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

In summary, to calculate the pressures and temperatures at points A, B, C, and D in the Otto Cycle, you will need to use the first law of thermodynamics and the adiabatic condition. By knowing the value of T at point B, you can calculate the pressure at point B and use that to calculate the value of PV^gamma at point B and, therefore, at point C. With this information, you can then solve for the pressures and temperatures at points A and D.
  • #1
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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  • #2
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 [itex]PV^\gamma = K[/itex] where [itex]\gamma[/itex] is the ratio Cp/Cv for air (1.4).

AM
 
  • #3
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 [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
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 [itex]PV^\gamma[/itex] for point B and, therefore, for point C.

AM
 

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

What is an ideal gas?

An ideal gas is a theoretical gas that follows the gas laws exactly at all temperatures and pressures. It is composed of particles that have no volume and do not interact with each other.

How do you solve for the temperature and pressure at points A, B, C, and D on an ideal gas graph?

To solve for the temperature and pressure at points A, B, C, and D, you can use the ideal gas law, which states that the pressure and volume of an ideal gas are directly proportional to its temperature and number of moles, and inversely proportional to its volume. This law can be represented by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. By rearranging this equation, you can solve for any of the variables at points A, B, C, and D.

Can you solve for the temperature and pressure at any point on an ideal gas graph?

Yes, you can solve for the temperature and pressure at any point on an ideal gas graph using the ideal gas law. However, it is important to note that the ideal gas law is only accurate for ideal gases at low pressures and high temperatures. Real gases may deviate from ideal gas behavior at high pressures and low temperatures.

How do temperature and pressure relate to each other on an ideal gas graph?

On an ideal gas graph, temperature and pressure are directly proportional to each other. This means that as temperature increases, pressure also increases, and vice versa. This relationship is described by the ideal gas law, which states that at a constant volume, the pressure of an ideal gas is directly proportional to its temperature.

What is the significance of points A, B, C, and D on an ideal gas graph?

Points A, B, C, and D represent different states of an ideal gas on a graph. Point A represents the initial state, where the pressure and temperature are known. Point B represents the final state, which can be calculated using the ideal gas law. Point C represents the temperature and pressure at a constant volume, and point D represents the temperature and pressure at a constant pressure. These points are important for understanding the behavior of an ideal gas and for solving problems involving ideal gases.

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