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Topic: internal energy  (Read 3393 times)

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Offline stupid_one

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internal energy
« on: August 20, 2009, 05:58:27 PM »
hi, i did a question but i m not sure if it is correct.

5moles of an ideal gas at initial temp. pressure and volume of 273K, 1atm, and 100L, is allowed to expand to a final pressure of 0.5atm in 2 different ways:
a-- isothermally and irreversibly against a constant external pressure of 0.5atm
b-- isothermally and reversibly to a final external pressure of 0.5atm

calculate(i) work done on or by the system, (ii) the heat exchanged between the system and the surroundings, (iii) the internal energy change of the system


Offline stupid_one

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contd
« Reply #1 on: August 20, 2009, 06:02:18 PM »
firstly i calculated the  :delta: V
V = nRT/ P
Vf = 226.9722L  Vi = 100L
for part A  i did the following
(i)
w= -Pex (Vf-Vi)
w= -(0.5atm) (126.9722)
w = -63.4861J
 
(ii)  :delta: U = q + w
since it is at a constant external pressure
 :delta: U =  0
therefore, q = 63.4861J

(iii)  :delta: U = 0

Offline stupid_one

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contd
« Reply #2 on: August 20, 2009, 06:10:26 PM »
for part b i did the following,
(i)
w = -Pex  :delta: V
w = -nRT In (Vf/ Vi)
w = -(5moles) (8.314Jmol-1K-1)(273K) In (226.9722/ 100)
w = -9301.971679J

(ii)
q = (Vf + PexVf) - (Vi+Pex Vi)
q = [226.9722 + (0.5atm)(226.9722)] -[100 + (0.5atm)(100)]
q = 340.4583 - 150
q = 190.4583

(iii)
 :delta: U = q - Pex  :delta: V
 :delta: U = 190.4583 - (0.5atm) (126.9722)
 :delta: U = 126.9722

AM I RIGHT??

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