# all2 - ChE 141 Thermodynamics Berend Smit 2 Chapter 1...

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Unformatted text preview: ChE 141: Thermodynamics Berend Smit January 22, 2009 2 Chapter 1 Moving molecules 1.1 Total energy Let us consider two particles 1 and 2 that interact with a potential u ( r 12 ) , where r 12 is the distance between particles 1 and 2: r 12 = q ( r 2- r 1 ) 2 . (1.1.1) Note the position of a particle can be positive or negative but the distance in equation (1.1.1) is always positive. For the total energy we can write U t = U k + U p where U k is the kinetic energy: U k = 1 2 m 1 v 2 1 + 1 2 m 2 v 2 2 where v 1 is the velocity of a particle 1: v 1 = ˙ r 1 = d r 1 d t For the potential energy we have U p = u ( r 12 ) To show that the total energy is conserved, we compute the time derivative of the total energy d U t d t = d U k d t + d U p d t (1.1.2) 4 Chapter 1. Moving molecules If we first look at the kinetic energy d U k d t = m 1 v 1 ˙ v 1 + m 1 v 1 ˙ v 1 . For the time derivative of the potential energy we have d U p d t = d u ( r 12 ) d r 12 d r 12 d t (1.1.3) For the second term in equation (1.1.3), we have to look at the time derivative of the distance between particles 1 and 2, as this is a function of two variables r 1 and r 2 : r 12 = r 12 ( r 1 ,r 2 ) we have to use partial differentials: d r 12 d t = ∂r 12 ∂r 1 r 2 ˙ r 1 + ∂r 12 ∂r 2 r 1 ˙ r 2 The first term in this equation is the derivative of the potential with respect to the position, which is related to the force. For two particles, however, we have to be a bit more precise. The force on particle 1 is defined as the negative of the change of the potential energy if we displace the position of particle 1 by d r : f 1 = - ∂u ( r 12 ) ∂r 1 r 2 = - d u ( r 12 ) d r 12 ∂r 12 ∂r 1 r 2 (1.1.4) and for the force on particle two: f 2 = - ∂u ( r 12 ) ∂r 2 r 1 = - d u ( r 12 ) d r 12 ∂r 12 ∂r 2 r 1 (1.1.5) If we collect the terms, we can write for equation (1.1.3) d U p d t = d u ( r 12 ) d r 12 " ∂r 12 ∂r 1 r 2 ˙ r 1 + ∂r 12 ∂r 2 r 1 ˙ r 2 # = - f 1 ˙ r 1- f 2 ˙ r 2 From Newton’s third law we have; f 1 = m 1 d 2 r d t 2 = m 1 ˙ v 1 giving for the time derivative of the potential energy: d U p d t = - m 1 v 1 ˙ v 1- m 2 v 2 ˙ v 2 1.2 Pressure 5 If we add this to the derivative for the kinetic energy, equation (1.1.2), we obtain d U t d t = Which shows that Newton’s law leads to the conservation of total energy. Newton’s second law, action = -reaction, follows from equations (1.1.4) and (1.1.5). From equation (1.1.1), it follows that ∂r 12 ∂r 1 r 2 = - ( r 2- r 1 ) p ( r 2- r 1 ) 2 and for the partial derivative of particle 2: ∂r 12 ∂r 2 r 1 = + ( r 2- r 1 ) p ( r 2- r 1 ) 2 Or f 1 = - d u ( r 12 ) d r 12 ∂r 12 ∂r 1 r 2 = + d u ( r 12 ) d r 12 ∂r 12 ∂r 2 r 1 = - f 2 (1.1.6) 1.2 Pressure Another consequence of Newton’s laws is the conservation of total momen- tum. For our two particle system we have: p t = m 1 v 1 + m 2 v 2 and for the time derivative, we can write d p t d t = m 1 ˙ v 1 + m 2 ˙ v 2 = m 1 f 1 m 1 + m 2 f 2 m 2...
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all2 - ChE 141 Thermodynamics Berend Smit 2 Chapter 1...

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