Can Ideal Gas Thermodynamics Explain This Circle Process?

In summary: The conversation is discussing a circle process that has four stages: adiabatic compression, isochoric heating, adiabatic spreading, and isochoric cooling. The process involves an ideal gas with a specific heat ratio (x=1.4) and starting conditions of volume (V=30dm3), pressure (P=2.026bar) and temperature (T=366K). In the first stage, the volume decreases to 20dm3, and in the second stage, the pressure increases to 4.13bar. The goal is to calculate the work and heating changes in the process, with the result being (delta) I>U=0; Q=-W=453J. However, the
  • #1
Amina
1
0
Some circle process contains next stages:
1.adiabatic compression
2. isochoric heating
3. adiabatic spreading
4. isochoric cooling
If we put ideal gas(x=1.4) in this process from starting conditions :
-Volume (V=30dm3)
-Pressure (P= 2.026bar)
-Temperature (T=366K)
* in first stage volume lows down on 20dm3,
*on the second stage pressure goes on 4.13bar.
CALCULATE VALUE OF ALL WORK AND ALL HEATING CHANGES IN THIS CIRCLE PROCESS!
Result should be (delta) I>U =0 ;Q= -W= 453J

This is what i used to calculate but didnt get right result:
stage 1)
P1V1=nRT1 ---->n
Cv=5/2 R
Q=0
W=(delta) U= n Cv (delta)T ; delta T(T2-T1)
T1/T2= (V2/V1) exp (x-1) ------> T2
stage 2)
W=0
P1/P2= (V2/V1)exp (x) ----->P2
T2/T3=P2/P3 ----->T3
Q=nCv (delta)T ;delta T= (T3-T2)
stage 3)
T3/T4= (V4/V3) exp(x-1) ----->T4
Q=0
W=(delta)U= n Cv (delta)T ;delta T= ( T4-T3)
stage 4)
W=0
Q=(delta)U= n Cv (delta)T ; delta T= (T1-T4)Did i messed smt with formulas? Calculating not problem, just this is a bit hard to visualise and set right formulas.. if someone can catch my mistake fast so i can try again, i don't want to bother anybary or take time. Thank you in front..
I m new here, sorry if didnt apsorb all rules at first.
 
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  • #2
Please show us some more details of your calculations. It's really hard to figure out where you went wrong without that.

Chet
 

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