Thermodynamics: Piston-Cylinder

  • Thread starter Thread starter aznkid310
  • Start date Start date
  • Tags Tags
    Thermodynamics
AI Thread Summary
The discussion revolves around solving a thermodynamics problem involving the expansion of nitrogen gas in a piston-cylinder arrangement. The initial temperature of the nitrogen needs to be determined, along with the work done during the expansion from 20 bar to 1 bar at 300K. Participants suggest using the conservation of energy equation and the ideal gas law to find the necessary values. It is noted that since the process is adiabatic, the heat transfer (Q) is zero, simplifying the calculation of work (W) to the change in internal energy. Various methods to calculate work, including using specific heat capacities and integrals, are discussed.
aznkid310
Messages
106
Reaction score
1

Homework Statement



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?

Homework Equations



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

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
 
Last edited:
Physics news on Phys.org
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
 
is it just the ideal gas law pV = nRT?
 
For part b, this is an adiabatic process, so Q = 0.

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

What should i do with this?
 
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
 
So would it be something like: U = (c_v)nRT, where c_v = (5/2)T

Then W = (c_v)nR([change in]T)
 
Yes, though c_v is simply 5/2.
 
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)
 
Back
Top