Work done in an adiabatic process

  • Thread starter Cade
  • Start date
  • #1
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Homework Statement


An adiabatic process is performed on 15 moles of an ideal gas. The initial temperature is 320 K and the initial volume is
0.80 m3. The final volume is 0.40 m3. The adiabatic constant for the gas is 1.44. How much work is done by the gas?

The answer is -32kJ

Homework Equations


W = Integrate[p dv, vInitial, vFinal] for adiabatic processes
pV = nRT

The Attempt at a Solution


If p(v^y) initial == p(v^y) final where y = 1.44, and v varies, then p also varies, and I don't know how to use a varying pressure with the formula for work for adiabatic processes.

I try to find the initial pressure by
pV = nRT
p*V == 15*8.31*320
With Vinitial, p = 49860Pa
With Vfinal, p = 99720Pa

And I'm stuck.
 

Answers and Replies

  • #2
122
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Here is my solution: for adiabatic process, you can write: PV^(gamma) = C => P = C / V^(gamma). Plug it into W = int(PdV). After that, replace C by P and V.
 
  • #3
92
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p*((0.8)^1.44) = c
Which gives
p = 1.37896c

Integrating 1.37896 c dV from 0.8 to 0.4 gives
-0.551582 c

If I use the original p (that I found) and v (given) for c,
-0.551582(p*v) = -0.551582(49860*0.8) = -22kJ, which isn't the answer. Where did I go wrong?
 
  • #4
122
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No.

We have [tex]PV^ \gamma = C ==> P = C/V^ \gamma[/tex]

[tex]W = \int PdV = \int CV^- ^\gamma dV = \frac{C V^ 1 ^- ^\gamma}{1 - \gamma}[/tex]
 
  • #5
92
0
Oh, I understand, thanks for your help!

The solution for future reference:
p*0.8 = 15*8.31*320
p -> 49860 Pa (Initial pressure)

p*(0.8^1.44) = c (0.8 is the initial volume)
c -> 36157.8

w = Integrate[c (v^-1.44), dv]
w = -(2.27273 c)/v^0.44
-> wf - wi = -32327.8 J
 

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