Thermodynamics of Helium using a P-V Diagram

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Homework Help Overview

The discussion revolves around the thermodynamics of helium gas, specifically analyzing its behavior through a P-V diagram during various processes including isothermal expansion, isobaric compression, and adiabatic compression. The original poster describes the initial conditions and the steps involved in the thermodynamic cycle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to sketch the P-V diagram and calculate the states of the gas at different points in the cycle. They express uncertainty about the calculations for the volume and temperature after the isobaric compression.

Discussion Status

Participants are engaged in clarifying the relationships between the states of the gas and are providing guidance on setting up equations based on the known states. Some participants have noted the presence of two unknowns that can be solved using simultaneous equations, while others have offered specific equations to assist in finding the unknowns.

Contextual Notes

The original poster mentions a shaky understanding of thermodynamics, indicating that they may be working under constraints related to their current level of knowledge and experience with the subject matter.

mouzis
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Homework Statement


Helium gas is initially at a pressure of 16 atm, a volume of 1L, and a temperature of 600K. It is expanded isothermally until its volume is 4L and then compressed at constant pressure until its volume and temperature are such that an adiabatic compression will return the gas to its original state. (a) Sketch this process on the P-V diagram. (b) Find the volume and temperature after the isobaric compression. (c) Find the work done during each cycle. (d) Find the efficiency of the cycle.



Homework Equations


(P1V1) / T1 = (P2V2) / T2

PV\gamma = constant

P1V1\gamma = P2V2\gamma

\gamma ≈ 1.67

The Attempt at a Solution



At this time i have only attempted a and b. I believe i have drawn the P-V diagram correctly, but do not know how to draw one here, so i will describe what i have drawn.

a) Point A is at a pressure of 16atm and a volume of 1L. There is an isothermal curve between point A and B in the direction from A to B. Point B is at a pressure of P and volume of 4L. There is an isobaric line from Point B at a higher volume to Point C where there is a lower volume. Point C is at a pressure of P(same as point B due to isobaric process) and a volume between 1L and 4L. There is an adiabatic curve from Point C to Point A.

b) I began by finding the pressure at Point B by using (P1V1) / T1 = (P2V2) / T2. From this i found that the pressure at point B was 4atm, making this the pressure at Point C as well. From there I've tried to use P1V1\gamma = P2V2\gamma in order to try and find the volume at point C, but haven't been able to find much. My understanding of thermodynamics is still shaky at this point so any help would be welcome.
 
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You have the relation for B→C and another for C→A and you know the complete states for A and B. Therefore you have two unknowns and two simultaneous equations. What's the problem?
 
When i use the relation B→C and solve for one of the unknown i find that Tc = 150Vc. I can't seem to perform the math correctly for the process from C→A using P1V1γ = P2V2γ. Thanks for the reply as well Simon.
 
Last edited:
Check:

State A: ##P_A##, ##V_A##, ##T_A## ... all known quantities.

State B: ##P_B##, ##V_B##, ##T_B=T_A## ... all known quantities

State C: ##P_C=P_B## ... known: need V and T.

From ##B \rightarrow C## you get ##V_C=\frac{V_B}{T_A}V_C## ... (1)

From ##C\rightarrow A## you get ##P_BV_C^\gamma = P_AV_A^\gamma## ... (2)

in equation 1 the unknowns are ##V_C## and ##T_C##.
in equation 2 the only unknown is ##V_C##

make ##V_C## the subject in equation (2) and evaluate. Then substitute into (1) to find ##T_C##
 
Okay i finally figured out why i couldn't solve for Vc. Just a simple calculation error. Thanks again for all the help
 
No worries.
 

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