Physics Help Forum

Physics Help Forum Feed Site Feed

  #1  
Old 05-23-2009, 04:36 PM
C.E C.E is offline
Member
 
Join Date: Apr 2009
Posts: 71
Thanks: 54
Thanked 6 Times in 6 Posts
C.E is on a distinguished road
Default past exam question- Urgent

Hi, can someone please help me with this past exam question? (My exam is really soon).

1. (a) Define the molar heat capacity at constant volume, CV, and hence show that
dU =nCV dT for n moles of gas.

I am able to define the molar heat capacity but do not know how to show that dU =nCV dT . I always thought that this was just an experimental result. Any suggestions?

(b). State Carnot’s Theorem. Using a schematic diagram and reductio ad absurdum
(proof by contradiction), show that the Theorem must be true or the 2
nd Law would be contravened.

I have no idea how to go about this part of the question so any help would be welcome.

(c). What is the mean free path for a hypothetical molecule of radius 0.15 nm at 300K and 10^5 Pa? How does this compare to the average molecular separation under this condition?

I was able to do to find the mean free path but I do not know what the average molecular seperation is or how to calculate it.

Last edited by C.E; 05-23-2009 at 04:53 PM.
Reply With Quote
Advertisement
 
  #2  
Old 05-23-2009, 09:49 PM
PHF Helper
 
Join Date: Feb 2009
Location: India
Posts: 365
Thanks: 41
Thanked 181 Times in 152 Posts
Parvez has a spectacular aura aboutParvez has a spectacular aura about
Default

I can answer the first one.
From second law-
dQ = dU +pdV. Since we are talking of the specific heat at constant volume, dV=0 and hence pdV = 0. The above equation reduces to-
dQ = dU .........................(1)

Now heat dQ required to raise the temperature of n mole of gas at constant volume by dT is given by-
dQ =nCvdT .......................(2)
From equation (1) and (2) we have
dU = nCvdT
Reply With Quote
The Following User Says Thank You to Parvez For This Useful Post:
C.E (05-24-2009)

Donate to PHF
  #3  
Old 05-24-2009, 04:58 AM
C.E C.E is offline
Member
 
Join Date: Apr 2009
Posts: 71
Thanks: 54
Thanked 6 Times in 6 Posts
C.E is on a distinguished road
Default

So (1) is for constant volume and case (2) is for a changing volume? Is this formula true in all cases of ideal gasses, i.e. in non-adiabatic and non- isothermal processes?
Reply With Quote
  #4  
Old 05-28-2009, 11:16 AM
Senior Member
 
Join Date: May 2009
Location: Mumbai,India
Posts: 102
Thanks: 6
Thanked 42 Times in 35 Posts
Akshay will become famous soon enough
Default

1 dq=du+dw is a fundamental law of nature. its the first law of thermodynamics

at constant volume dw=0
hence dq=du
hence dq=nCvdT
dq/dT=nCV
clearly dq/dT=nCV only for constant volume processes
for other quasistatic processes involving ideal gases,
dq=nCdT where C is the molar heat capacity of THAT PROCESS
at constant volume C=Cv
at constant pressure C=Cp
for adiabatic processesC=0 as dq=0
now for isothermal processes, C is not defined as theres no RISE IN TEMPERATURE since all the heat supplied is utilized by the system for doing work..
Reply With Quote
The Following 2 Users Say Thank You to Akshay For This Useful Post:
C.E (06-06-2009), physicsquest (05-28-2009)

Donate to PHF
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 09:50 PM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2010, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2009 Physics Help Forum

Physics Help Forum is a community of physics forums with an emphasis on physics help in all levels of physics.
Register to post your physics questions on the message board.