# Need help finding temperature-volume relation in solving Adiabatic problem

1. Apr 24, 2012

### defmar

A monatomic ideal gas (γ = 5/3) is contained within a perfectly insulated cylinder that is fitted with a movable piston. The initial pressure of the gas is 1.31 × 105 Pa. The piston is pushed so as to compress the gas, with the result that the Kelvin temperature doubles. What is the final pressure of the gas?

I know the equation I will need to solve this with is P_i*(V_i)^γ = P_f*(V_f)^γ
I'm stuck at figuring out how to plug that T_f = 2*T_i to find the volume. Any pointers?

2. Apr 25, 2012

### Andrew Mason

Substitute nRT/V for P into the equation.

AM

3. Apr 25, 2012

### defmar

In that case I'm getting an equation that reduces to (V_f/V_i)^γ = V_i/(2*V_f).

I don't know the initial or final volume or how to solve from here.

4. Apr 26, 2012

### Andrew Mason

Substituting nRT/V for P gives:

$$PV^\gamma = nRTV^{(\gamma-1)} = K$$

So if:

$$\frac{T_f}{T_i} = 2$$

then what is

$$\left(\frac{V_i}{V_f}\right)^{(\gamma-1)} = ?$$

AM

5. Apr 26, 2012

### defmar

I don't know. I would say either 2 or 1/2, but that'd just be me guessing. Or I can say it's (V_i/V_f)^(2/3) - but that doesn't help me either. I don't know how to make sense of this because the problem is none of constant volume, temperature or pressure.

6. Apr 26, 2012

### Andrew Mason

You are making it harder than it is. It is just algebra. I gave you the equation!

Since:

$$\frac{T_f}{T_i} = \left(\frac{V_i}{V_f}\right)^{(\gamma-1)}$$

and since:

$$\frac{T_f}{T_i} = 2$$

then

$$\left(\frac{V_i}{V_f}\right)^{(\gamma-1)} = 2$$

Solve that for Vi/Vf. How do you determine Pf/Pi from Tf/Ti and Vi/Vf?

AM

7. Apr 27, 2012

### defmar

Thank you so far. I took (V_i/V_f)^(2/3) = 2 ==> (V_i/V_f) = 2*SQRT(2).
I then attempted to apply both Charles and Gay-Lussac's law to come up with:
P_f/P_i = 2*SQRT(2) ==> P_f = P_i*2*SQRT(2), but this is not correct. I'm sure I'm missing something obvious and simple, but I'm oblivious to it yet.

8. Apr 27, 2012

### Andrew Mason

You may need to brush up a bit on algebra involving logarithms.

$$\left(\frac{V_i}{V_f}\right)^{\gamma-1} = 2$$

Take the log of both sides:

$$(\gamma-1)\ln{\left(\frac{V_i}{V_f}\right)} = \ln{2}$$

Since $\gamma = 5/3$

$$\ln{\left(\frac{V_i}{V_f}\right)} = 3\ln{2}/2$$

Take the antilog of both sides:

$$\left(\frac{V_i}{V_f}\right) = e^{1.04} = 2.83$$

Use PV = nRT to determine the final pressure.

AM

(P.S. One could also express the adiabatic condition in terms of T and P and work it out - just avoids the last step)

9. Apr 28, 2012

### defmar

Thank you. I knew I brain-farted something obvious. At this point I found it easier to plug the V_i/V_f value back into the P_i*(V_i)^γ = P_f*(V_f)^γ equation and solve for:
P_f = P_i*(V_i/V_f)^γ.

Thank you again