# Simplified Otto Cycle

1. Dec 4, 2008

### doggieslover

The idealized cycle shown is known as the Otto cycle. (Intro 1 figure) Suppose an engine is executing this Otto cycle, using a gas (not necessarily ideal) as its working substance. From state A to state B, the gas is allowed to expand adiabatically. (An adiabatic process is one in which no heat is added to, or given off by, the working gas.) The gas is then cooled at constant volume until it reaches state C, at which point it is adiabatically compressed to state D. Finally, it is heated at constant volume until it returns to state A.

The pressure and volume of the gas in state A are p_A and V_A respectively. The pressure and volume of the gas in state C are p_C and V_C respectively.

http://session.masteringphysics.com/problemAsset/1011140/12/STH_tc_2.jpg

Part C
What is DeltaU, the change in the gas's internal energy after a complete cycle?
Express your answer in terms of any needed variables from the problem introduction.

Okay I know that Q is zero, and deltaU = Q - W, so deltaU = -W, and W = p*deltaV, and a complete cycle means it returns to its original state.

So my answer should be p_A*V_A, but it's incorrect, it says the answer does not depend on those variables, what variables are they looking for then?

Help?

2. Dec 6, 2008

### alphysicist

No, I don't think that is correct. The heat flow is zero for the adiabatic parts of the cycle, but for the two times when it is heated and cooled at constant volume there is a heat flow. Do you see what to do now?

3. Dec 7, 2008

### Andrew Mason

It's a trick question. If the gas returns to its original state, has there been any change to its internal energy?

AM

4. Dec 7, 2008

### doggieslover

I know that Q is zero already, so now deltaU is zero too since there's no change in internal energy?

5. Dec 7, 2008

### Andrew Mason

The work done per cycle and the heat flow from the hot to cold reservoir is not zero. But the change in internal energy of the gas is zero over one complete cycle.

AM

6. Dec 7, 2008

### doggieslover

Then if deltaU = Q - W, deltaU = 0, Q = 0, then -W will be left alone, but my question is are there other ways to find what work is other than using pdeltaV?

7. Dec 7, 2008

### alphysicist

Q is not zero; it is only zero for two parts of this four-part cycle.

8. Dec 8, 2008

### Andrew Mason

Yes. If U = 0, $\Delta Q = W$. In other words, W = Qh-Qc.

You could work out Qh and Qc from the BC and DA parts of the path (constant volume). For example, if this was an ideal gas, from the ideal gas equation: $Q = nC_v\Delta T = VC_v\Delta P/R$

AM