# week3 - 3 More applications of derivatives 3.1 Exact&...

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Unformatted text preview: 3 More applications of derivatives 3.1 Exact & inexact differentials in thermodynamics So far we have been discussing total or “exact” differentials du = ∂u ∂x ¶ y dx + ∂u ∂y ¶ x dy, (1) but we could imagine a more general situation du = M ( x, y ) dx + N ( x, y ) dy. (2) If the differential is exact, M = ( ∂u ∂x ) y and N = ‡ ∂u ∂y · x . By the identity of mixed partial derivatives, we have ∂M ∂y ¶ x = ∂ 2 u ∂x∂y ¶ = ∂N ∂x ¶ y (3) Ex: Ideal gas pV = RT (for 1 mole), take V = V ( T, p ), so dV = ∂V ∂T ¶ p dT + ∂V ∂p ¶ T dp = R p dT- RT p 2 dp (4) Now the work done in changing the volume of a gas is dW = pdV = RdT- RT p dp. (5) Let’s calculate the total change in volume and work done in changing the system between two points A and C in p, T space, along paths AC or ABC . 1. Path AC: dT dp = T 2- T 1 p 2- p 1 ≡ Δ T Δ p so dT = Δ T Δ p dp (6) & T- T 1 p- p 1 = Δ T Δ p ⇒ T- T 1 = Δ T Δ p ( p- p 1 ) (7) so (8) dV = R p Δ T Δ p dp- R p 2 [ T 1 + Δ T Δ p ( p- p 1 )] dp =- R p 2 ( T 1- Δ T Δ p p 1 ) dp (9) dW =- R p ( T 1- Δ T Δ p p 1 ) dp (10) 1 A B C (p,T) (p 2 ,T 2 ) (p 1 ,T 1 ) T p Figure 1: Path in p,T plane for thermodynamic process. Now we can calculate the change in volume and the work done in the process: V 2- V 1 = Z AC dV = R ( T 1- Δ T Δ p p 1 ) 1 p fl fl fl fl p 2 p 1 = R T 2 p 1- T 1 p 2 p 1 p 2 (11) W 1- W 2 = Z AC pdV = R ( T 1- Δ T Δ p p 1 ) ln p fl fl fl fl p 2 p 1 = R T 2 p 1- T 1 p 2 p 2- p 1 ln p 2 p 1 (12) 2. Path ABC: Note along AB dT = 0, while along BC dp = 0. V 2- V 1 = Z ABC ∂V ∂T ¶ p dT + ∂V ∂p ¶ T dp = Z p 2 p 1 ∂V ∂p ¶ T dp + Z T 2 T 1 ∂V ∂T ¶ p dT = Z p 2 p 1- RT 1 p 2 dp + Z T 2 T 1 R p 2 dT = R T 2 p 1- T 1 p 2 p 1 p 2 (13) W 2- W 1...
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week3 - 3 More applications of derivatives 3.1 Exact&...

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