# An entropy change question?

1. Aug 3, 2010

### Faux Carnival

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

One mole of a monatomic gas first expands isothermally, then contracts adiabatically. The final pressure and volume of the gas are twice its initial pressure and volume. (Pf=2Pi and Vf=2Vi)

Find the net change in the entropy of the gas.

2. Relevant equations

3. The attempt at a solution

I found the entropy change for the first step. Let the volume between the two steps be Vm.

TVγ-1 = constant, so Vm = 22.5 Vi

ΔS = nR ln(Vm/Vi) = 1.73 nR

But I have no idea about the entropy change of the second step. What is the entropy change for an adiabatic process? Isn't dQ=0 ?

2. Aug 3, 2010

### Andrew Mason

There is no entropy change in a reversible adiabatic process, so you just have to determine the change in entropy of the isothermal process. You appear to have tried doing this by applying the adiabatic condition (after finding the final temperature) to get the volume at the end of the isothermal process, but I can't figure out how you get your answer. If you apply the adiabatic condition:

$$\left(\frac{V_m}{V_f}\right)^{\gamma-1} = \frac{T_f}{T_m} = 4$$

AM

3. Aug 3, 2010

### Faux Carnival

STEP 1 (ISOTHERMAL)

Pi Vi Ti ---> Punknown Vm Ti

Punknown Vm Ti ---> 2Pi 2Vi 2Ti

So, for the adiabatic process I applied TVγ-1 = constant.

Ti Vm2/3 = 2Ti (2Vi)2/3

Isn't this correct?

Another thing, is the change in entropy ΔS zero for adiabatic processes?

4. Aug 3, 2010

### Andrew Mason

It is not correct. What is the final temperature? (hint: apply the ideal gas law). The final temperature is not twice the initial temperature.

Yes. See my previous post.

AM