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Apr 5 2016 07:53am
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a Carnot engine with diatomic gas operates at pressure extremes of 2 and 20 atm and 500 and 800 kelvins find the pressures at the end of the expansion and compression
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Apr 5 2016 03:45pm
It's a Carnot cycle. It has two isothermal (T = constant) legs and two adiabatic (P = constant) legs. The expansion and compressions are isothermal going from one pressure extreme to the other. At the end of the expansion, the pressure goes down to the extreme (2 atm). At the end of the compression the pressure goes up to the extreme (20 atm).



This post was edited by Dontrunaway on Apr 5 2016 03:46pm
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Apr 5 2016 04:09pm
Quote (Dontrunaway @ Apr 5 2016 04:45pm)
It's a Carnot cycle. It has two isothermal (T = constant) legs and two adiabatic (P = constant) legs. The expansion and compressions are isothermal going from one pressure extreme to the other. At the end of the expansion, the pressure goes down to the extreme (2 atm). At the end of the compression the pressure goes up to the extreme (20 atm).

http://puu.sh/o7u7H.png



I should have clarified I'm looking for the pressure at the ends of the isothermal compression and expansion
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Apr 5 2016 04:11pm
Ah I made a mistake. Adiabatic isn't constant pressure Q_Q It's constant entropy.

This post was edited by Dontrunaway on Apr 5 2016 04:13pm
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Apr 5 2016 04:15pm
Quote (Dontrunaway @ Apr 5 2016 05:11pm)
Ah I made a mistake. Adiabatic isn't constant pressure Q_Q


I was trying to use PV=nRT so for the first isotherm P1V1 = P2V2 but theres too many unknowns to do it that way
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Apr 5 2016 04:41pm
You can also use the equation v1/v4 = v2/v3 source: http://www.roymech.co.uk/Related/Thermos/Thermos_Carnot.html

You know the temperature at all four endpoints, and you know the pressure at two of the endpoints. You should be able to use the volume ratios to get you to the other pressures.

Khan academy proves out the volumetric ratio but it's kinda not helpful. Still linking it in case you are curious. https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/entropy-chemistry-sal/v/proof-volume-ratios-in-a-carnot-cycle

This post was edited by Dontrunaway on Apr 5 2016 04:45pm
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Apr 5 2016 04:55pm
Disclaimer: It's been a while since I've done this, so if anyone else comes along who knows this, please double check.

A = Top left of the diagram posted
B = After isothermal expansion
C = Bottom right of diagram posted
D = After isothermal compression

Pa = 20
Ta = 800
Va = nRTa/Pa

Pb = ?
Tb = 800
Vb = ?

Pc = 2
Tc = 500
Vc = nRTc/Pc

Pd = ?
Td = 500
Vd = nRTd/Pd

B -> C and D -> A are adiabatic and reversible, implying that PV^x = Constant and TV^(x-1) = Constant, where x = 7/5 for a diatomic ideal gas.

TbVb^(x - 1) = TcVc^(x - 1)

Solve this for Vb and then solve for Pb

Pb = nRTb/Vb


TdVd^(x-1) = TaVa^(x-1)

Solve this for Vd and then solve for Pd

Pd = nRTd/Vd



This post was edited by RzChaos on Apr 5 2016 04:58pm
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Apr 5 2016 04:58pm
Quote (Dontrunaway @ Apr 5 2016 05:41pm)
You can also use the equation v1/v4 = v2/v3 source: http://www.roymech.co.uk/Related/Thermos/Thermos_Carnot.html

You know the temperature at all four endpoints, and you know the pressure at two of the endpoints. You should be able to use the volume ratios to get you to the other pressures.

Khan academy proves out the volumetric ratio but it's kinda not helpful. Still linking it in case you are curious. https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/entropy-chemistry-sal/v/proof-volume-ratios-in-a-carnot-cycle


Quote (RzChaos @ Apr 5 2016 05:55pm)
A = Top left of the diagram posted
B = After isothermal expansion
C = Bottom right of diagram posted
D = After isothermal compression

Pa = 20
Ta = 800
Va = nRTa/Pa

Pb = ?
Tb = 800
Vb = ?

Pc = 2
Tc = 500
Vc = nRTc/Pc

Pd = ?
Td = 500
Vd = nRTd/Pd

B -> C and D -> A are adiabatic and reversible, implying that PV^x = Constant and TV^(x-1) = Constant, where x = 7/5 for a diatomic ideal gas.

(800)Vb^(x - 1) = 500Vc^(x - 1)

Solve this for Vb and then solve for Pb

Pb = nR*(800)/Vb


(500)Vd^(x-1) = 800Va^(x-1)

Solve this for Vd and then solve for Pd

Pd = nR*(500)/Vd


ahh thank you both very much
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Apr 5 2016 05:33pm
Quote (RzChaos @ Apr 5 2016 05:55pm)
Disclaimer: It's been a while since I've done this, so if anyone else comes along who knows this, please double check.

A = Top left of the diagram posted
B = After isothermal expansion
C = Bottom right of diagram posted
D = After isothermal compression

Pa = 20
Ta = 800
Va = nRTa/Pa

Pb = ?
Tb = 800
Vb = ?

Pc = 2
Tc = 500
Vc = nRTc/Pc

Pd = ?
Td = 500
Vd = nRTd/Pd

B -> C and D -> A are adiabatic and reversible, implying that PV^x = Constant and TV^(x-1) = Constant, where x = 7/5 for a diatomic ideal gas.

TbVb^(x - 1) = TcVc^(x - 1)

Solve this for Vb and then solve for Pb

Pb = nRTb/Vb


TdVd^(x-1) = TaVa^(x-1)

Solve this for Vd and then solve for Pd

Pd = nRTd/Vd


wait so when I solve for Vb and Vd I still have Vc and Va in the answer so my final answer will have those volume variables in it?
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Apr 5 2016 06:23pm
You're given T and pressure for positions C and A, you should be able to calculate them....recognizing that mols is constant, so you can drop the n in the calculation.

This post was edited by RzChaos on Apr 5 2016 06:24pm
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