Tutorial 3 Notes - CHM2132 Tutorial notes for Sept 25th,...

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1 CHM2132 Tutorial notes for Sept 25 th , 29 th . 1) A sample consisting of 15.0 g of nitrogen is confined in a container at 220 kPa and 200 K and then allowed to expand adiabatically (a) reversibly to 110 kPa, (b) against a constant pressure of 110 kPa. We are given the following information: M = 15.0 g of N 2 p 1 = 220 kPa T 1 = 200 K We need to calculate the final temperature for: a) Reversible, adiabatic expansion to 110 kPa Since the reaction is reversible and adiabatic we can use the equation of the adiabat to find the final temperature: p 1 V 1 ! = p 2 V 2 We were given initial and final pressure, but we need to calculate temperature. To put this equation in a more useful form for this question we will substitute the ideal gas law for the volumes: p 1 nRT 1 p 1 ! " # $ % = p 2 nRT 2 p 2 ! " # $ % To simplify this it may be easier to represent both sides in log form. Concentrating on the left hand side: log p 1 + log T 1 p 1 ! " # $ % = log p 1 + log T 1 p 1 ! " # $ % = log p 1 + log T 1 ( log p 1 = 1 ( ( ) log p 1 + log T 1 We can simplify the equation further by dividing by γ : 1 ! " ( ) log p 1 + log T 1 = 1 ! ( ) log p 2 + log T 2 Remembering that = C p C V : 1 ! ( ) = 1 ! C p C V # $ % ( C p C V = C V ! C p C V # $ % ( ) C V C p = ! R C p
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2 Putting this back into the equation of the adiabat: ! R C p log p 1 + log T 1 = ! R C p log p 2 + log T 2 log T 2 = ! R C p log p 1 + R C p log p 2 + log T 1 T 2 = T 1 p 2 p 1 " # $ % R C p Plugging in the values: T 2 = 200 K 110 kPa 220 kPa ! " # $ % 8.314 J K -1 mol -1 29.125 J K -1 mol -1 = 200 K 0.5 ( ) 0.285 = 164 K b) What is the final temperature for the adiabatic expansion against a constant pressure of 110 kPa? It is no longer possible to use the equation of the adiabat to calculate the final
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This note was uploaded on 01/22/2011 for the course CHM 2132 taught by Professor Giorgi during the Fall '08 term at University of Ottawa.

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Tutorial 3 Notes - CHM2132 Tutorial notes for Sept 25th,...

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