HW2 - Homework 2: Due at the beginning of class 09/26/2011...

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Homework 2: Due at the beginning of class 09/26/2011 Problem 1. In class, we learn about the reversible isothermal transformation of ideal gas. There is another important path of the volume change where the system is thermally isolated from the surrounding. i.e. dq = 0. This is called the “ adiabatic ” path. Let’s derive the expression of a reversible adiabatic expansion of ideal gas. (a) Begin with dw dq dU + = to get the relation between dT and dV . ( Hint ) Replace dU with C V dT where C V is the heat capacity at constant volume (remember dT C V = for ideal gas in general conditions). We only consider PV work here, thus PdV dw = . Replace P with V nRT . (b) Integrate both sides from (T i ,V i ) to (T f , V f ) to show that f i c i f V V T T = Ο Ο Π Ξ Μ Μ Ν Λ where c is a constant to make the expression simpler. Express c with the constants described above (c) Both at the initial (i) and final (f) states above, you can apply ideal gas equations to change the above result to the relation between (P i ,V i ) to (P f , V f ). Show that P f P i = V i V f ! " # # $ % & & γ What is ? Express it with c . (d) With the above results, obtain the following relation. Ο Ο Π Ξ Μ Μ Ν Λ = Ο Ο Π Ξ Μ Μ Ν Λ f i i f T T P P 1 Problem 2. The results in Prob.1 imply significant differences between the reversible adiabatic process and the reversible isothermal one. Assume all the processes are done reversibly. (a) If you expand the ideal monatomic gas adiabatically from V i = 1L to V f = 2L, the temperature of the gas will change. Calculate i f T T for this volume change. You need to find the value of c
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This note was uploaded on 09/25/2011 for the course CH 353M taught by Professor Lim during the Fall '08 term at University of Texas.

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HW2 - Homework 2: Due at the beginning of class 09/26/2011...

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