Isothermal Question

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Homework Statement


A 2.00-mol sample of helium gas initially at 300K and 0.400 atm is compressed isothermally to 1.20atm. Assume the helium behaves as an ideal gas, find (a) the final volume of the gas, (b) the work done on the gas, and (c) the energy temperature?


Homework Equations


PV = nRT

[tex]W = nRTln(\frac{Pi}{Pf})[/tex]

[tex]\delta{U} = Q - W[/tex]


The Attempt at a Solution


This is what i have done can anyone tell me what is wrong, because it is not the same as the book's answer.
a)
[tex]Vf = \frac{2 * 8.314*10^3 * 300}{1.20}[/tex]

[tex]Vf = 4157000m^3[/tex]


b)
[tex]W = 2 * 8.314 * 10^3 * 300 * ln(\frac{1.20}{0.400})[/tex]
W = 5480317.541

c)
[tex]\delta{U} = -5480317.541[/tex] meaning this is compression

The correct answers for the following are:

a) 0.0410 [tex]m^3[/tex]
b) 5.48 kJ
c) -5.48 kJ

P.S
 

Answers and Replies

  • #2
Mapes
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It's the classic error: not checking your units.
 
  • #3
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could you tell me specifically which values have the wrong units?
 
  • #4
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Check your units for "R"
 
  • #5
175
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how do you convert mol into kmol??
 
  • #6
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ok would 2mol be 0.200 kmol because 1 mol is 1000mols?
 
  • #7
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yes because 1000 moles= 1 kmol
 
  • #8
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after i changed the units i still didn't get the correct answer:

This is what i changed:

[tex]Vf = \frac{.200 * 8.314*10^3 * 300}{1.20}[/tex]
[tex]Vf = 415700m^3[/tex]
 
  • #9
Andrew Mason
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after i changed the units i still didn't get the correct answer:

This is what i changed:

[tex]Vf = \frac{.200 * 8.314*10^3 * 300}{1.20}[/tex]
[tex]Vf = 415700m^3[/tex]
This is an enormous volume and obviously wrong!

What are your units for pressure?

What is the initial volume? Since T is unchanged, what is the final volume if the pressure is 3 x greater?

AM
 
  • #10
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It's .002 Kmole, change 1.2 atm to units of pressure in N/m^2 and 1 J = 1 N*m
 
  • #11
Andrew Mason
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I am not sure why you are using Kmol. It is much easier to use moles and R = 8.314 (J/mol - K) and pressure as J/m^3 or N/m^2

AM
 
  • #12
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My reference uses R= 8.314 kJ/kmole-K same as 8.314 J/mole-K pick your poison but keep the units consistent
 
  • #13
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[tex]Vf = \frac{.200 * 8.314*10^3 * 300}{1.20 * 1.013*10^5}[/tex]
[tex]Vf = 0.0410m^3[/tex]
i changed the units of atm to Pa and i got the correct answer.
 
  • #14
Andrew Mason
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i changed the units of atm to Pa and i got the correct answer.
I trust that you used .002Kmol not .2Kmol. as you have shown.

AM
 
  • #15
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oh yeh oops typo.
 

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