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Thermodynamics: Piston-Cylinder

 
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Jan9-09, 10:01 PM   #1
 

Thermodynamics: Piston-Cylinder


1. The problem statement, all variables and given/known data

5kg of nitrogen gas contained in a piston-cylinder arrangement expands in a process, such that pv1.4=p1v11.4, where p1 and v1 are the initial pressure and specific volume, from a pressure of 20 bar to a pressure of 1 bar and a temperature of 300K.

a. What is the initial temperature of the nitrogen?

b. How much work is done by the nitrogen as it expands?

2. Relevant equations

I'm not sure which relationships to use or how to even get started for both parts


3. The attempt at a solution

For part b, we will use the conservation of energy:

U2 + Ke2 + Pe2 = U1 + Ke1 + Pe1 + Q - W

So change in Pe and change in Ke is zero?

Thus, W = U2 - U1 - Q
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Jan11-09, 07:39 AM   #2
 
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When the gas has "a pressure of 1 bar and a temperature of 300K", what is its volume? Hint: use the equation the relates pressure, volume, and temperature for a gas.

After you get that, you can use

p v1.4 = p1 v11.4
Jan11-09, 04:11 PM   #3
 
is it just the ideal gas law pV = nRT?
Jan11-09, 04:32 PM   #4
 
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Thermodynamics: Piston-Cylinder


Yes.
Jan11-09, 06:49 PM   #5
 
For part b, this is an adiabatic process, so Q = 0.

Thus, W = -(change in internal energy)?

What should i do with this?
Jan11-09, 08:09 PM   #6
 
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There are a couple of ways to calculate W.

It's probably easiest to calculate the internal energy for a diatomic gas at the two temperatures involved (you'll need the right equation for that). Since Q=0, the change in energy gives us W.

Alternatively you can try doing the integral

W = ∫ P dV
Jan12-09, 01:12 AM   #7
 
So would it be something like: U = (c_v)nRT, where c_v = (5/2)T

Then W = (c_v)nR([change in]T)
Jan12-09, 06:47 AM   #8
 
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Yes, though c_v is simply 5/2.
Jan17-09, 10:42 PM   #9
 
Just as a heads up, the finished integral for work for a polytropic process turns out to be pretty simple.

for p(v^n)=c

P=pressure
v=specific volume

W/m=(P2v2-P1v1)/(1-n)
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