Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

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[2]-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/   
    
Associated Topics:
High School Physics/Chemistry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/