# Homework Help: Ideal Gas Question.

1. Nov 13, 2008

### Crosshash

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

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

3. 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

Last edited: Nov 13, 2008
2. Nov 13, 2008

### Andrew Mason

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

3. Nov 13, 2008

### Crosshash

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. Nov 13, 2008

### Andrew Mason

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