Ideal Gas Question.

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Homework Statement


1) Consider one-mole of gas in a heat engine undergoing the Otto Cycle
a) The gas absorbs heat, at constant volume between 120'C and 300'C
b) The gas expands adiabatically from V1 to V2 = 5V1
c) The gas cools, at constant volume to Td at point D where the pressure is 1At
d) The gas is then adiabatically compressed from V2 to V1 returning to the original temperature of 120'C
You may assume Cv = 5R/2 and Cp = 7R/2

What are the pressures and temperatures (in Kelvin) at points A,B,C,D?


Homework Equations



Ideal gas law pV = nRT
Charle's law = V1/T1 = K and V1/T1 = V2/T2
Boyle's law = P1V1 = P2V2


The Attempt at a Solution



Temperatures
A = 120 + 273.15 = 393.15K
B = 300 + 273.15 = 573.15K
C =
D =

Pressure
A =
B =
C = 1At
D =

I'm guessing i'm going to have to use Boyle's and Charle's laws in order to fill in the blanks.

V1 / 573.15 = 5V1 / T2

T2 / 573.15 = 5V1 / V1

T2 / 573.15 = 4V1

But here's the problem, it doesn't appear like this is going to solve anything. I reach this barrier for the other blanks as well. Could someone possibly edge me in the right direction please?

Thanks
 
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Answers and Replies

  • #2
Andrew Mason
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Homework Statement


1) Consider one-mole of gas in a heat engine undergoing the Otto Cycle
a) The gas absorbs heat, at constant volume between 120'C and 300'C
b) The gas expands adiabatically from V1 to V2 = 5V1
c) The gas cools, at constant volume to Td at point D where the pressure is 1At
d) The gas is then adiabatically compressed from V2 to V1 returning to the original temperature of 120'C

What are the pressures and temperatures (in Kelvin) at points A,B,C,D?


Homework Equations



Ideal gas law pV = nRT
Charle's law = V1/T1 = K and V1/T1 = V2/T2
Boyle's law = P1V1 = P2V2
Boyle's law only works if T is the same. It isn't. Charle's law only works if P is constant. It isn't.

You have to use the first law of thermodynamics: dQ = dU + dW and the adiabatic condition [itex]PV^\gamma = K[/itex] where [itex]\gamma[/itex] is the ratio Cp/Cv for air (1.4).

AM
 
  • #3
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Boyle's law only works if T is the same. It isn't. Charle's law only works if P is constant. It isn't.

You have to use the first law of thermodynamics: dQ = dU + dW and the adiabatic condition [itex]PV^\gamma = K[/itex] where [itex]\gamma[/itex] is the ratio Cp/Cv for air (1.4).

AM
So basically, if I can find out what K is from the Adiabatic condition, then I should be able to calculate the values of P and V for the other points?

Except I don't have a point which has both P and V values.
 
  • #4
Andrew Mason
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So basically, if I can find out what K is from the Adiabatic condition, then I should be able to calculate the values of P and V for the other points?

Except I don't have a point which has both P and V values.
Sure you do. You know that V = nRT/P. So if V is constant (ie. A to B) nRT/P is constant. You know T so you can work out P at point B.

So you can calculate [itex]PV^\gamma[/itex] for point B and, therefore, for point C.

AM
 

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