Solving Ideal Gas Q: Temps & Press at A,B,C,D

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SUMMARY

The discussion focuses on calculating the pressures and temperatures at points A, B, C, and D in an Otto Cycle involving one mole of gas. Key equations utilized include the Ideal Gas Law (pV = nRT), Charles's Law (V1/T1 = V2/T2), and Boyle's Law (P1V1 = P2V2). The first law of thermodynamics and the adiabatic condition (PV^γ = K, where γ = Cp/Cv = 1.4 for air) are essential for solving the problem. The temperatures at points A and B are established as 393.15K and 573.15K, respectively, while the pressure at point C is given as 1At.

PREREQUISITES
  • Understanding of the Ideal Gas Law (pV = nRT)
  • Familiarity with the first law of thermodynamics (dQ = dU + dW)
  • Knowledge of adiabatic processes and the adiabatic condition (PV^γ = K)
  • Basic concepts of the Otto Cycle in thermodynamics
NEXT STEPS
  • Calculate the pressure at point B using the Ideal Gas Law with known temperature.
  • Determine the value of K from the adiabatic condition at point B.
  • Use the calculated K to find pressures and volumes at points C and D.
  • Explore the implications of the first law of thermodynamics in heat engine cycles.
USEFUL FOR

Students studying thermodynamics, mechanical engineers focusing on heat engines, and anyone interested in the principles of the Otto Cycle and gas laws.

Crosshash
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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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Crosshash said:

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 PV^\gamma = K where \gamma is the ratio Cp/Cv for air (1.4).

AM
 
Andrew Mason said:
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 PV^\gamma = K where \gamma 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.
 
Crosshash said:
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 PV^\gamma for point B and, therefore, for point C.

AM
 

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