How Do You Analyze a PV Diagram for a Diatomic Ideal Gas Cycle?

  • Thread starter Thread starter joemama69
  • Start date Start date
  • Tags Tags
    Graph
AI Thread Summary
The discussion focuses on analyzing a PV diagram for a diatomic ideal gas undergoing three distinct processes: heating at constant volume, isothermal expansion, and isobaric compression. Participants are tasked with calculating heat (Q), work (W), and change in internal energy (ΔE) for each leg of the cycle, using the initial pressure (Po) and volume (Vo) as reference points. Key equations discussed include the ideal gas law (PV=nRT) and relationships between Q, W, and ΔE for each leg. There is emphasis on expressing variables in terms of gamma, Po, and Vo, particularly during the isothermal process where work done is calculated using the natural logarithm of volume ratios. The conversation highlights the need to manipulate these equations correctly to derive the necessary thermodynamic quantities.
joemama69
Messages
390
Reaction score
0

Homework Statement


A diatomic ideal gas originally at pressure Po and volume Vo is heated at
constant volume until its pressure increases by a factor of 3. Next it is
isothermally expanded until its pressure returns to Po. Finally it is
isobarically compressed to its initial volume.

a) Show the process on a PV diagram.

b) Find Q, W, and delta E for each of the 3 legs [All entries should be in terms of the “givens”—
Po and Vo.]


Homework Equations





The Attempt at a Solution



Part A) Note the very crude attachment.

Part B) I think i have this generaly correct, but I am having trouble using the given terms

Leg 1 (AB)

Q = E = nC_v(delta T), W = 0

Leg 2 (BC)

Q = W, E = 0

Leg 3 (CA)

Q = E + W, W = P(delta V)
 

Attachments

Physics news on Phys.org
joemama69 said:
Q = E = nC_v(delta T), W = 0

That's right. Just express n, Cv, and delta-T in terms of gamma, P0, and V0. Remember that PV=nRT.

Leg 2 (BC)

Q = W, E = 0

Do you know how to find the work done by an isothermal process?

Leg 3 (CA)

Q = E + W, W = P(delta V)

Yeah. Now express delta-V in terms of V0.
 
"That's right. Just express n, Cv, and delta-T in terms of gamma, P0, and V0. Remember that PV=nRT."

but PV isn't constant, how can is use the pv equation
 
PV isn't constant, but P1V1-P0V0 = nR(T1-T0).
 
ok so i just solve that equatio for n and delta-T and plug those in...

n = (PV - initial PV)/R(detaT)... delta T = (PV-initial PV)/nR...

C_v is constand volume so is that just C_v = V_o
 
leg 2...


Q = W = nRT(ln V/V_o), E = 0

leg 3...

Q = E + W, W = P(delta V) = P(v-v_o), can E be expressed in P & V
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top