Molar Heat Capacity of a Gas
Date: 11/17/1999 at 01:23:06 From: Wiley Ang Subject: Thermodynamics/Physics Dear Dr. Math I recently took part in the National Physics Competition in my country but I got stuck on the following question: The volume of n moles of a monatomic perfect gas with an initial volume V and an initial temperature T is halved at a constant rate. At the same time, its temperature is doubled at a constant rate. (a) Prove that the work done to the gas is given by W = nRT*(3*log-1) (b) Obtain an expression for the heat absorbed during the process (c) Prove that the molar heat capacity of this gas during the process is 3.49 J/mol/K. Can you please help me to solve it? Thank you.
Date: 11/17/1999 at 06:35:07 From: Doctor Mitteldorf Subject: Re: Thermodynamics/Physics Dear Wiley, Since the question defines T and V as the (constant) initial temperature and volume, let's use t and v for the variable quantities as the process occurs. The phrase "constant rate" assures us that change in temperature is proportional to change in volume. Since the overall change in volume is V/2 (from V down to V/2) and the overall change in temperature is T (from T up to 2T), we can say that delta t = (2T/V)*delta v. We can write this as (t-T)=(2T/V)*(V-v) So the temperature follows a straight line defined by t = T+2T(1-v/V)=T(3-2v/V) We're interested in the pressure as the gas is compressed, because it is the pressure that determines the work. P = nRt/v. Let's substitute our expression for t in terms of v: P = nRt/v = nRT(3-2v/V)/v = 3nRT/v - 2nRT/V Work is the integral of pressure with respect to volume, so V W = Int[3nRT/v - 2nRT/V] V/2 If you solve this integral, you should get the result in (a). For part (b), you want to know the heat absorbed by the gas during the entire process. First find the energy at the end minus the energy at the beginning: that is 3/2 nR(2T) at the end and 3/2 nRT at the beginning. Then subtract the work that was done, calculated in (a). The difference is the heat that was added. Converting the answer from (b) to a number should give you the answer for (c). Will you write back and let me know if it works? - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
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