Thermo cycles with only T and mol

  • Thread starter Thread starter bmartz
  • Start date Start date
  • Tags Tags
    Cycles Thermo
AI Thread Summary
The discussion revolves around calculating the efficiency of a three-step thermodynamic cycle involving an ideal diatomic gas. The first step involves heating the gas at constant volume, while the second step is an isothermal expansion, and the third step is a constant pressure contraction back to the original volume. The user is attempting to determine the total work and heat flow but is struggling with the isothermal expansion due to a lack of initial or final pressure or volume values. They consider using unknown variables for pressure or volume to simplify the calculations, hoping that these will cancel out in the end. The conversation highlights the challenges of applying thermodynamic equations without complete data.
bmartz
Messages
4
Reaction score
0

Homework Statement


A three-step cycle is undergone by 3.8 mol of an ideal diatomic gas:
(1) the temperature of the gas is increased from 210 K to 550 K at constant volume;
(2) the gas is then isothermally expanded to its original pressure;
(3) the gas is then contracted at constant pressure back to its original volume.
Throughout the cycle, the molecules rotate but do not oscillate. What is the efficiency of the cycle?


Homework Equations


Q=nCvΔT
Q=nCpΔT
Cv=5/2 R
Cp=7/2 R
W = nRTln(V2/V1)
ε=Wnet/QH

The Attempt at a Solution


I know I need to find the total work and heat flow in. For step one it is easy enough, W=0 and Q=nCvΔT.
I'm stuck at step 2. I know that since it is isothermal ΔEint is 0, so Q=W. I am trying to calculate by:
W = nRTln(V2/V1)
But without any initial or final pressure or volume values I am lost.
 
Physics news on Phys.org
Assume the problem as stated gives all the required information. Can you just let either the initial pressure or volume equal to an unknown with the hope that it will cancel out in the end? Just a thought.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top