1. Mar 1, 2015

### Incand

1. The problem statement, all variables and given/known data
Air is compressed at room temperature from atmospheric pressure to $\frac{1}{15}$ of the initial volume. Calculate the temperature at the end of compression assuming the process is reversible and adiabatic.

2. Relevant equations
$pV^\gamma = constant \Longleftrightarrow T \cdot V^{\gamma -1} = constant$

3. The attempt at a solution
First its not mentioned anywhere in the question this process is for an ideal gas but since the chapter only derived them for ideal gases I'm assuming it is.
So we got
$V_2 = 0.15V_1 = \frac{3}{20}V_1$
$p_1 = 1atm$
$T_1 = 293K$
and for an ideal gas
$\gamma = 5/3$
If i insert the volume and temperature into the equation i get
$T_1V_1^{\gamma -1} = T_2(0.15V_1)^{\gamma -1} \Longleftrightarrow T_2 = T_1(\frac{20}{3})^{2/3} = 1038K$
The answer is supposed to be $870K$ and maybe i am supposed to use the pressure that was given somehow?

Last edited: Mar 1, 2015
2. Mar 1, 2015

### Staff: Mentor

This is correct for a monatomic ideal gas.

3. Mar 1, 2015

### Incand

So I know that you can approximate $\gamma$ with the degrees of freedom. So since air is mostly made up of Nitrogen and oxygen and they're both come in pairs I would get $5$ degrees of freedom so i would get $\gamma = \frac{7}{5}$.
$T_2 = T_1(20/3)^{2/5) = 628K$
I think $\gamma$ is about correct, if i look up the heat capacity ratio for air at $20^ \circ C$ its about $1.4$.
Am i supposed to use the pressure anywhere?

4. Mar 1, 2015

### Staff: Mentor

That looks right to me. Are you sure about the data of in the problem? Could there be an error in the numerical answer you were given?

5. Mar 1, 2015

### Incand

You're right. I misread $0.15$ when it should be $1/15$. Thanks for all the help!