

κT=−1V(∂V∂p)T

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Chapter 2 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 2 - Prob. 2A.1ST Ch. 2 - Prob. 2B.1ST Ch. 2 - Prob. 2B.2ST Ch. 2 - Prob. 2C.1ST Ch. 2 - Prob. 2C.2ST Ch. 2 - Prob. 2D.1ST Ch. 2 - Prob. 2D.2ST Ch. 2 - Prob. 2A.1DQ Ch. 2 - Prob. 2A.2DQ Ch. 2 - Prob. 2A.3DQ

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